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Nonparametric Analysis of Finite Mixtures

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Nonparametric Analysis of Finite Mixtures

\address{Cowles Foundation for Research in Economics, Yale University, New Haven, CT 06520.} \email{[email removed]} \address{Cowles Foundation for Research in Economics, Yale University, New Haven, CT 06520.} \email{[email removed]}

abstractFinite mixture models are useful in applied econometrics. They can be used to model unobserved heterogeneity, which plays major roles in labor economics, industrial organization and other fields. Mixtures are also convenient in dealing with contaminated sampling models and models with multiple equilibria. This paper shows that finite mixture models are nonparametrically identified under weak assumptions that are plausible in economic applications. The key is to utilize the identification power implied by information in covariates variation. First, three identification approaches are presented, under distinct and non-nested sets of sufficient conditions. Observable features of data inform us which of the three approaches is valid. These results apply to general nonparametric switching regressions, as well as to structural econometric models, such as auction models with unobserved heterogeneity. Second, some extensions of the identification results are developed. In particular, a mixture regression where the mixing weights depend on the value of the regressors in a fully unrestricted manner is shown to be nonparametrically identifiable. This means a finite mixture model with function-valued unobserved heterogeneity can be identified in a cross-section setting, without restricting the dependence pattern between the regressor and the unobserved heterogeneity. In this aspect it is akin to fixed effects panel data models which permit unrestricted correlation between unobserved heterogeneity and covariates. Third, the paper shows that fully nonparametric estimation of the entire mixture model is possible, by forming a sample analogue of one of the new identification strategies. The estimator is shown to possess a desirable polynomial rate of convergence as in a standard nonparametric estimation problem, despite nonregular features of the model.

Introduction

In empirical economics it is often crucially important to control for unobserved heterogeneity, and mixture models provide convenient ways to deal with it. This paper studies identification problems in the presence of unobserved heterogeneity under weak assumptions, by exploring identification in nonparametric finite mixture models. We then propose a fully nonparametric estimation method.

A generic mixture model takes the following form. Consider a probability distribution function $F_\alpha(\cdot)$, indexed by a random variable $\alpha$ that takes values on a sample space $\mathcal A$. $\alpha$ is sometimes called a mixing variable or a latent variable. It can be interpreted as a term representing unobserved heterogeneity. Let $G$ denote the probability distribution for $\alpha$. Define

equation[equation omitted — 79 chars of source]

The researcher observes $w$ distributed according to $F$. In other words, the mixture distribution $F(\cdot)$ is generated by mixing the component probability measures $F_\alpha(\cdot), \alpha \in \mathcal A$ according to the mixing distribution $G(\cdot)$. In an important special case where $G$ is discretely distributed and the space $\mathcal A$ is finite, ((ref)) becomes

equation[equation omitted — 102 chars of source]

For example, suppose there are $J$ types of economic agents that have type specific distributions $F_j(z), j = 1,...,J$. If type $j$ is drawn with probability $\lambda_j$, the resulting data obeys the finite mixture model ((ref)). The $F$ defined in ((ref)) is called a finite mixture distribution function. This is the main concern of the current paper. Since the paper presents various results with different models, a brief discussion of the overall nature of our contributions might be in order, as we now summarize in the following three points:

\ \

{\it (i) Relation to other identification results.} \quad As we mention below, currently available nonparametric identification strategies for finite mixtures often require either (A) multiple observations (a leading example being panel data) or (B) exclusion restriction and/or specific conditions on the shapes of the component distribution functions $F_\alpha, \alpha \in \mathcal A$. All the results in this paper concern identification in cross-section settings (i.e. the econometrician never observes an individual with a particular realization of the mixing/latent variable $\alpha$ more than once), therefore our identification strategy has little in common with the ones that belong to Category (A). Some papers in Category (B) assume exclusion restrictions, then invoke an identification-at-infinity type argument by focusing on observations at the tails of the component distributions. This paper does not rely on exclusion restrictions (and it even allows the mixture weights to depend on covariates in the model discussed in Section (ref)). Some other papers in Category (B) rely on symmetry of $F_\alpha$, which we do not assume either.

\ \

{\it (ii) Source of identification.} \quad The primal identification power in this paper comes from what may be called “componentwise shift-restriction" when a covariate is observed. That is, under an independence assumption, each component distribution generates a set of cross restrictions over a family indexed by the covariate values. Here the term “shift-restriction" is adopted from \citeasnoun{klein2002shift}, who consider semiparametric estimation of ordered response models (hence their paper is not about mixtures) though the identification strategy in the current paper not directly related to theirs: it is crucial to observe that in our case {\it for each component distribution} we obtain continuous limit analogues of shift-restrictions defined for a (possibly finite) set of covariate values. These componentwise shift-restrictions --- and equally importantly, the fact that after aggregating such latent distributions, the resulting mixture distribution function {\it lacks} the shift-restriction property under a “non-parallel condition" described later --- deliver fully nonparametric identification.

\ \

{\it (iii) On identification/estimation strategies.} \quad The “componentwise shift-restriction" described above can be usefully exploited after taking Fourier/Laplace transforms of the model. We then take limits in the Fourier/Laplace domains. As noted in {\it (i)} above, this is quite different from the approach based on exclusion restrictions together with nonparametric estimators with observations at the tails of the component distributions. Moreover, basing identification on the upper and lower tails generally limits the number of identifiable components, typically to the case with $J=2$, whereas our approach can be used to identify models with arbitrary $J$ (Section (ref)). The number of components $J$ itself will be identified in our approach as well. Alternatively, if we impose a large support restriction on covariates we can in principle establish identification in a straightforward manner. This would be a variant of the identification-at-infinity argument, and our approach does not share this feature either. As we shall see in Section (ref) it is possible to estimate the entire mixture model fully nonparametrically with standard polynomial convergence rates under mild assumptions. This desirable property is achieved without focusing on observations at the tails of the component distributions, nor a large support condition on the covariate.

\ \

We now mention some literature on the use of mixture models in general, followed by existing methods of identification for (finite) mixtures. As noted before, mixtures are commonly used in models with unobserved heterogeneity, especially in labor economics and industrial organization. See, for example, \citeasnoun{cameron1998life}, \citeasnoun{keane-wolpin}, \citeasnoun{berry1996airline}, \citeasnoun{arcidiacono2011conditional}, and \citeasnoun{aguirregabiria2013identification} for applications of finite mixture models in these fields. They are also used extensively in duration models with unobserved heterogeneity; see \citeasnoun{heckman1984method}, \citeasnoun{heckman1994econometric} and \citeasnoun{van2001duration}. A somewhat different use of mixtures can be found in models of regime changes, which can be viewed as finite mixture models. \citeasnoun{porter1983study}, for example, uses a switching simultaneous equations for an empirical IO model (see also \citeasnoun{ellison1994theories} and \citeasnoun{lee1984switching}). Some models with multiple equilibria can be regarded as mixtures as well (e.g. \citeasnoun{berry2006identification}, \citeasnoun{echenique2009testing}). Finally, contaminated models, as analyzed by \citeasnoun{horowitz1995identification} and \citeasnoun{manski2003partial} can be formulated as mixture models.

The most common estimation method for mixture models is parametric maximum likelihood (ML). In the notation introduced in ((ref)), ML requires parameterizing $F_\alpha(\cdot)$ and $G(\cdot)$ so that they are known up to a finite number of parameters. The EM algorithm often provides a convenient way to calculate the ML estimator for a mixture model.

This paper considers nonparametric identification problems in finite mixture models. The goal of the paper is to show that it is possible to treat the component distributions of a mixture model in a flexible manner. It should be noted that \citeasnoun{jewell1982mixtures} and \citeasnoun{heckman1984method} provide important identification results for mixture models in semiparametric settings. Again in the notation in ((ref)), these authors treat the component distributions $F_\alpha(\cdot)$ parametrically, (so that it is parameterized as $F_\alpha(\cdot,\theta)$, say, by a finite dimensional parameter $\theta$) while treating $G$ nonparametrically. They develop nonparametric ML estimators (NPMLE) for this type of models. Note that NPMLE, in actual applications, yields nonparametric estimates for $G$ that are typically discrete distributions with only a few support points. This fact may suggest that considering finite mixture distributions from the outset, as this paper does, is likely to be flexible enough for practical purposes.

Identification problems of finite mixtures have attracted much attention in the statistics literature. Teicher's pioneering work (Teicher \citeyear*{teicher1961identifiability}, \citeyear*{teicher1963identifiability}) initiated this research area. \citeasnoun{rao1992identifiability} provides a nice summary of this topic. See, also, \citeasnoun{lindsay1995mixture} for a comprehensive treatment of mixture models including their identification issues. Many results known in this area assume parametric component distributions. Indeed, as \citeasnoun{hall2003nonparametric} put it, “(v)ery little is known of the potential for consistent nonparametric inference in mixtures without training data.” Nevertheless, a number of papers have appeared on this subject, especially after the first version of the current paper was circulated. These include approaches based on multiple outcomes (e.g. \citeasnoun{bonhomme2016non}, \citeasnoun{bonhomme2016estimating}, \citeasnoun{d2015identification}, \citeasnoun{kasahara2009nonparametric}), or identification results based on exclusion restrictions, with/without tail restrictions on component distributions (e.g. \citeasnoun{adams2016finite}, \citeasnoun{compiani2016using}, \citeasnoun{henry2014partial}, \citeasnoun{henry2010identifying}, \citeasnoun{hohmann2013semiparametric}, \citeasnoun{jochmans2017inference}), or methods based on symmetry restrictions (e.g. \citeasnoun{butucea2014semiparametric}, \citeasnoun{hohmann2013two}).

The main result of the present paper is that nonparametric treatment of the component distributions of a finite mixture model is possible in a cross-sectional setting, if appropriate covariates are available.

Mixture Model with Covariates

Consider random vectors $z$ and $x$. Suppose the conditional distribution of $z$ given $x$ is given by a finite mixture model of the following form:

equation[equation omitted — 146 chars of source]

The main goal is to identify the mixing probability weights $\lambda_j, j = 1,...,J$ and the conditional component distributions $F_j(z|x)$ from the conditional mixture distribution $F(\cdot|x)$, using nonparametric restrictions. Sections (ref) - (ref) consider the case where $J = 2$. The above expression then becomes:

equation[equation omitted — 106 chars of source]

The case with $\lambda = 0$ is ruled out as we seek identification only up to labeling. Section (ref) considers an extension to the case with $J \geq 3$.

Regression

This section develops basic nonparametric identification results for ((ref)). Suppose $z$ and $x$ reside in $\mathbb{R}$ and $\mathbb{R}^k$, respectively. Define $$ m_j(x) = \int_\mathbb{R} z dF_j(z|x), j = 1,2, $$ i.e. the mean regression functions of the component distributions. Let $F_{\epsilon|x}^j, j = 1,2$ denote the distribution functions of the random variables \[ \epsilon_j = z_j - m_j(x), \quad j = 1,2. \] Note that by construction $\int \epsilon dF_{\epsilon|x}^j(\epsilon) = 0, j = 1,2$. With this notation $F_j(z|x) = F_{\epsilon|x}^j(z - m_i(x)), j = 1,2$, and the model ((ref)) can be written as

equation[equation omitted — 122 chars of source]

Our goal in this section is then to identify the elements of the right hand side of ((ref)) nonparametrically from the knowledge of $F(\cdot|x)$ evaluated at various $x$. Note that the model ((ref)) is further interpreted as a switching regression model:

equation[equation omitted — 265 chars of source]

Models as described above are conventionally estimated using parametric ML. That is, the researcher specifies (1) parametric functions for $m_1(x)$, $m_2(x)$, e.g. $m_1(x) = \beta_1^\top x$, $m_2(x) = \beta_2^\top x$, and (2) parametric distribution functions for $F_{\epsilon|x}^1$ and $F_{\epsilon|x}^2$, e.g. $\epsilon_1|x \sim N(0,\sigma_1^2)$, $\epsilon_2|x \sim N(0,\sigma_2^2)$. Examples of such methods can be found in \citeasnoun{quandt1972new} and \citeasnoun{kiefer1978discrete}; see also \citeasnoun{hamilton1989new} for application of ML in a time series context. The EM algorithm is often used in computing the ML estimator.

While the parametric approach is attractive and practical, the consistency of ML depends crucially on whether the parametric model is correctly specified or not. For example, even if $m_1$ and $m_2$ have the correct form, misspecifications in $F_{\epsilon|x}^1$ and $F_{\epsilon|x}^2$ would result in a failure of consistency. This is quite different from standard (possibly nonlinear) regression models, for which many distribution free estimators are available. This may discourage applied researchers from using mixture models. It also raises a more fundamental question: Is the model ((ref)) identified under weaker, non/semi-parametric assumptions? The results in this section provide a positive answer to this question.

Before discussing how nonparametric identification is possible, it may be helpful to see that a certain nonparametric restriction fails to generate identification in the model. Arguably the most common identification assumption for the standard regression model (without mixtures) is the conditional mean restriction. In our case, by the construction of $F_{\epsilon|x}^1$ and $F_{\epsilon|x}^2$ we have $\int_\mathbb{R} \epsilon dF_{\epsilon|x}^1(\epsilon) = 0$ and $\int_\mathbb{R} \epsilon dF_{\epsilon|x}^2(\epsilon) = 0$. The question is whether the knowledge of the conditional mixture distribution $F(z|x)$ at various $x$, combined with these “restrictions,” uniquely determine $F_{\epsilon|x}^1$, $F_{\epsilon|x}^2$, $m_1$, $m_2$, and $\lambda$. The answer is negative; at each $x$, we can split the mixture distribution $F(z|x)$ into increasing and right continuous $\mathbb{R}_+$-valued functions $a(z)$ and $b(z)$, say, so that $F(z|x) = a(z) + b(z)$. If we let $\lambda = \int da(z)$, $m_1(x)=\frac{1}{\lambda}\int z da(z),$ $m_2(x) = \frac{1}{1 - \lambda}\int z db(z)$, $F_{\epsilon|x}^1(\epsilon) = a(\epsilon + m_1(x))/\lambda$ and $F_{\epsilon|x}^2(\epsilon) = b(\epsilon + m_2(x))/(1-\lambda)$ they would satisfy all the available restrictions and information at all $x$. Even if $m_1$ and $m_2$ are completely parameterized, the model is not identified; “splitting” of $F(z|x)$ is not unique.

While it is straightforward to see the above identification failure, it highlights the fact the conditional mean zero condition allows “too many” ways to split the mixture distribution, thereby failing to deliver identification. Fortunately, however, there exists an alternative nonparametric restriction which identifies the model ((ref)). In what follows we focus on independence restrictions, i.e. independence of $(\epsilon_1,\epsilon_2)$ from $x$.

remNote that it suffices to assume that the independence restriction holds (i) for just one element of the $k$-vector of covariates (wlog we assume that it is the first element) (ii) over a small subset of the support of the element. The dependence property between $\epsilon$'s and the elements of $x$ other than the first is completely left unspecified. In this sense the independence requirement should be interpreted as a conditional independence assumption. With a rich set of controls such a requirement might be regarded reasonable. Note this point applies to all the other identification results in this paper as well.

First identification result

Our first result is concerned with cases where at least one element of the vector of covariates $x = (x^1,...,x^k)^\top$ is continuous. Assume that the first $k^*$ elements $x^1,...,x^{k^*}$ are continuous covariates. We establish nonparametric identifiability at $x = x_0$ utilizing local variation in one of the $k^*$ continuous covariates. It is convenient to assume that the first element $x^1$ is such an element, which is assumed to be prior knowledge both for identification and estimation. The following notation is useful in considering local variations of $x^1$: for a point $x_0 = (x_0^1,...,x_0^k)^\top \in \mathbb{R}$, define $$ N^1(x_0,\delta) = \{(x^1,x_0^2,...,x_0^k)^\top \in \mathbb{R}^k| x^1 \in (x_0^1 - \delta,x_0^1 + \delta)\}. $$

assFor some $\delta > 0$, \begin{enumerate}[(i)] • $\epsilon_1|x \sim F_1$ and $\epsilon_2|x \sim F_2$ at all $x \in N^1(x_0,\delta)$ where $F_1$ and $F_2$ do not depend on the value of $x$, • If $0 < \lambda < 1$, $m_1(x_0) - m_1(x) \neq m_2(x_0) - m_2(x)$, for all $x \in N^1(x_0,\delta)$, $x \neq x_0$, • $m_1$ and $m_2$ are continuous in $x^1$ at $x_0$. \end{enumerate}

With the notation above, ((ref)) is written as:

equation[equation omitted — 97 chars of source]

Note that the mixing distribution is allowed to be degenerate, i.e. $J = 1$. As a convention let $\lambda = 1$ if the mixing model is degenerate. That is, with degeneration ((ref)) becomes

equation[equation omitted — 41 chars of source]

The parameter space of $\lambda$ is therefore $(0,1]$.

We first discuss identification of the functions $m_1(\cdot), m_2(\cdot)$ in a neighborhood of the point $x_0 \in \mathbb{R}^k$. To this end a set of regularity conditions for nonparametric identification are stated in terms of moment generating functions. Let $$ M_i(t) = \int_\mathbb{R} e^{t\epsilon}dF_i(\epsilon), \quad i = 1,2, $$ for all $t$ such that this integral exists. $M_1$ and $M_2$ are the moment generating functions of the disturbance terms $\epsilon_1$ and $\epsilon_2$. Define $$ D(x) := m_2(x) - m_1(x) $$ on $\mathbb{R}^k$ and $$ h(c,t) : = e^{tD(x_0)(1 + c)} \frac {M_2(t)} {M_1(t)}, \quad c \in \mathbb{R}, t \in \mathbb{R}. $$ The following imposes a very weak regularity condition on the behavior of these moment generating functions.

ass\begin{enumerate}[(i)] • The domains of $M_1(t)$ and $M_2(t)$ are $(-\infty,\infty)$, and • For some $\varepsilon >0$ either $h(\pm\varepsilon,t) = O(1)$ or $1/ h(\pm\varepsilon,t) = O(1)$, or both hold as $t \rightarrow +\infty$. Moreover, the same holds as $t \rightarrow - \infty$. \end{enumerate}
remNote that the requirement (ref) for the asymptotic behavior of the ratio $\frac {M_2(t)} {M_1(t)}$ is very weak and reasonable, as it allows the ratio to grow, decline or remain bounded as $t$ diverges.

Let $M(t|x)$ denote the moment generating function of $z$ conditional on $x$, that is, $$ M(t|x) := \int_\mathbb{R} e^{tz}dF(z|x), $$ whose domain, by ((ref)) and Assumption (ref)((ref)), is $(-\infty,\infty)$, and also let $$ R(t,x) := \frac{M(t|x)}{M(t|x_0)}. $$ Note that these functions are observable. The domain of these functions are $\mathbb{R}^k \times (-\infty,\infty)$ by Assumption (ref)((ref)).

lemSuppose Assumptions (ref) and (ref) hold. Then there exists $\delta' \in (0,\delta)$ such that for every $x' \in N^1(x_0,\delta')$ \begin{enumerate}[(i)] • $\lim_{t \rightarrow \infty}\frac 1 t \log R(t,x') = m_1(x') - m_1(x_0) $ or $ \lim_{t \rightarrow \infty}\frac 1 t \log R(t,x') = m_2(x') - m_2(x_0)$, and • $\lim_{t \rightarrow -\infty}\frac 1 t \log R(t,x') = m_1(x') - m_1(x_0) $ or $ \lim_{t \rightarrow -\infty}\frac 1 t \log R(t,x') = m_2(x') - m_2(x_0)$ hold. \end{enumerate}
proof[Proof of Lemma (ref)] First consider the case with $0 < \lambda < 1$. By the continuity condition (Assumption (ref)((ref))), there exist a $\delta' \in (0,\delta)$ such that \begin{equation} |m_2(x') - m_2(x_0)| < \frac{ \epsilon|D(x_0)|} 2 \qquad and \qquad |m_1(x') - m_1(x_0)| < \frac{ \epsilon|D(x_0)|} 2 \end{equation} for all $x' \in N^1(x_0,\delta')$. By ((ref)) we have \begin{equation} M(t|x) = \lambda e^{tm_1(x)}M_1(t) + (1 - \lambda) e^{tm_2(x)}M_2(t). \end{equation} Now we prove part (ref), i.e., the result with $t \rightarrow \infty$. Suppose $h(\pm\epsilon,t) = O(1)$ holds. Write \begin{align*} \frac 1 t \log R(x',t) &= \frac 1 t \log \left ( \frac { \lambda e^{tm_1(x')}M_1(t) + (1 - \lambda) e^{tm_2(x')}M_2(t)} { \lambda e^{tm_1(x_0)}M_1(t) + (1 - \lambda) e^{tm_2(x_0)}M_2(t)} \right ) \\ &= m_1(x') - m_1(x_0) + \frac 1 t \log \left ( \frac{\lambda + (1 - \lambda) e^{t[m_2(x') - m_1(x')] \frac{M_2(t)}{M_1(t)} }}{\lambda + (1 - \lambda) e^{t[m_2(x_0) - m_1(x_0)] \frac{M_2(t)}{M_1(t)} }} \right ) \end{align*} Note that ((ref)) guarantees that $|m_2(x') - m_1(x')|$ is less than $|D(x_0)|(1 + \epsilon)$. We have $$ \lim_{t \rightarrow \infty}\frac 1 t \log R(x',t) = m_1(x') - m_1(x_0). $$ If $1/h(\pm\epsilon,t) = O(1)$ instead, then write $$ \frac 1 t \log R(x',t) = m_2(x') - m_2(x_0) + \frac 1 t \log \left ( \frac {\lambda e^{t[m_1(x') - m_2(x')] \frac{M_1(t)}{M_2(t)} } + (1 - \lambda) } {\lambda e^{t[m_1(x_0) - m_2(x_0)] \frac{M_1(t)}{M_2(t)} } + (1 - \lambda) } \right ) $$ and again by $|m_2(x') - m_1(x')| < |D(x_0)|(1 + \epsilon)$ we obtain $$ \lim_{t \rightarrow \infty}\frac 1 t \log R(x',t) = m_2(x') - m_2(x_0). $$ If both hold, then it has to be the case that $D(x_0)= 0$. If, on top of that, $D(x') = m_2(x') - m_1(x') > 0$ then $$ \lim_{t \rightarrow -\infty}\frac 1 t \log R(x',t) = m_1(x') - m_1(x_0) $$ and $$ \lim_{t \rightarrow \infty}\frac 1 t \log R(x',t) = m_2(x') - m_2(x_0). $$ The analysis of the case with $D(x') = m_2(x') - m_1(x') < 0$ is similar. The proof of part (ref) is similar. If $\lambda = 1$ (i.e. the mixing distribution is degenerate) we have $$ \frac 1 t \log R(x',t) = m_1(x') - m_1(x_0) $$ thus the claim trivially holds.

Lemma (ref) suggests that the slopes of $m_1$ and $m_2$ are identified as far as the following condition holds. To state it, define $$ \mathbb{E}[z|x] = \int z d F(z|x) $$ and $$ \lambda_c := \frac { \mathbb{E}[z|x] - \mathbb{E}[z|x_0] - (1 + c) \lim_{t \rightarrow - \infty} \frac 1 t \log R(t,x) } { \lim_{t \rightarrow + \infty} \frac 1 t \log R(t,x) - (1 + c)\lim_{t \rightarrow - \infty} \frac 1 t \log R(t,x)}. $$ Note these are well defined under Assumption (ref) ((ref)). The constant $\delta$ in the following condition will be specified in the statements of Lemmas (ref) and (ref).

conditionEither \begin{enumerate}[(i)] • $ \lim_{t \rightarrow \infty}\frac 1 t \log R(t,x) \neq \lim_{t \rightarrow -\infty}\frac 1 t \log R(t,x) \text{ for some } x \in N^1(x_0,\delta) $ or • $\lim_{c \downarrow 0} \lambda_{c} = 1$ holds. \end{enumerate}

With this condition we have:

lemSuppose Assumptions (ref), (ref) and Condition (ref) hold. Then there exists $\delta' \in (0,\delta)$ such that $F(\cdot|x), x \in N^1(x_0,\delta)$ uniquely determines the value of $\lambda$, and moreover, $$(m_1(x) - m_1(x_0),m_2(x) - m_2(x_0)) \text{ if } \lambda \in (0,1) $$ up to labeling and $$m_1(x) - m_1(x_0)\text{ if } \lambda = 1$$ for all $x$ in $N^1(x_0,\delta')$ as well.
proofFirst consider the case with $\lambda \in (0,1)$. Suppose Condition (ref)(ref) fails, i.e. $ \lim_{t \rightarrow \infty}\frac 1 t \log R(t,x) = \lim_{t \rightarrow -\infty}\frac 1 t \log R(t,x) \text{ for every } x \in N^1(x_0,\delta'). $ In view of Lemma (ref) these limits are either equal to $m_1(x) - m_1(x_0)$ or $m_2(x) - m_2(x_0)$. Wlog suppose it is the former. Note \begin{equation} \mathbb{E}[z|x] = \lambda m_1(x) + (1 - \lambda) m_2(x), \end{equation} therefore \begin{align*} \mathbb{E}[z|x] - \mathbb{E}[z|x_0] & = \lambda[(m_1(x) - m_1(x_0)) - (m_2(x) - m_2(x_0))] + (m_2(x) - m_2(x_0)) \\ &= (1 - \lambda)[(m_2(x) - m_2(x_0)) - (m_1(x) - m_1(x_0))] + (m_1(x) - m_1(x_0)). \end{align*} Using this \begin{align*} \lambda_{c} &= \frac{(1 - \lambda)[(m_2(x) - m_2(x_0)) - (m_1(x) - m_1(x_0))] + (m_1(x) - m_1(x_0))- (1 + c)(m_1(x) - m_1(x_0))}{(m_1(x) - m_1(x_0)) - (1+c)(m_1(x) - m_1(x_0))} \\ &= -\frac{(1 - \lambda) [(m_2(x) - m_2(x_0)) - (m_1(x) - m_1(x_0))] }{c (m_1(x) - m_1(x_0) ) } + 1. \end{align*} Thus Condition (ref)(ref) does not hold either. In sum, if $\lambda \neq 1$ then Condition (ref) reduces to its first part, i.e. Condition (ref)(ref). Lemma (ref) and Condition (ref)(ref) imply either $$ \lim_{t \rightarrow + \infty} \frac 1 t \log R(t,x) = m_1(x) - m_1(x_0), \quad \lim_{t \rightarrow - \infty} \frac 1 t \log R(t,x) = m_2(x) - m_2(x_0)] $$ or $$ \lim_{t \rightarrow + \infty} \frac 1 t \log R(t,x) = m_2(x) - m_2(x_0), \quad \lim_{t \rightarrow - \infty} \frac 1 t \log R(t,x) = m_1(x) - m_1(x_0)]. $$ Either way the slopes are identified. If the former holds, then \begin{align*} \lambda_{c} &= \frac{(1 - \lambda)[(m_2(x) - m_2(x_0)) - (m_1(x) - m_1(x_0))] + (m_1(x) - m_1(x_0))- (1 + c)(m_2(x) - m_2(x_0))}{(m_1(x) - m_1(x_0)) - (1+c)(m_2(x) - m_2(x_0))} \\ & \rightarrow \lambda \end{align*} as $c \downarrow 0$, which identifies $\lambda$. In the latter case re-labeling delivers the result, with $\lambda$ replaced by $1-\lambda$. Next, consider the case with $\lambda = 1$. Then Condition (ref)(ref) cannot hold; as noted before $$ \frac 1 t \log R(x',t) = m_1(x') - m_1(x_0) $$ (which identifies the slope). On the other hand \begin{align*} \lambda_{\delta} &= \frac{ (m_1(x) - m_1(x_0))- (1 + \delta)(m_1(x) - m_1(x_0))} {(m_1(x) - m_1(x_0)) - (1+\delta)(m_1(x) - m_1(x_0))} \\ & = 1, \end{align*} so indeed Condition (ref)(ref) is consistent with $\lambda = 1$. Moreover this shows that the limit of $\lambda_{\delta}$ once again identifies $\lambda$.
remCondition (ref) is the main regularity restriction for our first identifiability result. Importantly, it is testable, as both $R(t,x)$ and $\lambda_\delta$ are observable.
remA sufficient condition.

The next Lemma gives a full identification result. Let ${\mathcal F}(\mathbb{R}^p)$ denote the space of distribution functions on $\mathbb{R}^p$ for some $p \in \mathbb{N}$. Define $$ \bar{\mathcal F}(\mathbb{R}^p) = \{F: \int u F(du) = 0, F \in {\mathcal F}(\mathbb{R}^p)\}, $$ the set of distribution functions with mean zero. The parameter space of $(F_1(\cdot),F_2(\cdot))$ is given by $\bar {\mathcal F}(\mathbb{R})^2$. Also, for a set $\mathcal C \subset \mathbb{R}^k$ let $\mathcal V(\mathcal C)$ denote the space of all real valued functions on $\mathcal C$.

lemSuppose Assumptions (ref), (ref) and Condition (ref) hold. Then there exists $\delta' \in (0,\delta)$ such that $F(\cdot|x), x \in N^1(x_0,\delta)$ uniquely determines $(\lambda,F_1(\cdot),F_2(\cdot),m_1(\cdot),m_2(\cdot))$ in the set $(0,1] \times \bar{\mathcal F}(\mathbb{R})^2 \times {\mathcal V(N^1(x_0,\delta'))}^2$ up to labeling.
proof[Proof of Lemma (ref)] Define $\dot M(t,x) = \frac \partial {\partial t} M(t,x)$, $\ddot M(t,x) = \frac {\partial^2} {(\partial t)^2}M(t,x)$, $\ddot M_i(t) =\frac {\partial^2} {(\partial t)^2} M_i(t), i = 1,2$ and $\ddot M_i(t) = \frac {\partial^2} {(\partial t)^2}M_i(t), i = 1,2$, whose existences follow from Assumption (ref) ((ref)). Note \begin{equation} \dot M(0|x) = \int z dF(z|x) = \lambda m_1(x) + (1 - \lambda) m_2(x). \end{equation} Using this, \begin{align*} \dot M(0|x_0) - \dot M(0|x) & = \lambda[(m_1(x_0) - m_1(x)) - (m_2(x_0) - m_2(x))] + (m_2(x_0) - m_2(x)) \\ &= (1 - \lambda)[(m_2(x_0) - m_2(x)) - (m_1(x_0) - m_1(x))] + (m_1(x_0) - m_1(x)). \end{align*} By this and Assumption (ref)((ref)), if $0 < \lambda < 1$, $\lambda$ is identified from \begin{equation} \lambda = \frac {[\dot M(0|x_0) - \dot M(0|x)] - \lim_{t \rightarrow - \infty} \frac 1 t \log \frac {M(t|x_0)} {M(t|x)} } { \lim_{t \rightarrow + \infty} \frac 1 t \log \frac {M(t|x_0)} {M(t|x)} - \lim_{t \rightarrow - \infty} \frac 1 t \log \frac {M(t|x_0)} {M(t|x)}} \end{equation} evaluated at an arbitrary $x \in N^1(x_0,\delta')$ (note $\delta'$ is defined in Lemma (ref)), since $\lim_{t \rightarrow + \infty} \frac 1 t \log \frac {M(t|x_0)} {M(t|x)}$ and $\lim_{t \rightarrow - \infty} \frac 1 t \log \frac {M(t|x_0)} {M(t|x)}$ identify the factors $[m_1(x_0) - m_1(x)]$ and $ [m_2(x_0) - m_2(x)]$ by Lemma (ref) (here and in what follows, we assume that $m_2(x_0) - m_1(x_0) < 0$; if $m_2(x_0) - m_1(x_0) > 0$, $\lambda$ should be replaced by $(1-\lambda)$). The right hand side of ((ref)), however, is not well-defined ($ = 0/0$) if the mixing distribution is degenerate, i.e. $\lambda = 1$. To avoid the discontinuity, let $$ \lambda_\delta = \frac {[\dot M(0|x_0) - \dot M(0|x)] - (1 + \delta) \lim_{t \rightarrow - \infty} \frac 1 t \log \frac {M(t|x_0)} {M(t|x)} } { \lim_{t \rightarrow + \infty} \frac 1 t \log \frac {M(t|x_0)} {M(t|x)} - (1 + \delta)\lim_{t \rightarrow - \infty} \frac 1 t \log \frac {M(t|x_0)} {M(t|x)}}, $$ which approaches to $\lambda$ as $\delta \rightarrow 0$ whether $\lambda < 1$ or not. Thus $\lambda$ is determined by $$ \lambda = \lim_{\delta \rightarrow 0} \lambda_\delta. $$ Next, to show that $m_1(x_0)$ and $m_2(x_0)$ are identified, note the basic relationship of the first and second order moments: $$ \ddot M(0|x) = \lambda [m_1(x)^2 + \ddot M_1(0)] + (1 - \lambda)[m_2(x)^2 + \ddot M_2(0)]. $$ Therefore \begin{align*} \ddot M(0|x_0) - \ddot M(0|x) = & \lambda [m_1(x_0)^2 - m_1(x)^2] + (1 - \lambda)[m_2(x_0)^2 - m_2(x)^2] \\ = & \lambda (2m_1(x_0) - [m_1(x_0) - m_1(x)])[m_1(x_0) - m_1(x)] \\ & + (1 -\lambda)(2m_2(x_0) - [m_2(x_0) - m_2(x)])[m_2(x_0) - m_2(x)]. \end{align*} Let $$ C(x) = \left \{\ddot M(0|x_0) - \ddot M(0|x) + \lambda [m_1(x_0) - m_1(x)]^2 + (1 - \lambda) [m_2(x_0) - m_2(x)]^2 \right \}/2, $$ then $$ C(x) = [m_1(x_0) - m_1(x)]\lambda m_1(x_0) + [m_2(x_0) - m_2(x)](1 - \lambda) m_2(x_0). $$ Notice that $C(x)$ is already identified over $N^1(x_0,\delta')$ from the above argument and Lemma (ref). Together with ((ref)), \begin{equation} \begin{bmatrix} C(x) \\ \dot M(0|x_0) \end{bmatrix} = \begin{pmatrix} [m_1(x_0) - m_1(x)] & [m_2(x_0) - m_2(x)] \\ 1 & 1 \end{pmatrix} \begin{pmatrix} \lambda & 0 \\ 0 & (1 - \lambda) \end{pmatrix} \begin{bmatrix} m_1(x_0) \\ m_2(x_0) \end{bmatrix}, \end{equation} for all $x \in N^1(x_0,\delta')$. By Assumptions (ref)((ref)), this can be uniquely solved for $m_1(x_0)$ and $m_2(x_0)$ (if $\lambda = 1$, the above equation can be solved directly to determine $m_1(x_0)$; another way to proceed in the degenerate case is to solve ((ref)) using the Moore-Penrose generalized inverse, which identifies $m_1(x_0)$ and yields the solution that $m_2(x_0) = 0$). As the slopes are already obtained in Lemma (ref), the levels of $m_1$ and $m_2$ over $N^1(x_0,\delta)$ are also identified. The only components remaining are $F_1$ and $F_2$. By evaluating ((ref)) at $x_0$ and $x \in N^1(x_0,\delta')$, $x \neq x_0$, obtain \begin{equation} \begin{bmatrix} M(t|x_0) \\ M(t|x) \end{bmatrix} = E(x_0,x,t) \Lambda \begin{bmatrix} M_1(t) \\ M_2(t) \end{bmatrix}, \end{equation} where \[ E(x,x',t) = \begin{pmatrix} e^{tm_1(x)} & e^{tm_2(x)} \\ e^{tm_1(x')} & e^{tm_2(x')} \end{pmatrix}, \Lambda = \begin{pmatrix} \lambda & 0 \\ 0 & (1 - \lambda) \end{pmatrix}. \] If the mixing distribution is non-degenerate, \begin{align*} Det(E(x_0,x,t))& = e^{t[m_1(x_0) + m_2(x)]} - e^{t[m_1(x) + m_2(x_0)]} \\ & = e^{t[m_1(x_0) + m_2(x)]}\left(1 - e^{t\{[m_1(x) - m_1(x_0)] - [m_2(x) - m_2(x_0)] }\right ) \\ & \neq 0 \end{align*} for all $x \in N^1(x_0,\delta)$, $x \neq x_0$, $t \neq 0$, because of Assumption (ref)((ref)), guaranteeing the invertibility of $E(x_0,x',t)$. Moreover, $$ E(x_0,x,t) = e^{t[m_1(x_0) + m_2(x_0)]} \begin{pmatrix} e^{-tm_2(x_0)} & e^{-tm_1(x_0)} \\ e^{t\left\{[m_1(x) - m_1(x_0)] - m_2(x_0)\right \}} & e^{t\left\{[m_2(x) - m_2(x_0)] - m_1(x_0)\right \}} \end{pmatrix}. $$ Therefore $E(x_0,x,t)$ for all $x \in N^1(x_0,\delta')$ and $t$ are identified from the above argument and Lemma (ref). Evaluate ((ref)) at an arbitrary $x \in N^1(x_0,\delta')$ and solve it to determine $M_1(\cdot)$ and $M_2(\cdot)$. If $\lambda = 1$, solve ((ref)) directly to identify $M_1$ (or, alternatively, use the Moore-Penrose generalized inverse as before). Since distribution functions are uniquely determined by their Laplace transforms (see, for example, \citeasnoun{feller1968introduction}, p.233), $F_1(\cdot)$ and $F_2(\cdot)$ are uniquely determined. This completes the proof.
remTo show the above lemma, some regularity conditions on the nature of $m_1$, $m_2$, $F_1$ and $F_2$ (e.g. Assumptions (ref)((ref)), (ref)((ref))-((ref))) are imposed. Note that such restrictions are not imposed on the parameter set $(0,1] \times \bar{\mathcal F}(\mathbb{R})^2 \times \mathcal V( N^1(x_0,\delta') )^2$. The space of candidate parameters being searched over generally contains parameter values that violate, say, the non-parallel regression function condition as in Assumption (ref)((ref)). The only restrictions imposed on the parameter space are the independence restriction, which enables us to have $\bar{\mathcal F}(\mathbb{R})^2$ as the space of the distributions of $\epsilon$'s, and the mean zero property of $\epsilon$'s, which holds by construction. Lemmas (ref) and (ref) claim that as far as the true parameter value $(\lambda, F_1(\cdot), F_2(\cdot), m_1(\cdot), m_2(\cdot))$ satisfies the regularity conditions like Assumptions (ref)((ref)), (ref)((ref))-((ref))), it is uniquely determined in the unrestricted parameter space $(0,1] \times \bar{\mathcal F}(\mathbb{R})^2 \times \mathcal V( N^1(x_0,\delta') )^2$. This point should be clear from the proof. It is of course much easier to establish nonparametric identification by restricting the parameter space we search over, for example, by making the parameter space for $m_1$ and $m_2$ the space of pairs of functions that are non-parallel. Such a result is not satisfactory from a practical point of view: imposing conditions such as Assumption (ref)((ref)) in estimation is difficult in practice. This is the reason why this paper considers the more challenging problem which removes unnecessary restrictions on the parameter space.
remNote that Lemmas (ref) and (ref) do not require $\lambda <1$. That is, if the true model has $J=1$, the model is still correctly identified (to be a model with just one “type” of individuals).
remSome of the assumptions made above are crucial. The main source of identification is the independence assumption (Assumption (ref)((ref))), as discussed before. Also Assumption (ref)((ref)) is essential. If we have $m_1$ and $m_1$ that are completely parallel everywhere, it is easy to see that the “shift restriction” implied by independence loses its identifying power.
remOn the other hand, some of the assumption made here are “regularity conditions”. First, Assumption (ref)((ref)) imposes a rather strong assumption requiring that the moment generating functions $M_1$ and $M_2$ of $F_1$ and $F_2$ exist over $\mathbb{R}$. Second, Assumption (ref)((ref)) imposes a very mild condition: see Remark (ref). Assumption (ref) is important for this result, and as discussed earlier, it is testable. It is satisfied by a large class of parameters, and interestingly, it even includes the case where $F_1$ and $F_2$ are completely identical.

Second identification result

This section propose an alternative approach for identifying (ref). One advantage of this second identification result is that it is based on characteristic functions, so their existence is not an issue. Like the first identification result, the key sufficient condition, which differs from the MGF based condition in the previous section, is testable, Nonparametric identification holds under the following alternative set of sufficient conditions.

assThere exist three points $x_a, x_b, x_c$ in $\mathbb{R}^k$ such that \begin{enumerate}[(i)] • $\epsilon_1|x \sim F_1$ and $\epsilon_2|x \sim F_2$ at all $x = x_a, x = x_b, x = x_c$, where $F_1$ and $F_2$ do not depend on $x$, • $m_1(x_a) - m_1(x_b) \neq m_2(x_a) - m_2(x_b)$, $m_1(x_a) - m_1(x_c) \neq m_2(x_a) - m_2(x_c)$, and $m_1(x_b) - m_1(x_c) \neq m_2(x_b) - m_2(x_c)$. \end{enumerate}

Assumption (ref) is similar to Assumption (ref), though here the continuity of $m_1$ and $m_2$ is not an issue. Next assumption imposes regularity conditions of the characteristic functions of $F_1$ and $F_2$, defined by $$ \phi_i(t) := \int_{\mathbb{R}} e^{it\epsilon} dF_i(\epsilon), \quad i = 1,2. $$

ass$\lim_{t \rightarrow \infty} \left| \frac {\phi_1(t)} {\phi_2(t)} \right | \rightarrow 0$ or $ \left| \frac {\phi_2(t)} {\phi_1(t)} \right | \rightarrow 0$ or $\lambda = 1$.

It is interesting to compare Assumption (ref) with Condition (ref). The former gives a sufficient condition in terms of the characteristic function, whereas the latter the moment generating function. It holds, for example, if $F_1$ and $F_2$ are the CDFs of $N(0,\sigma_1^2)$, $N(0,\sigma_2^2)$, $\sigma_1^2 \neq \sigma_2^2$. \citeasnoun{teicher1963identifiability} uses an assumption similar to this. Assumption (ref) rules out the case with $F_1 \equiv F_2$, which is allowed by Assumption (ref). Fortunately, just like Condition (ref), the new condition Assumption (ref) is verifiable through the observables, as is clear from the next lemma. This means which of the two identification strategies to be used can be determined by the observable features of the data. To state this more precisely, let $\phi(t|x)$ denote the characteristic functions of the conditional mixture distribution $F(z|x)$, that is, $$ \phi(t|x) := \int_\mathbb{R} e^{itz}dF(z|x), $$ and for $x_0 \in \mathbb{R}^k$ define $$ \rho(x,t) := \frac{\phi(t|x)}{\phi(t|x_0)}, \quad x \in \mathbb{R}^k. $$

conditionThere exists $\epsilon > 0$ such that $$ \lim_{t \rightarrow \infty}|\rho(x,t)| = 1 $$ and $$ \lim_{t \rightarrow \infty} \frac {-i} a \mathrm{Log}\left( \frac{\rho(x,t+a)}{\rho(x,t)} \right) = \mathrm{const.} $$ for every $x \in N^1(x_0,\epsilon)$ and $a \in (0,\epsilon]$ where the constant in the second condition may depend on $x$ and $\mathrm{Log}(z)$ denotes the principal value of the complex logarithm of $z \in \mathbb{C}$.
lemIf $m_1$ and $m_2$ are non-parallel on $N^1(x_0,\epsilon)$, Assumption (ref) and Condition (ref) are equivalent.
proofDefine $\delta(x) := m_2(x) - m_1(x_0)$. Note \begin{eqnarray} \rho(x,t) &=& e^{i t [m_1(x) - m_1(x_0)]}\frac{1 + \frac{1 - \lambda}{\lambda} e^{i t \delta(x)} \frac{\phi_2(t)}{\phi_1(t)} }{1 + \frac{1 - \lambda}{\lambda} e^{i t \delta(x_0)} \frac{\phi_2(t)}{\phi_1(t)} } \\ &=& e^{i t [m_2(x) - m_2(x_0)]}\frac{\frac{1 - \lambda}{\lambda} e^{-i t \delta(x)} \frac{\phi_1(t)}{\phi_2(t)} + 1}{\frac{1 - \lambda}{\lambda} e^{-i t \delta(x_0)} \frac{\phi_1(t)}{\phi_2(t)} + 1} \end{eqnarray} The treatment of the case with $\lambda = 1$ is trivial, thus we maintain that $\lambda \in (0,1)$ in the rest of the proof. It is enough to prove the necessity, since the sufficiency follows from (ref) and (ref), with the constant in the second condition being either $m_1(x) - m_1(x_0)$ or $m_2(x) - m_2(x_0)$. So suppose the necessity fails, i.e. Condition (ref) holds but also \begin{equation} \limsup_{t \to \infty} \left| \frac {\phi_1(t)} {\phi_2(t)} \right | = C, C \in (0,\infty] \end{equation} and \begin{equation} \limsup_{t \to \infty} \left| \frac {\phi_2(t)} {\phi_1(t)} \right | = C', C' \in (0,\infty]. \end{equation} hold. Then if either $C$ or $C'$ is finite (so suppose $C$ is) then there exists a sequence $\{t_k\}_{k=1}^\infty$ such that $\lim_{k \rightarrow \infty} t_k= \infty$ and $\lim_{k \rightarrow \infty} \left |\frac {\phi_1(t_k)} {\phi_2(t_k)} \right | = C$. But then with the first part of Condition (ref) and (ref) we have to have $$ \lim_{k \rightarrow \infty} \left |\frac{\frac{1 - \lambda}{\lambda} e^{-i t_k \delta(x)} \frac{\phi_1(t_k)}{\phi_2(t_k)} + 1}{\frac{1 - \lambda}{\lambda} e^{-i t_k \delta(x_0)} \frac{\phi_1(t_k)}{\phi_2(t_k)} + 1} \right| = 1, x \in N^1(x_0,\epsilon). $$ which holds only if $$ \lim_{k \rightarrow \infty} \left[ \mathrm{Arg}\left( \left( \frac{\phi_1(t_k)}{\phi_2(t_k)} \right ) ^2\right) - \left(t_k [\delta(x) - \delta(x_0)] + 2 \pi \left \lfloor{\frac{1}{2} - \frac{t_k [\delta(x) - \delta(x_0)] }{2 \pi}}\right \rfloor \right) \right ] = 0 $$ at every $x \in N^1(x_0,\epsilon)$. Under the non-parallel hypothesis this is impossible. Finally, if both $C$ and $C'$ are infinite, then there exits two sequences $\{t_k\}_{k=1}^\infty$ and $\{s_k\}_{k=1}^\infty$ such that $\lim_{k \rightarrow \infty} t_k= \infty$, $\lim_{k \rightarrow \infty} s_k= \infty$, $\lim_{k \rightarrow \infty} \left |\frac {\phi_1(t_k)} {\phi_2(t_k)} \right | = \infty$ and $\lim_{k \rightarrow \infty} \left |\frac {\phi_2(s_k)} {\phi_1(s_k)} \right | = \infty$. With (ref) and (ref), these imply that for sufficiently small $a$ $$ \lim_{k \rightarrow \infty} \frac {-i} a \mathrm{Log}\left( \frac{\rho(x,t_k+a)}{\rho(x,t_k)} \right) = m_1(x) - m_1(x_0) $$ and $$ \lim_{k \rightarrow \infty} \frac {-i} a \mathrm{Log}\left( \frac{\rho(x,s_k+a)}{\rho(x,s_k)} \right) = m_1(x) - m_1(x_0) $$ hold simultaneously, which contradicts the second part of Condition (ref).

Finally, assume

ass$\sigma_1^2 := \int \epsilon^2 d F_1(\epsilon)$ and $\sigma_2^2 := \int \epsilon^2 d F_2(\epsilon)$ are finite.

Note that the next lemma holds if the set of the regressors values includes at least three points. It therefore allows, for example, two regressors cases where one regressor is binary and the other is continuous.

lemUnder Assumption (ref) (or Condition (ref)), as well as Assumptions (ref) and (ref), $F(\cdot|x)$ at $x = x_a, x_b$ and $x_c$ uniquely determine $(\lambda, m_1(x_a), m_1(x_b), m_1(x_c), m_2(x_a), m_2(x_b), m_2(x_c), F_1(\cdot), F_2(\cdot))$ in the set $\mathbb{R}^7 \times \bar{\mathcal F}(\mathbb{R})^2$ up to labeling.
proof[Proof of Lemma (ref)] The proof proceeds in two steps. Step 1 considers the slopes of $m_1$ and $m_2$. Using the results in Step 1, Step 2 establishes the identification of all the parameters. \ \ (Step 1) By ((ref)) \begin{equation} \phi(t|x) = \lambda e^{itm_1(x)}\phi_1(t) + (1 - \lambda)e^{itm_2(x)}\phi_2(t). \end{equation} Suppose there exists an alternative set of parameters $$(\lambda^*, m_1^*(x_a), m_1^*(x_b), m_1^*(x_c), m_2^*(x_a), m_2^*(x_b), m_2^*(x_c), F_1^*(\cdot), F_2^*(\cdot))$$ in $\mathbb{R}^7 \times \bar{\mathcal F}(\mathbb{R})^2$ such that \begin{equation} F(z|x) = \lambda^* F_1^*(z - m_1^*(x)) + (1 - \lambda^*)F_2^*(z - m_2^*(x)), \qquad x = x_a, x_b, x_c. \end{equation} Let $\phi_1^*$ and $\phi_2^*$ denote the characteristic functions of $F_1^*$ and $F_2^*$. Then \begin{align} &\lambda e^{itm_1(x_a)}\phi_1(t) + (1 - \lambda)e^{itm_2(x_a)}\phi_2(t) = \lambda^* e^{itm_1^*(x_a)}\phi_1^*(t) + (1 - \lambda^*)e^{itm_2^*(x_a)}\phi_2^*(t), \\ &\lambda e^{itm_1(x_b)}\phi_1(t) + (1 - \lambda)e^{itm_2(x_b)}\phi_2(t) = \lambda^* e^{itm_1^*(x_b)}\phi_1^*(t) + (1 - \lambda^*)e^{itm_2^*(x_b)}\phi_2^*(t), \\ &\lambda e^{itm_1(x_c)}\phi_1(t) + (1 - \lambda)e^{itm_2(x_c)}\phi_2(t) = \lambda^* e^{itm_1^*(x_c)}\phi_1^*(t) + (1 - \lambda^*)e^{itm_2^*(x_c)}\phi_2^*(t). \end{align} Let $\alpha$ and $\beta$ be arbitrary two indices from the index set $\{a,b,c\}$. For a function $f:\mathbb{R}^k \rightarrow \mathbb{R}$, let $\Delta_{\alpha \beta}f$ denote the differences of the values of $f$ at $x_\alpha$ and $x_\beta$, that is, $\Delta_{\alpha\beta}f = f(x_\alpha) - f(x_\beta)$. Define the following function of $t$ that also depends on functions $f_1:\mathbb{R}^k \rightarrow \mathbb{R}$, $f_2:\mathbb{R}^k \rightarrow \mathbb{R}$ and indices $\alpha$ and $\beta$: \begin{align*} H(t;f_1,f_2,\alpha,\beta) & = e^{itf_2(x_\alpha)}\left(1 - e^{it(\Delta_{\alpha\beta} (f_1 - f_2)}\right) \\ & = e^{itf_2(x_\alpha)} \left(1 - e^{it\{[f_1(x_\alpha) - f_1(x_\beta)]-[f_2(x_\alpha) - f_2(x_\beta)]\}}\right). \end{align*} Now, multiply ((ref)) by $e^{it\Delta_{ab}m_2^*}$ then subtract both sides from ((ref)) to obtain \begin{equation} \lambda H(t;m_2^*,m_1,a,b)\phi_1(t) + (1 - \lambda)H(t;m_2^*,m_2,a,b)\phi_2(t) = \lambda^*H(t;m_2^*,m_1^*,a,b) \phi_1^*(t). \end{equation} Repeat this with ((ref)) and $e^{it\Delta_{ab}m_2^*}$ replaced by ((ref)) and $e^{it\Delta_{ac}m_2^*}$: \begin{equation} \lambda H(t;m_2^*,m_1,a,c)\phi_1(t) + (1 - \lambda)H(t;m_2^*,m_2,a,c)\phi_2(t) = \lambda^*H(t;m_2^*,m_1^*,a,c) \phi_1^*(t). \end{equation} ((ref)) and ((ref)) imply \begin{align} \lambda H(t;m_2^*,m_1,a,b) H(t;m_2^*,m_1^*,a,c)\phi_1(t) + (1 - \lambda) H(t;m_2^*,m_2,a,b)H(t;m_2^*,m_1^*,a,c)\phi_2(t) \\ = \lambda^*H(t;m_2^*,m_1^*,a,b)H(t;m_2^*,m_1^*,a,c) \phi_1^*(t), \nonumber \end{align} and \begin{align} \lambda H(t;m_2^*,m_1^*,a,b)H(t;m_2^*,m_1,a,c)\phi_1(t) + (1 - \lambda)H(t;m_2^*,m_1^*,a,b)H(t;m_2^*,m_2,a,c)\phi_2(t) \\ = \lambda^*H(t;m_2^*,m_1^*,a,b)H(t;m_2^*,m_1^*,a,c) \phi_1^*(t), \nonumber \end{align} yielding \begin{align*} \lambda H(t;m_2^*,m_1,a,b)H(t;m_2^*,m_1^*,a,c) \phi_1(t) + (1 - \lambda) H(t;m_2^*,m_2,a,b)H(t;m_2^*,m_1^*,a,c)\phi_2(t) \\ = \lambda H(t;m_2^*,m_1^*,a,b)H(t;m_2^*,m_1,a,c) \phi_1(t) + (1 - \lambda)H(t;m_2^*,m_1^*,a,b)H(t;m_2^*,m_2,a,c)\phi_2(t), \nonumber \end{align*} or \begin{align} &\lambda \left [ H(t;m_2^*,m_1^*,a,b)H(t;m_2^*,m_1,a,c) - H(t;m_2^*,m_1,a,b)H(t;m_2^*,m_1^*,a,c) \right ] \phi_1(t) \\ &= (1 - \lambda) \left [ H(t;m_2^*,m_1^*,a,b)H(t;m_2^*,m_2,a,c) - H(t;m_2^*,m_2,a,b)H(t;m_2^*,m_1^*,a,c) \right ] \phi_2(t). \nonumber \end{align} Divide both sides of ((ref)) by $e^{m_1^*(x_a)}$ and rewriting: \begin{align*} &\lambda e^{itu_1}\left [(1 - e^{it u_{11}})(1 - e^{it u_{12}}) - (1 - e^{itu_{13}})(1 - e^{itu_{14}}) \right ] \phi_1(t) \\ &= (1 - \lambda) e^{itu_2}\left [(1 - e^{it u_{21}})(1 - e^{it u_{22}}) - (1 - e^{itu_{23}})(1 - e^{itu_{24}}) \right ] \phi_2(t) \qquad for all t \end{align*} where $u_1 = m_1(x_a)$, $u_2 = m_2(x_a)$, $u_{11} = \Delta_{ab}(m_2^* - m_1^*)$, $u_{12} = \Delta_{ac}(m_2^* - m_1)$, $u_{13} = \Delta_{ab}(m_2^* - m_1)$, $u_{14} = \Delta_{ac}(m_2^* - m_1^*)$, $u_{21} = \Delta_{ab}(m_2^* - m_1^*) = u_{11}$ , $u_{22} = \Delta_{ac}(m_2^* - m_2)$, $u_{23} = \Delta_{ab}(m_2^* - m_2)$, $u_{24} = \Delta_{ac}(m_2^* - m_1^*) = u_{14}$. First, consider the non-degenerate case, i.e. $\lambda \neq 1$. Define $$ L_1(t) = (1 - e^{it u_{11}})(1 - e^{it u_{12}}) - (1 - e^{itu_{13}})(1 - e^{itu_{14}}) $$ and $$ L_2(t) = (1 - e^{it u_{21}})(1 - e^{it u_{22}}) - (1 - e^{itu_{23}})(1 - e^{itu_{24}}), $$ then \begin{equation} L_1(t) = e^{it(u_2 - u_1)}\frac {1 - \lambda} \lambda \frac {\phi_2(t)} {\phi_1(t)} L_2(t) \qquad for all t. \end{equation} We now use the condition \begin{equation} \lim_{t \rightarrow \infty} \frac {\phi_2(t)}{\phi_1(t)} = 0 \end{equation} from Assumption (ref) (the treatment of the case with $\lim_{t \rightarrow \infty} \frac {\phi_1(t)}{\phi_2(t)} = 0$ is essentially identical). The following argument shows that \begin{equation} L_1(t) = 0 \qquad for all t. \end{equation} Suppose ((ref)) is false, i.e. suppose the set $A = \{t: L_1(t) \neq 0, \quad t \in \mathbb{R}\}$ is non-empty. Pick an arbitrary point $t_0$ from $A$. Then there exists an $\epsilon > 0$ such that $|L_1(t_0)| \geq \epsilon > 0$. But since $\lim_{t \rightarrow \infty} e^{it(u_2 - u_1)}\frac {1 - \lambda} \lambda \frac {\phi_2(t)} {\phi_1(t)} L_2(t) = 0$ under ((ref)), together with ((ref)), there exists $t_1(\epsilon) \in \mathbb{R}$ such that \begin{equation} |L_1(t)| < \frac \epsilon 2 \qquad for all t > t_1(\epsilon). \end{equation} Because of the definition of $t_0$, it must be the case that $t_0 \leq t_1(\epsilon)$. Now, since $L_1(\cdot)$ is a sum of periodic functions, it is almost periodic (see, e.g. \citeasnoun{dunford1958linear})). Therefore there exists a positive number $l(\epsilon)$ such that for all $\tau \in \mathbb{R}$ one can find a $\xi(\tau,\epsilon,l(\epsilon)) \in [\tau,\tau+ l(\epsilon)]$ such that \begin{equation} |L_1(t) - L_1(t + \xi(\tau,\epsilon,l(\epsilon)))| < \frac \epsilon 2 \qquad \text{for all } t \in \mathbb{R}. \end{equation} In particular, evaluating ((ref)) at $t = t_0$ and $\tau = -t_0 + t_1(\epsilon)$; \begin{equation} |L_1(t_0) - L_1(t_0 + \xi^*)| < \frac \epsilon 2 \end{equation} where $\xi^* = \xi(-t_0+t_1(\epsilon),\epsilon,l(\epsilon)))$. But $\xi^* \in [-t_0 + t_1(\epsilon), -t_0 + t_1(\epsilon) + l(\epsilon)]$, therefore $t_0 + \xi^* \leq t_0 - t_0 + t_1(\epsilon) = t_1(\epsilon)$. By ((ref)), \begin{equation} |L_1(t_0 + \xi^*)| < \frac \epsilon 2 \end{equation} Using the triangle inequality, ((ref)) and ((ref)), conclude that \begin{align*} |L_1(t_0)| & \leq |L_1(t_0) - L_1(t_0 + \xi^*)| + |L_1(t_0 + \xi^*)| \\ & < \epsilon. \end{align*} But the $\epsilon$ was originally defined so that $|L_1(t_0)| \geq \epsilon$, contradicting the last inequality. Since the choice of $t_0 \in A$ was arbitrary, ((ref)) is now proved. Next, as $\lambda \neq 0$, ((ref)) and ((ref)) imply that $$ \phi_2(t) L_2(t) = 0 \qquad \text{for all } t. $$ But by the basic properties a characteristic function, $\phi_2(\cdot)$ is continuous and $\phi_2(1) = 0$. Therefore for a $d > 0$, $\phi_2(t) \neq 0$ for all $t \in [-d,d]$. It follows that $L_2(t) = 0$ for all $t \in [-d,d]$. Moreover, $L_2(t)$ is analytic on the entire complex plane, and $[-d,d]$ obviously has an accumulation point, therefore by the identity theorem of analytic functions, $L_2(t) = 0$ for all $t \in \mathbb{R}$. In summary, $L_1(t) = L_2(t) = 0$ for all $t \in \mathbb{R}$, or: \begin{equation} (1 - e^{it u_{11}})(1 - e^{it u_{12}}) - (1 - e^{itu_{13}})(1 - e^{itu_{14}}) = 0 \end{equation} and \begin{equation} (1 - e^{it u_{21}})(1 - e^{it u_{22}}) - (1 - e^{itu_{23}})(1 - e^{itu_{24}}) = 0 \end{equation} for all $t$. These conditions in turn identify the slopes of $m_1$ and $m_2$, as shown by the subsequent argument. Consider the following set of conditions \begin{align} \Delta_{ab}(m_2^* - m_1^*) = \Delta_{ab}(m_2^* - m_1) \text{ and } \Delta_{ac}(m_2^* - m_1^*) = \Delta_{ac}(m_2^* - m_1) \tag{C1}, \\ \Delta_{ab}(m_2^* - m_1^*) = \Delta_{ac}(m_2^* - m_1^*) \text{ and } \Delta_{ab}(m_2^* - m_1) = \Delta_{ac}(m_2^* - m_1) \tag{C2}, \\ \Delta_{ab}(m_2^* - m_1^*) = \Delta_{ab}(m_2^* - m_2) \text{ and } \Delta_{ac}(m_2^* - m_1^*) = \Delta_{ac}(m_2^* - m_2) \tag{C3}, \\ \Delta_{ab}(m_2^* - m_1^*) = \Delta_{ac}(m_2^* - m_1^*) \text{ and } \Delta_{ab}(m_2^* - m_2) = \Delta_{ac}(m_2^* - m_2) \tag{C4}. \end{align} Then by ((ref)) and ((ref)), if $u_{jk} \neq 0$ for all $j = 1,2, k = 1,2,3,4$, one of the following four cases has to be true: (D1): ((ref)) and ((ref)) hold; (D2): ((ref)) and ((ref)) hold; (D3): ((ref)) and ((ref)) hold; (D4): ((ref)) and ((ref)) hold. First, consider (D1). ((ref)) and ((ref)) imply $\Delta_{ab}m_1^* = \Delta_{ab}m_1$ and $\Delta_{ab}m_1^* = \Delta_{ab}m_2$, respectively, thereby yielding $\Delta_{ab}m_1 = \Delta_{ab}m_2$, which violates Assumption (ref)((ref)). Next, turn to (D2). From ((ref)) get $\Delta_{ab} m_1 = \Delta_{ab} m_1^*$ and $\Delta_{ac}m_1 = \Delta_{ac}m_1^*$, therefore $\Delta_{bc}m_1 = \Delta_{bc}m_1^*$. But ((ref)) also implies $\Delta_{bc} m_2^* = \Delta_{bc} m_1^*$ and $\Delta_{bc}m_2^* = \Delta_{bc}m_2$, hence $\Delta_{bc}m_1 = \Delta_{bc}m_2$, violating Assumption (ref)((ref)). Since (D3) is identical to (D2) except for the switched roles of $m_1$ and $m_2$, it also violates Assumption (ref)((ref)). Finally, (D4) also leads to a violation of Assumption (ref)((ref)), because the second equations of ((ref)) and ((ref)) yield $\Delta_{bc} m_1 = \Delta_{bc} m_2$. As (D1)-(D4) are impossible, some of the $u_{jk}$'s should be non-zero. To consider the cases with some non-zero $u_{jk}$, it is useful to introduce the following classification (note that for $i = 1,2$, if $u_{ij} = 0$ for $j = 1$ or 2 (3 or 4), then $u_{ij} = 0$ for $j = 3$ or 4 (1 or 2), Case (i): $u_{11} = 0$ Case (ii): $u_{12} = u_{13} = 0$ Case (iii): $u_{14} = 0$ Case (iv): $u_{21} = 0$ Case (v): $u_{22} = u_{23} = 0$ Case (vi): $u_{24} = 0$ First consider Case (i). Then $ H(t,m_2^*,m_1^*,a,b) = e^{itm_1^*(x_a)}(1 - e^{it(\Delta(m_2^* - m_1^*))}) = 0. $ Therefore ((ref)) becomes \begin{equation} \lambda H(t;m_2^*,m_1,a,b)\phi_1(t) + (1 - \lambda)H(t;m_2^*,m_2,a,b)\phi_2(t) = 0, \end{equation} or \begin{equation} (1 - e^{it\Delta_{ab}(m_2^* - m_1)}) + \frac {1 - \lambda} \lambda \frac {\phi_2(t)} {\phi_1(t)}e^{-itm_1(x_a)}H(t;m_2^*,m_2,a,b) = 0. \end{equation} Let $t \rightarrow \infty$, then again by ((ref)), the third term goes to zero. Since the first term is periodic, it must be the case that $u_{13} = 0$ for all $t \in \mathbb{R}$. Since $1 - \lambda \neq 0$ in the current analysis of the non-degenerate case, $H(t;m_2^*,m_2,a,b)\phi_2(t) = 0$ for all $t$, or, $$ (1 - e^{itu_{23}})\phi_2(t) = 0 \qquad \text{for all }t. $$ As argued before, this means $$ 1 - e^{itu_{23}} = 0 \qquad \text{for } t \in [-d,d] $$ for some $d > 0$. But this is possible iff $u_{23} = 0$. In sum, $u_{11} = 0$ automatically implies that $u_{13} = u_{23} = 0$ as well. But the latter condition means $\Delta_{ab}(m_2^* - m_1) = 0$ and $\Delta_{ab}(m_2^* - m_2) = 0$, which in turn imply $\Delta_{ab}(m_1 - m_2) = 0$, thereby violating Assumption (ref)((ref)). Next, consider Case (ii). This case means that \begin{align} \Delta_{ab}m_2^* &= \Delta_{ab}m_1, \\ \Delta_{ac}m_2^* &= \Delta_{ac}m_1.\nonumber \end{align} On top of this, ((ref)) has to hold at the same time. First, suppose all $u_{2k}, k = 1,2,3,4$ in ((ref)) are non-zero. Then ((ref)) and/or ((ref)) has to hold. Suppose ((ref)) holds. Then \begin{align} \Delta_{ab}m_1^* &= \Delta_{ab}m_2, \\ \Delta_{ac}m_1^* &= \Delta_{ac}m_2.\nonumber \end{align} ((ref)) and ((ref)) imply that slopes of $m_1^*$ and $m_2^*$ have to coincide with those of $m_2$ and $m_1$, respectively, proving a part of the identification result. Next, suppose ((ref)) holds. In particular, the second equation of ((ref)), together with ((ref)) means that $\Delta_{ab}(m_1 - m_2) = \Delta_{ac}(m_1 - m_2)$, or $\Delta_{bc}m_1 = \Delta_{bc}m_2$, violating Assumption (ref)((ref)). To complete the analysis of Case (ii), now suppose some of $u_{2k}, k = 1,2,3,4$ in ((ref)) are zero. If $u_{21} = 0$, then $u_{11} = 0$, but we have already shown that the latter condition leads to a violation of Assumption (ref)((ref)). Next, suppose $u_{22} = 0$, i.e., $\Delta_{ac}m_2^* = \Delta_{ac}m_2$. But with the second equation of ((ref)), $\Delta_{ac}m_1 = \Delta_{ac}m_2$, again violating Assumption (ref)((ref)). If $u_{23} = 0$ or $u_{24} = 0$, it means at least one of $u_{21}$ and $u_{22}$ must be zero, so the above argument covers the cases. This completes the analysis of Case (ii); in sum, Case (ii) implies ((ref)) and ((ref)). Case (iii) is identical to Case (i), with the roles of the indices $b$ and $c$ switched, therefore it violates Assumption (ref)((ref)). Case (iv) is identical to Case (i). Note that case (v) is identical to Case (ii) with the role of the functions $m_1$ and $m_2$ reversed. But the treatment of Case (ii) only uses Equations ((ref)) and ((ref)), which are equivalent to ((ref)) and ((ref)), respectively, after switching $m_1$ and $m_2$. Therefore the above treatment of Case (ii) applies with $m_1$ and $m_2$ reversed; that is, Case (v) implies that \begin{align} \Delta_{ab}m_1^* &= \Delta_{ab}m_1, \\ \Delta_{ac}m_1^* &= \Delta_{ac}m_1.\nonumber \end{align} and \begin{align} \Delta_{ab}m_2^* &= \Delta_{ab}m_2, \\ \Delta_{ac}m_2^* &= \Delta_{ac}m_2.\nonumber \end{align} Finally, Case (vi) is identical to Case (vi). The above arguments prove that if the mixture model is non-degenerate, the only possible cases are either (A): ((ref)) and ((ref)) hold, or (B): ((ref)) and ((ref)) hold. That is, the slopes of $m_1$ and $m_2$ are identified, up to labeling. Next consider the case where the mixture model is degenerate, i.e. $\lambda = 1$. Then ((ref)) is now written as \begin{equation} F_1(z - m_1(x)) = \lambda^* F_1^*(z - m_1^*(x)) + (1 - \lambda^*)F_2^*(z - m_2^*(x)). \end{equation} Define ${\sigma_1^*}^2 = \int \epsilon^2 F_1^*(d\epsilon)$ and ${\sigma_2^*}^2 = \int \epsilon^2 F_2^*(d\epsilon)$. Taking the conditional variance of both sides given $x$, \begin{align*} \sigma_1^2 &= \lambda^*(m_1^*(x)^2 + {\sigma_1^*}^2) + (1 - \lambda^*) (m_2^*(x)^2 + {\sigma_2^*}^2) - [\lambda^*m_1^*(x) + (1 - \lambda^*)m_2^*(x)]^2 \\ &=\lambda^*(1 - \lambda^*)[m_1^*(x) - m_2^*(x)]^2 + \lambda^*{\sigma_1^*}^2 + (1 - \lambda^*) {\sigma_2^*}^2 \qquad \text{ at } x = x_a, x_b \text{ and } x_c. \end{align*} This equation is used to establish identification for the degenerate case. In particular, it admits two solutions: \begin{align} &\lambda^* = 1, \quad {\sigma_1^*}^2 = \sigma_1^2, \\ &[m_1^*(x_a) - m_2^*(x_a)]^2 = [m_1^*(x_b) - m_2^*(x_b)]^2 = [m_1^*(x_c) - m_2^*(x_c)]^2. \end{align} ((ref)) obviously leads to full identification: integrating both sides of ((ref)) gives $m_1^*(x) = m_1(x)$, and this trivially determines $F_1^*(z) = F_1(z)$ for all $z$. ((ref)) implies that, for at least one pair of points, $(x,x')$, say, out of the three points $\{x_a,x_b,x_c\}$, the following holds: \begin{equation} m_1^*(x) - m_2^*(x) = m_1^*(x') - m_2^*(x'). \end{equation} Unlike the case with $\lambda < 1$, this does not fully determine the slopes of $m_1$ and $m_2$ over $\{x_a,x_b,x_c\}$; it will be done in (Step 2). \ \ (Step 2) \ \ We now argue that $\lambda$ is identified whether the model is degenerate or not. Let $m_j^*(x), j = 1,2, x = x_1, x_b, x_c$ be (arbitrary) six numbers that satisfy ((ref)). By (Step 1), in the case $\lambda \neq 1$, they have to satisfy ((ref)) and ((ref)), or, ((ref)) and ((ref)). Similarly, in the case $\lambda = 1$, they have to satisfy ((ref)) (the case with ((ref)) is trivial). For an arbitrary pair of points $(x,x')$ from the three support points ${x_a,x_b,x_c}$, define $$ \lambda(x,x') = \lim_{\delta \downarrow 0} \frac{\int z F^*(dz|x) - \int z F^*(dz|x') - (1 + \delta)(m_2^*(x) - m_2^*(x'))}{(m_1^*(x) - m_1^*(x')) - (1 + \delta)(m_2^*(x) - m_2^*(x'))}. $$ Then $\lambda$ is uniquely determined from the values $m_j^*(x), j = 1,2, x = x_1, x_b, x_c$ by \begin{equation} \max_{(x,x') = (x_a,x_b), (x_a,x_c), (x_b,x_c)} \lambda(x,x'), \end{equation} using an argument as in the proof of Lemma (ref), up to labeling It holds whether $\lambda < 1$ or not. (Note that the maximization in the line above is unnecessary if $\lambda \neq 1$, since $\lambda(x,x')$ is identical for all pairs $(x, x')$ in that case.) Let $(\bar x,\bar x')$ be a maximizer of ((ref)), which is possibly not unique. Now, evaluating ((ref)) at $(\bar x,\bar x')$ and $(\bar x',\bar x)$, instead of $(x,x_0)$ and solving for $m_1$ and $m_2$, obtain $m_1(\bar x)$, $m_2(\bar x)$, $m_1(\bar x')$ and $m_2(\bar x')$ ($m_1(\bar x)$ and $m_1(\bar x')$ in the degenerate case). To identify $F_1$ and $F_2$, use \begin{equation} \begin{bmatrix} \phi(t|\bar x) \\ \phi(t|\bar x') \end{bmatrix} = G(\bar x,\bar x',t) \Lambda \begin{bmatrix} \phi_1(t) \\ \phi_2(t) \end{bmatrix}, \end{equation} where \[ G(x,x',t) = \begin{pmatrix} e^{itm_1(x)} & e^{itm_2(x)} \\ e^{itm_1(x')} & e^{itm_2(x')} \end{pmatrix}, \Lambda = \begin{pmatrix} \lambda & 0 \\ 0 & (1 - \lambda) \end{pmatrix}, \] instead of ((ref)) in the proof of Lemma (ref). Then \begin{align*} \text{Det}(G(\bar x,\bar x,t))& = e^{it[m_1(\bar x) + m_2(\bar x')]} - e^{it[m_1(\bar x') + m_2(\bar x)]} \\ & = e^{it[m_1(\bar x) + m_2(\bar x')]}\left(1 - e^{it\{[m_1(\bar x) - m_1(\bar x')] - [m_2(\bar x') - m_2(\bar x)] }\right ) \\ & \neq 0 \quad \text{ for all } t \neq \frac {2\pi j}{[m_1(\bar x) - m_1(\bar x')] - [m_2(\bar x') - m_2(\bar x)]}, \quad j \in \mathbb{Z}. \end{align*} under Assumption (ref)((ref)) if $\lambda < 1$, therefore $G(\bar x,\bar x,t)$ is invertible (and all of its elements are identified). This determines $\phi_1(t)$ and $\phi_2(t)$ for all $t \neq 0 \text{ for all } t \neq \frac {2\pi j}{[m_1(\bar x) - m_1(\bar x')] - [m_2(\bar x') - m_2(\bar x)]}, j \in \mathbb{Z}$ (as before, solve ((ref)) directly or using the Moore-Penrose inverse in the degenerate case to determine $\phi_1$). Since $\phi_1(t)$ and $\phi_2(t)$ are continuous, they are identified on $\mathbb{R}$. This identifies $F_1$ and $F_2$. The foregoing argument shows that $\lambda$, $F_1(\cdot)$, $F_2(\cdot)$ and $m_1(x)$ and $m_2(x)$ evaluated at two points (i.e. $\bar x$ and $\bar x'$ defined right after ((ref))) out of the three support points $\{x_a,x_b,x_c\}$, are identified Note that $m_1$ and $m_2$ at the third point (= $\tilde x$, say) is identified by the relation $$ \phi(t|\tilde x) = \lambda e^{itm_1(\tilde x)}\phi_1(t) + (1 - \lambda)e^{itm_2(\tilde x)}\phi_2(t) \quad \text{ for all }t. $$ Let \begin{align*} g_\tau (t) & = \frac {\phi(t + \tau|\tilde x)}{\lambda \phi_1(t + \tau)} \\ & = e^{i(t + \tau)m_1(\tilde x)} + \frac{1 - \lambda}{\lambda} e^{i(t+ \tau)m_2(\tilde x)} \frac {\phi_2(t + \tau)}{\phi_1(t + \tau)}, \end{align*} Then under ((ref)) the second term converges to zero as $\tau \rightarrow \infty$, and if we write, for all $c$, \begin{align*} h(c) & = \lim_{\tau \rightarrow \infty} \frac{g_\tau (t+c)}{g_\tau (t)} = e^{ic m_1(\tilde x)}, \end{align*} then $m_1(\tilde x)$ is uniquely determined by the formula $m_1(\tilde x) = \frac{ -i h'(c)}{h(c)} $. If the model is non-degenerate, $m_2(\tilde x)$ is identified from $e^{itm_2(\tilde x)} = \frac {\phi(t|\tilde x) - \lambda e^{itm_1(\tilde x)}\phi_1(t)}{(1 - \lambda)\phi_2(t)}$.
remOnce identification is achieved at some values of $x$, as implied by Lemmas (ref) and (ref), the complete knowledge of $M_1$ and $M_2$ is available. Since the identity for conditional characteristic functions or conditional moment generating functions as in ((ref)) holds for all $t$, it can be used to determine $m_1$ and $m_2$ even at points where they fail to satisfy the non-parallel condition (i.e. Assumption (ref)((ref)) or (ref)((ref))). Suppose $F(\cdot|x)$ is known on a set ${\mathcal X} \in \mathbb{R}^k$. Assume that, for example, Assumptions (ref)((ref)) and (ref) hold. Then $F(\cdot|x), x\in \mathcal X$ uniquely determines $(\lambda,F_1(\cdot),F_2(\cdot),m_1(x),m_2(x))$ for all $x \in \mathcal X$ up to labeling, unless $\lambda = 1 - \lambda = \frac 1 2$ and $F_1(z) = F_2(z)$ for all $z \in \mathbb{R}$.

Third identification result

We now propose an identification strategy that has an approach similar to the first identification result, though differs from it in some important ways. It uses one sided limit (e.g. $t$ tending to {\it positive} infinity) of MGFs and also characteristic functions. Unlike our first result, it for instance addresses the case where $F_1$ and $F_2$ are CDFs of $N(0,\sigma_1^2)$, $N(0,\sigma_2^2)$, $\sigma_1^2 \neq \sigma_2^2$. Moreover, the identification strategy for the distribution functions avoids Laplace inversion, a problematic step in practice. For these reasons it is the identification strategy in this section that will be used to construct our estimator in Section (ref).

Recall our definition of the function $h(\cdot,\cdot)$ (see Assumption (ref)) in the statements of the following assumption.

ass\begin{enumerate}[(i)] • The domains of $M_1(t)$ and $M_2(t)$ include $[0,\infty)$ and for some $\varepsilon >0$ either $h(\pm\varepsilon,t) = O(1)$ or $1/ h(\pm\varepsilon,t) = O(1)$ or both hold as $t \rightarrow +\infty$, or • The domains of $M_1(t)$ and $M_2(t)$ include $(-\infty,0]$ and for some $\varepsilon >0$ either $h(\pm\varepsilon,t) = O(1)$ or $1/ h(\pm\varepsilon,t) = O(1)$ or both hold as $t \rightarrow - \infty$. \end{enumerate}

Note that this assumption does not demand the MGFs $M_1$ and $M_2$ to be defined on the whole real line, sometimes a restrictive assumption.

lemSuppose Assumptions (ref), (ref) and (ref) hold. Then there exists $\epsilon \in (0,\delta)$ such that for every $x \in N^1(x_0,\epsilon)$ and $a \in (0,\epsilon]$ \begin{enumerate}[(i)] • $\lim_{t \rightarrow \infty}\frac 1 t \log R(t,x') = m_1(x') - m_1(x_0) $ or $ \lim_{t \rightarrow \infty}\frac 1 t \log R(t,x') = m_2(x') - m_2(x_0)$ if Assumption (ref)(ref) holds, and $\lim_{t \rightarrow -\infty}\frac 1 t \log R(t,x') = m_1(x') - m_1(x_0) $ or $ \lim_{t \rightarrow -\infty}\frac 1 t \log R(t,x') = m_2(x') - m_2(x_0)$ if Assumption (ref)(ref) holds instead. • $\lim_{t \rightarrow \infty} \frac {-i} a \mathrm{Log}\left( \frac{\rho(x,t+a)}{\rho(x,t)} \right) = m_1(x') - m_1(x_0) $ or $\lim_{t \rightarrow \infty} \frac {-i} a \mathrm{Log}\left( \frac{\rho(x,t+a)}{\rho(x,t)} \right) = m_2(x') - m_2(x_0)$. \end{enumerate}
proofThe proof of Part (ref) is essentially in the proof of Lemma (ref). For Part (ref), note that the ratios on the right hand side of (ref) and by (ref) converge to $1$ as $s \to \infty$. Since $$\rho(s,x) = e^{i s \nabla} \frac{\frac{\lambda}{1-\lambda} e^{i s (m_1(x_1) - m_2(x_1))} \frac{\phi_1(s)}{\phi_2(s)} + 1}{\frac{\lambda}{1-\lambda} e^{i s (m_1(x_0) - m_2(x_0))} \frac{\phi_1(s)}{\phi_2(s)} + 1},$$ and under Assumption (ref) the ratio on the right hand side converges to $1$ as $s \to \infty$. Therefore we have \begin{align*} \lim_{s \rightarrow \infty} \frac{-i}{a} Log \left( \frac{\rho(x,s+a)}{\rho(x,s)} \right) & = \frac{-i}{a} Log (e^{ia\nabla})\\ & = \frac{1}{a} \left( a \nabla + 2 \pi \left \lfloor{\frac{1}{2} - \frac{a \nabla}{2 \pi}}\right \rfloor \right) , \end{align*} where $\mbox{ Log}$ corresponds to the principal value of the log. This limit is a piecewise continuous function of $a$, constant equal to $\nabla$ only when $a$ is small enough to guarantee $a\nabla \in (-\pi,\pi).$ And if $\lambda=1$, $\frac{\phi(s|x_1)}{\phi(s|x_0)}= e^{i s \Delta}$ so that $\lim_{s \rightarrow \infty} \frac{-i}{a} \mbox{ Log} \left( \frac{\phi(s+a|x_1)}{\phi(s+a|x_0)} \left( \frac{\phi(s|x_1)}{\phi(s|x_0)} \right)^{-1} \right) = \frac{1}{a} \left( a \Delta + 2 \pi \left \lfloor{\frac{1}{2} - \frac{a \Delta}{2 \pi}}\right \rfloor \right) .$ By assumption, $m_1(x_1) - m_1(x_0) \neq m_2(x_1) - m_2(x_0)$ that is, $\Delta \neq \nabla$ therefore if the former limit is equal to $\Delta$, one knows $\lambda =1$ and there is no $m_2$.

The constant $\delta$ in the following condition is specified in Assumption (ref).

conditionEither \begin{enumerate}[(i)] • there exists $\epsilon \in (0,\delta)$ such that $ \lim_{t \rightarrow \infty}\frac 1 t \log R(t,x) \neq \lim_{s \rightarrow \infty} \frac{-i}{a} \mbox{\rm Log} \left( \frac{\rho(x,s+a)}{\rho(x,s)} \right) \text{ for every} x \in N^1(x_0,\epsilon)$ and $a \in (0,\epsilon]$ if Assumption (ref)(ref) holds or • there exists $\epsilon \in (0,\delta)$ such that $ \lim_{t \rightarrow \infty}\frac 1 t \log R(t,x) \neq \lim_{s \rightarrow \infty} \frac{-i}{a} \mbox{\rm Log} \left( \frac{\rho(x,s+a)}{\rho(x,s)} \right) \text{ for every } x \in N^1(x_0,\epsilon)$ and $a \in (0,\epsilon]$ if Assumption (ref)(ref) holds or • $\lim_{\delta \downarrow 0} \lambda_{\delta} = 1$ holds. \end{enumerate}

The above condition is verifiable with information in the observables as $\rho$, $R$ and $\lambda_{\delta}$ are all observed.

lemSuppose Assumptions (ref), (ref), (ref) and Condition (ref) hold. Then there exists $\delta' \in (0,\delta)$ such that $F(\cdot|x), x \in N^1(x_0,\delta)$ uniquely determines the value of $\lambda$, and moreover, $$(m_1(x) - m_1(x_0),m_2(x) - m_2(x_0)) \text{ if } \lambda \in (0,1) $$ up to labeling and $$m_1(x) - m_1(x_0)\text{ if } \lambda = 1$$ for all $x$ in $N^1(x_0,\delta')$ as well.
proofSimilar to the proof of Lemma (ref).

Note that we once again needed the non-parallel regression function condition. Once the increments of the regression functions are identified, their levels as well as the mixture weight $\lambda$ are obtained using the same procedure as in the first identification result. Thus we have:

lemSuppose Assumptions (ref), (ref), (ref) and Condition (ref) hold. Then there exists $\delta' \in (0,\delta)$ such that $F(\cdot|x), x \in N^1(x_0,\delta)$ uniquely determines $(\lambda,m_1(\cdot),m_2(\cdot))$ in the set $(0,1] \times \mathcal V( N^1(x_0,\delta') )^2$ up to labeling.

To identify the distribution functions $(F_1(.),F_2(.))$, we now propose another method which will be used to construct our estimator and avoids Laplace inversion. The main benefit of this is that it let us nonparametrically estimate the distribution functions without resorting to empirical MGF inversion, which is hard to handle in terms of obtaining polynomial rates of convergence. We will use previous identification of $\lambda$ and $m_1$ and $m_2$ evaluated at two points only, $x_1$ and $x_0$.

The idea is the following. Equation ((ref)) gives $F(z|x) = \lambda F_1(z - m_1(x)) + (1 -\lambda)F_2(z - m_2(x)), \forall (x,z) \in \mathbb{R}^{k+1},$ implying $ \forall (x,y) \in \mathbb{R}^{k+1},$

equation[equation omitted — 100 chars of source]

Applying Equation ((ref)) to $(x_0,y)$ and $(x_1,y)$ and taking the difference, we obtain

align[align omitted — 173 chars of source]

which means that $\forall y \in \mathbb{R},$ $ F_2(m_1(x_1)-m_2(x_1) + y) - F_2(m_1(x_0)-m_2(x_0) + y) $ is identified. Using recursively identification of this increment and the fact that the conditional cumulative distribution function $F_2$ converges to $1$ at infinity, we obtain identification of $F_2(z),$ $\forall z \in \mathbb{R}.$ Writing $g(x)=m_1(x)-m_2(x)$ and $\delta(x,x')=g(x)-g(x'),$ we assume that $\delta(x_1,x_0)>0.$ Note that $\delta(x_1,x_0) = \Delta - \nabla.$ Now, apply, for a given $z \in \mathbb{R},$ Equation ((ref)) to $y=z - g(x_0)$ to obtain $$F_2(z + \delta(x_1,x_0)) - F_2(z) = \frac{1}{1 -\lambda} (F(z + m_1(x_1) - g(x_0) |x_1) - F(z + m_2(x_0) |x_0)),$$ and, more generally, $\forall j \in \mathbb{N},$

align*[align* omitted — 224 chars of source]

Using $\lim_{j \to \infty} F_2(z + (j+1) \delta(x_1,x_0)) = 1,$ the identifying equation for $F_2(.)$ is

align[align omitted — 195 chars of source]

where the infinite sum is a convergent series of positive terms.

Finally the equation $F(z|x)= \lambda F_1(z - m_1(x)) + (1 - \lambda) F_2(z - m_2(x))$ identifies $F_1(.)$ as $$F_1(z) = \frac{1}{\lambda} \left[ F(z + m_1(x)) - (1 - \lambda) F_2(z + m_1(x) - m_2(x) \right]. $$

lemSuppose Assumptions (ref), (ref), (ref) and Condition (ref) hold. Then there exists $\delta' \in (0,\delta)$ such that $F(\cdot|x), x \in N^1(x_0,\delta)$ uniquely determines $(F_1(\cdot), F_2(\cdot))$ in the set $\bar{\mathcal F}(\mathbb{R})^2$.

A model with “fixed effects”

The model we have focused on so far assumes that heterogeneity is exogenously determined. With $J=2$, a draw $(z,x)$ is generated from the first type of population or from the second with {\it fixed} probabilities $\lambda$ and $1 - \lambda$. This section relaxes this assumption. We assume that the binary probability distribution over the two types/population can depend on $x$ in a completely unrestricted, nonparametric manner. In terms of the switching regression formulation, this means:

equation[equation omitted — 257 chars of source]

where $x$ and $\epsilon_1$ ($\epsilon_2$) are, as before, assumed to be independent. Equivalently, we can write

equation[equation omitted — 103 chars of source]

The goals is now to identify the 5-tuple of functions $(\lambda(\cdot),m_1(\cdot),m_2(\cdot),F_1(\cdot),F_2(\cdot))$ from the joint distribution of $(z,x)$.

This model is of a particular interest in terms its implications. As in the rest of the paper, we often interpret the difference between $(m_1(\cdot),F_1(\cdot) )$ and $(m_1(\cdot),F_1(\cdot) )$ as a representation of unobserved heterogeneity. In a standard panel data regression model often such heterogeneity is represented by a scalar, and when it is assumed to be independent of the regressor it would be representing random effects, whereas if it is allowed to be correlated with the regressor in an arbitrary manner it becomes a fixed effects model. In certain applications fixed effects models are highly desirable. Panel data often offers approaches to deal with fixed effects, a leading case being a linear model with additive scalar-valued fixed effects. The model (ref) (or equivalently (ref)) is in this sense analogous to these fixed effects models. Unobserved heterogeneity in (ref) is function-valued (i.e. $m$ and $F$), as opposed to, say, an additive scalar. Its distribution, represented by $\lambda(x)$, is dependent on $x$ in a fully unrestricted way, accommodating arbitrary correlation between the unobserved heterogeneity and the regressor, so it resembles a panel data fixed effects model in this aspect. In this section we show that (ref) is nonparametrically identified, without requiring panel data, when the finite mixture modeling of unobserved heterogeneity is appropriate. Moreover, unlike in the standard panel data fixed effects model, the distribution of unobserved heterogeneity conditional on $x$ is identified fully nonparametrically. This means we identify the entire model, enabling the researcher to calculate desired counterfactuals.

We replace Assumption (ref) with

assFor some $\delta > 0$, \begin{enumerate}[(i)] • $\epsilon_1|x \sim F_1$ and $\epsilon_2|x \sim F_2$ at all $x \in N^1(x_0,\delta)$ where $F_1$ and $F_2$ do not depend on the value of $x$, • If $0 < \lambda(x_0) < 1$, $m_1(x_0) - m_1(x) \neq m_2(x_0) - m_2(x)$, for all $x \in N^1(x_0,\delta)$, $x \neq x_0$, • $\lambda$, $m_1$ and $m_2$ are continuous in $x^1$ at $x_0$. \end{enumerate}

We maintain Assumption (ref), which, as noted before, is a weak regularity condition. Define $$ K_{+\infty,t}(x) := R(t,x)\exp \left( - t\lim_{s \rightarrow +\infty} \frac 1 s \log(R(s,x)) \right), $$ $$ K_{-\infty, t}(x) := R(t,x)\exp \left( - t\lim_{s \rightarrow -\infty} \frac 1 s \log(R(s,x)) \right), $$ $$ K_{+\infty}(x) := \lim_{t \rightarrow +\infty} K_{+\infty,t}(x) $$ and $$ K_{-\infty}(x) := \lim_{t \rightarrow -\infty} K_{ -\infty,t }(x). $$

Note that the limits in these definitions are well-defined over a neighborhood of $x_0$.

We replace Condition (ref) with:

conditionEither \begin{enumerate}[(i)] • $ \lim_{t \rightarrow \infty}\frac 1 t \log R(t,x) \neq \lim_{t \rightarrow -\infty}\frac 1 t \log R(t,x) \text{ for some } x \in N^1(x_0,\delta). $ or • $K_{+\infty,t}(x) = 1$ for every $t \in \mathbb{R}$ and $x \in N^1(x_0,\delta)$ holds for some $\delta > 0$. \end{enumerate}
lemSuppose Assumptions (ref), (ref) and Condition (ref) hold. Then there exists $\delta' \in (0,\delta)$ such that $F(\cdot|x), x \in N^1(x_0,\delta)$ uniquely determines $\lambda(x)$, and moreover, $$(m_1(x) - m_1(x_0),m_2(x) - m_2(x_0)) \text{ if } \lambda(x)\lambda(x_0) \in (0,1) $$ up to labeling and $$m_1(x) - m_1(x_0)\text{ if } \lambda(x)\lambda(x_0) = 1$$ for all $x$ in $N^1(x_0,\delta')$ as well.
proofSee appendix.

The next result shows that the model that allows $\lambda$ to be arbitrarily dependent on $x$ is nonparametrically identified. Note that the mixture can be degenerate (i.e. $\lambda(x) = 1$) for some values of $x$, and this can be also inferred from the observables. As in the previous identification results presented in Lemmas (ref), (ref) and (ref), its main sufficient condition (i.e. Condition (ref)) is verifiable in terms of observables.

lemSuppose Assumptions (ref), (ref) and Condition (ref) hold. Then $F(\cdot|x), x \in N^1(x_0,\delta)$ uniquely determines $(\lambda(x_0),F_1(\cdot),F_2(\cdot),m_1(x_0),m_2(x_0))$ in the set $(0,1] \times \bar{\mathcal F}(\mathbb{R})^2 \times \mathbb{R}^2$ up to labeling.
proofGiven Lemma (ref) the only remaining task is to identify the levels of $m_1$ and $m_2$ at $x_0$, $F_1$ and $F_2$. Using the notation introduced in the proof of Lemma (ref), with an additional definition $$ \dot \lambda(x) = \lambda(x) - \lambda(x_0), $$ write $$ \mathbb{E} [z|x] - \mathbb{E}[z|x_0] = \lambda(x)[\dot m_1(x) - \dot m_2(x)] + \dot \lambda(x)[m_1(x_0) - m_2(x_0)]. $$ If $\lambda(x) \neq \lambda(x_0)$ then we can proceed as in the proof of Lemma (ref) to show that $m_1(x_0)$ and $m_2(x_0)$ are identified. Accordingly, consider the case $\lambda(x) \neq \lambda(x_0)$. Define $$ c(x) := \frac{ \mathbb{E} [z|x] - \mathbb{E}[z|x_0] - \lambda(x)[\dot m_1(x) - \dot m_2(x)] }{ \dot \lambda(x) }, $$ which is observable by Lemma (ref), then $c(x) = m_1(x_0) - m_2(x_0)$, and we obtain $$ \begin{bmatrix} c(x) \\ \mathbb{E} [z|x] \end{bmatrix} = \begin{pmatrix} 1 & 1 \\ \lambda(x) & 1 - \lambda(x) \end{pmatrix} \begin{bmatrix} m_1(x_0) \\ m_2(x_0) \end{bmatrix}. $$ Since the determinant of the matrix on the right hand side is unity, once again $m_1(x_0)$ and $m_2(x_0)$ are identified. Finally, we proceed as in as in the proof of Lemma (ref) to identify $F_1$ and $F_2$, though here the 2-by-2 matrix in the following display does not factorize: \begin{equation} \begin{bmatrix} M(t|x) \\ M(t|x_0) \end{bmatrix} = \begin{pmatrix} \lambda(x)e^{tm_1(x)} & (1 - \lambda(x) ) e^{tm_1(x)} \\ \lambda(x_0) e^{tm_1(x)} & ( 1 - \lambda(x_0) ) e^{tm_2(x_0)} \end{pmatrix} \begin{bmatrix} m_1(x_0) \\ m_2(x_0) \end{bmatrix}, \quad \rm for every t \in \mathbb{R}. \end{equation} Nevertheless, its determinant is, if $\lambda(x_0) \neq 1$ \begin{eqnarray*} \lambda(x)(1 - \lambda(x_0) ) e^{t[m_1(x) + m_2(x_0)]} &-& \lambda(x_0)(1 - \lambda(x) ) e^{t[m_1(x_0) + m_2(x)]} \\ =&& \lambda(x_0)(1 - \lambda(x_0) ) e^{t[m_1(x_0) + m_2(x)]} \left\{ \frac{\lambda(x)}{\lambda(x_0)} e^{t[\dot m_1(x) - \dot m_2(x) ]} - \frac{1 - \lambda(x)}{1 - \lambda(x_0)} \right \} \end{eqnarray*} which is non-zero for almost all $t$ under the non-parallel condition. Therefore (ref) uniquely determines $M_1$ and $M_2$, hence $F_1$ and $F_2$. The treatment of the case with $\lambda(x_0) = 1$ is straightforward.

Instrumental Variables

The identification results developed in the preceding sections can be used to identify nonparametric finite mixture regression with endogenous regressors. Suppose we observe a triple of random variables $(y,w,x)$ taking its value in $\mathcal Y \times \mathcal W \times \mathcal X$ where $\mathcal Y \subset \mathbb{R}$, $\mathcal W \in \mathbb{R}^p$ and $\mathcal X \in \mathbb{R}^k$. Also let $$ z := \binom y w. $$ In a manner similar to Section (ref), consider a switching regression model:

equation[equation omitted — 248 chars of source]

Unlike in the previous sections, however, we no longer assume that $\eta$'s and $w$ are uncorrelated or independent. Instead, we assume

equation[equation omitted — 87 chars of source]

that is $$ \mathbb{E} [\eta_1|x] = \mathbb{E} [\eta_2|x] = 0. $$ Here and thereafter the notation $F_i(\star_1,\star_2,...)$ and $F_i(\star_1,\star_2,...|\star)$ denote the joint distribution of $\star_1,\star_2,...$ and the conditional distribution of $\star_1,\star_2,...$ given $\star$ when the joint distribution is given by $F_i, i = 1,2$. Consider linear operators

equation[equation omitted — 105 chars of source]

and assume that these operators are invertible.

The main goal is to identify $g_1$ and $g_2$. Here $x$ plays the role of instrumental variables. As before, define $m_1(x) = \int z dF_1(z|x)$ and $m_2(x) = \int z dF_2(z|x)$. Note that $m_1: \mathbb{R}^k \rightarrow \mathbb{R}^{p+1}$ and $m_2: \mathbb{R}^k \rightarrow \mathbb{R}^{p+1}$. For $j = 1,...,p+1$, let $m_{j,1}(\cdot)$ and $m_{j,2}(\cdot)$ denote the $j-$th elements of $m_1(\cdot)$ and $m_2(\cdot)$, respectively. Define the $p+1$-dimensional vectors of random variables $\epsilon_j = z - m_i(x), (z,x) \sim F_i(z,x), j = 1,2$. Consistent with the previous notation let $F_i(\epsilon_i|x), i = 1,2,$ denote the conditional distribution of $\epsilon_1$ and $\epsilon_2$ under $F_1$ and $F_2$.

By construction, $$ \int \epsilon dF_i(\epsilon|x) = 0, i = 1,2. $$ If we further assume that $F_i(\epsilon|x), j = 1,2$ do not depend on $x$, an appropriate extension of the theory developed in Section (ref) can be used to identify $m_{p+1,1}(x)$, $m_{p+1,2}(x)$, $F_1(z|x)$ and $F_2(z|x)$, which in turn, also identify the operators $T_1$ and $T_2$. By (ref), (ref) and (ref) we have $$ m_{p+1,1}(x) = T_1[g_1](x), \quad m_{p+1,2}(x) = T_2[g_2](x). $$ Then by their invertibility $g_1$ and $g_2$ are identified as . To formalize this idea, consider the following assumptions:

assFor some $\delta > 0$, \begin{enumerate}[(i)] • $\epsilon_1|x \sim F_1^{\epsilon}$ and $\epsilon_2|x \sim F_2^\epsilon$ at all $\mathcal X$ where $ F_1^{\epsilon}$ and $ F_1^{\epsilon}$ do not depend on the value of $x$; • $m_{j1}(x_0) - m_{j1}(x) \neq m_{j2}(x_0) - m_{j2}(x)$, for all $x \in N^1(x_0,\delta)$, $x \neq x_0$ and for all $j$, $j = 1,...,p+1$; • $m_1$ and $m_2$ are continuous at $x_0$. \end{enumerate}

To state a multivariate extension of Assumption (ref), define the multivariate moment generating function $$ M_i({\bf t}) = \int e^{{\bf t}^\top\eta} dF_i(\eta), \quad i = 1,2, \quad {\bf t} \in \mathbb{R}^{p+1}. $$ Let ${\bf e}_j$ denote the unit vector whose $j-$th element is 1. Accommodating the identification strategy in Section (ref) require some modification as follows. Define $ {D(x)} := m_2(x) - m_1(x) $ as before, though now $D: \mathbb{R}^k \rightarrow \mathbb{R}^p $ is vector-valued. Also let $$ h_j(c,t) : = e^{ {t {\bf e}_j}' D(x_0)(1 - c)} \frac {M_2(t{\bf e}_j)} {M_1(t{\bf e}_j)}, \quad c \in \mathbb{R}_{++ }, t \in \mathbb{R} $$ and $$ R({\bf t},x) := \frac{M({\bf t} | x)}{M({\bf t} | x_0)}, \qquad {\bf t} \in \mathbb{R}^{p+1}. $$

ass\begin{enumerate}[(i)] • The domains of $M_1({\bf t})$ and $M_2({\bf t})$ are $(-\infty,\infty)^{p+1}$; • For some $\varepsilon >0$ either $h_j(\pm\varepsilon,t) = O(1)$ or $1/ h_j(\pm\varepsilon,t) = O(1)$, or both hold as $t \rightarrow +\infty$ for each $j \in \{1,...,p+1\}$. Moreover, the same holds as $t \rightarrow - \infty$ \end{enumerate}
conditionEither \begin{enumerate}[(i)] • $ \lim_{t \rightarrow \infty}\frac 1 t \log R(t{\bf e}_j,x) \neq \lim_{t \rightarrow -\infty}\frac 1 t \log R(t{\bf e}_j,x) \text{ for each } j \in \{1,...,p\} \text{ and for some } x \in N^1(x_0,\delta) $ or • $\lim_{c \downarrow 0} \lambda_{c} = 1$ holds. \end{enumerate}

By modifying the proofs of Lemmas (ref) and (ref) appropriately to deal with $\mathbb{R}^p$-valued random variables, we can show that $(\lambda,F_1^{\epsilon},F_2^\epsilon,m_1(x_0),m_2(x_0))$ is identified under Assumptions (ref) and (ref). Then, as noted in Remark (ref), $F(z|x), x \in {\mathcal X}, z \in \mathbb{R}^p$ uniquely determines $(\lambda,F_1^\epsilon,F_2^\epsilon,m_1(x),m_2(x))$ in $\mathbb{R} \times {\bar {\mathcal F}(\mathbb{R}^p)^2 \times \mathbb{R}^{2p}}$ for all $x \in \mathcal X$ up to labeling. Therefore each component distribution of $z$ is obtained by $$ F_1(z|x) = F_1^\epsilon(z - m_1(x)), \quad F_2(z|x) = F_2^\epsilon(z - m_2(x)). $$ We now have:

thmSuppose Assumptions (ref), (ref) and Condition (ref) hold. Then $g_1(\cdot)$ and $g_2(\cdot)$ are identified.
remIt is possible to further introduce flexibility into the model (ref) by allowing unrestricted dependence between unobserved heterogeneity and the instrument $x$. This can be achieved by making $\lambda$ in (ref) an arbitrary function of $x$. Applying the results in Section (ref) to identify $m_{p+1,1}(x)$, $m_{p+1,2}(x)$, $F_1(z|x)$ and $F_2(z|x)$ and proceeding as above, we recover $g_1$ and $g_2$ nonparametrically.

Mixtures with arbitrary $J$

Previous sections studied the identifiability for mixtures with $J = 2$. It is desirable, however, to be able to deal with mixtures with many components in some applications, especially when mixtures are used to represent unobserved heterogeneity. This section shows that nonparametric identification can be established for general $J$, possibly greater than 2, and moreover we show that the number $J$ itself is also identifiable.

The basic setup in this section is analogous to the one considered in Section (ref), though the conditional distribution of $z \in \mathbb{R}$ given $x \in \mathbb{R}^n$ consists of $J$ components, $J \in \mathbb{N}$, as in ((ref)). As before, define $$ m_j(x) = \int_\mathbb{R} z dF_j(z|x), j = 1,2,...,J. $$ Define also \[ \epsilon_j = z_j - m_j(x), \quad j = 1,2,...,J. \] Later we impose independence between $\epsilon_j, j = 1,...,J$ and $x$, which enables us to write $F(z|x)$ as

equation[equation omitted — 76 chars of source]

For later use, define $M_j(t) = \int e^{t\epsilon}F_j(d\epsilon), j = 1,...,J$. This section shows that the parameter $(\{\lambda_j\}_{j=1}^J,\{F_j(\cdot)\}_{j=1}^j,\{m_j(\cdot)\}_{j=1}^J)$ is identifiable under suitable conditions.

At an intuitive level, the argument developed in Section (ref) still offers a valid picture behind the identifiability result here. The independence of $\epsilon$ from $x$ leads to a shift restriction: the shapes of the distributions of $\{\epsilon_j\}_{j=1}^J$ have to remain invariant along the $J$ regression functions. This restriction, with other conditions, nails down the true parameters uniquely. Moving from $J = 2$ to $J \geq 3$, however, involves rather different theoretical arguments as developed subsequently. Recall that Section (ref) presented alternative conditions that guarantee the identifiability of two-component mixture models, as summarized by Lemma (ref), Lemma (ref) and Lemma (ref). This section proves the nonparametric identifiability of ((ref)) under conditions that are similar to the ones used in Lemma (ref), which seems least prohibiting of the three to generalize. Even so, this generalization calls for multistep identification argument with recursive procedures, as will be seen shortly.

To see how the treatment of general mixtures differs from the $J=2$ case, consider the case $J=3$. Instead of Equation ((ref)), we now have

equation[equation omitted — 172 chars of source]

Wlog, suppose $m_1(x_0) > m_2(x_0) > m_3(x_0)$ at a point $x_0$ in $\mathbb{R}^k$. Take a point $x'$ in the neighborhood of $x_0$ and consider the case $m_1(x') - m_1(x_0) \geq 0$ (if this term is negative, the roles of $m_1$ and $m_3$ get interchanged). The method used in the proof of Lemma (ref) to identify the $J=2$ model still works for the slopes of $m_1$ and $m_3$. Following the proof, take the ratio of the conditional moment generating functions at $x_0$ and a point in its neighborhood, $x'$, say, then take its logarithm followed by a normalization by $t$:

align*[align* omitted — 766 chars of source]

Suppose the ratios of $M_1(t)$, $M_2(t)$ and $M_3(t)$ do not explode exponentially, and $m_1$, $m_2$ and $m_3$ are continuous so that $m_2(x') - m_1(x_0)$ and $m_3(x') - m_1(x_0)$ are negative. Then as $t$ approaches to infinity, the above expression approaches to the slope $m_1(x') - m_1(x_0)$ if it is non-negative (though it yields the identical result if the slope is negative as well, as seen in the proof of Lemma (ref)). Similarly, by taking the limit $t \rightarrow - \infty$, the slope of $m_3$ is identified. This argument, however, leaves the slope of the middle term $m_2$ undetermined. And in the general case of $J \geq 3$, $J-2$ slopes remain to be determined. The approach in Lemma (ref) does fall short of achieving its goal when applied to models with $J \geq 3$.

It is, however, possible to identify the slope of $m_2$ by proceeding as follows. Suppose, evaluated at $x$, the regression functions satisfy the inequality $m_1(x) > m_2(x) > m_3(x)$. Pick a point $y$ in a neighborhood of $x$. Multiply ((ref)) by $e^{-t{[m_1(x) - m_1(y)]}}$ to obtain:

equation[equation omitted — 200 chars of source]

This purges $x$ out of the first term on the right hand side. Note $[m_1(x) - m_1(y)]$ can be identified by applying the argument in Lemma (ref) to the $J = 3$ model ((ref)), as demonstrated above. Therefore the left hand side of the above equation is known.

The above step enables us to eliminate all unknown parameters associated with the first mixture component. To see this, suppose $m_j, j = 1,2,3$ are differentiable in at least one of the $k$ elements of $x = (x^1,x^2,...,x^k)$. In what follows we assume that it is differentiable in the first element $x^1$ without loss of generality. As before, we assume that this is a prior knowledge. Let $D_x$ denote the partial differentiation operator with respect to the first component of x, i.e. $D_xf(x) = \frac \partial {\partial x^1}f(x)$. Differentiating both sides of the above equation by $x^1$ and rearranging,

align[align omitted — 265 chars of source]

Note that operating $D_x$ eliminates the unknown function $M_1(t)$ out of the right hand side of ((ref)). We now have $$ \frac {\partial}{\partial t} \log\left(D_x\left [e^{-t{[m_1(x) - m_1(y)]}}M(t|x)\right ]\right) = \frac {A_1}{A_2}, $$ say, where

align*[align* omitted — 484 chars of source]

and $$ A_2 = 1 + \frac {\lambda_3} {\lambda_2} \frac {[D_xm_3(x) - D_xm_1(x)]}{ [D_xm_2(x) - D_xm_1(x)]} e^{t[m_3(x) - m_2(x)]}\frac {M_3(t)}{M_2(t)}. $$ Note that the factor $D_xm_2(x) - D_xm_1(x)$ is non-zero if the two regression functions are not parallel at $x$, which makes the division by the factor valid. As far as $\frac {M_3}{M_2}$ and $\frac {D_xM_3}{M_2}$ do not explode exponentially, all the terms above except for the second and third terms of $A_1$ and the first term of $A_2$ converge to zero as $t \rightarrow \infty$. It follows that

equation[equation omitted — 266 chars of source]

The only unknown component in the above equation is $\frac {\frac {\partial}{\partial t} M_2(t)}{M_2(t)}$, but this term depends only on $t$, so it can be differenced out: repeat the above argument with replacing $x \in \mathbb{R}^k$ with a point $z \in \mathbb{R}^k$ so close to $x$ that $m_1(z) > m_2(z) > m_3(z)$. This yields $$ \lim_{t \rightarrow \infty} \left \{\frac {\partial}{\partial t} \log\left(D_z\left [e^{-t{[m_1(z) - m_1(y)]}}M(t|z)\right ]\right) \right \} = \{m_2(z) - [m_1(z) - m_1(y)]\} + \frac {\frac {\partial}{\partial t} M_2(t)}{M_2(t)}. $$ The slope of $m_2$ is $$ m_2(x) - m_2(z) = \lim_{t \rightarrow \infty} \frac {\partial}{\partial t} \log \left(\frac {D_x\left [e^{-t{[m_1(x) - m_1(y)]}}M(t|x)\right]}{D_z\left [e^{-t{[m_1(z) - m_1(y)]}}M(t|z)\right ]}\right) + (m_1(x) - m_1(z)). $$ The terms such as $m_1(x) - m_1(z)$ on the right hand side are identified by the method developed in Lemma (ref), as noted earlier. The equation above shows the identifiability of the slope of $m_2$.

We have already noted that the identifiability of the slope of $m_3$ basically follows from Lemma (ref). It is nevertheless instructive to present an alternative way to identify it by carrying on the foregoing analysis one step further. This will illustrate the basic idea behind our general identification theory for $J \in \mathbb{N}$.

Let us return to Equation ((ref)), changing the notation and writing $x_a$ for $x$, $x_b$ for $y$. As before, $\Delta_{ab}f$ stands for $f(x_a) - f(x_b)$. The first step is to purge $x_a$ from the first term on the right hand side, as we did in Equation ((ref)), as follows:

align*[align* omitted — 337 chars of source]

which yields

align[align omitted — 424 chars of source]

Notice that again this eliminates an unknown moment generating function, this time $M_2(t)$. Differentiating the above expression with respect to $t$ and following the line of argument presented above, the slope of $m_3$ is given by $$ \Delta_{ac}m_3 = \lim_{t \rightarrow \infty} \frac {\partial}{\partial t} \log \left[ \left( \frac { \frac {e^{-t[\Delta_{ab}m_2 - \Delta_{ab}m_1]}}{t[D_{x_a}m_2(x_a) - D_{x_a}m_1(x_a)]} D_{x_a}\left [e^{-t{\Delta_{ab}m_1}}M(t|x_a)\right ] } { \frac {e^{-t[\Delta_{cb}m_2 - \Delta_{cb}m_1]}}{t[D_{x_c}m_2(x_c) - D_{x_c}m_1(x_c)]} D_{x_c}\left [e^{-t{\Delta_{cb}m_1}}M(t|x_c)\right ] } \right) \right] + \Delta_{ac}m_2. $$

Let us now turn to the identifiability of the general model ((ref)) for a generic $J,$ at a point $x_a \in \mathbb{R}^k$. The general setting is the same as in Section (ref): the first $k*$ elements $x^1,...,x^{k*}$ of the vector of covariates $x$ are continuous covariates, and we will again use local variations in $x^1$.

assFor some $\delta > 0$, \begin{enumerate}[(i)] • $\epsilon_j|x \sim F_j, j = 1,...,J$ at all $x \in N^1(x_a,\delta)$ where $F_j, j = 1,...,J$ do not depend on the value of $x$; • $m_j, j = 1,...,J$ are continuous in $x^1$at $x_a$; • $m_j, j = 1,...,J$ are J times differentiable on $B(x_a,\delta)$ at least in one of the $k^*$ continuous covariates of $x$; \end{enumerate}

Though Condition (ref) imposes $J$-th order differentiability in one argument for simplicity of presentation, this is not essential: it is sufficient to assume that there exists at least one multi-index $\alpha := (\alpha_1,...,\alpha_k) \in \mathbb{Z}^k, \alpha_1 + \cdots \alpha_k = J$ such that the derivative $D^\alpha m(x) = \left(\frac{\partial}{\partial x^1}\right)^{\alpha_1} \cdots \left(\frac{\partial}{\partial x^k}\right)^{\alpha_k} m(x) $ is well-defined for every $x$ in $B(x_a,\delta)$. See Remark (ref) for further discussions.

The independence assumption ((ref)) enables us to write the observable conditional distribution in the form ((ref)). The continuity assumption ((ref)) was also assumed in Lemma (ref). The differentiability condition ((ref)) may not be essential for the proof of the Lemma, though replacing derivatives in the proof with differences leads to extremely complex case-by-case analysis. Note that differentiability in only one element of $x$ suffices. Without loss of generality in what follows we assume that the $m_j, j=1,...,J$ are differentiable in the first element $x^1$. Recall that $D_1$ is the differentiation operator with respect to $x^1$. From now on, we will use the notation $$m_{k,j}(x)=m_k(x)-m_j(x).$$

ass\begin{enumerate}[(i)] • $\displaystyle\min_{k \neq j} |m_{k,j}(x_a)| > \Delta $, $\Delta>0$; • $D_1 m_{j}(x_a), j = 1,...,J$ takes $J$ distinct values in $\mathbb{R}$; • The domains of $M_1(t)$ and $M_2(t)$ are $(-\infty,\infty)$; • For some $\epsilon > 0$ , $ \displaystyle\lim_{t \rightarrow \infty} e^{t(\epsilon - \Delta) } \frac{M_j(t)}{M_k(t)} = 0$ and $ \displaystyle\lim_{t \rightarrow \infty} e^{t(\epsilon - \Delta) } \frac{\frac {\partial}{\partial t} M_j(t)}{M_k(t)} = 0$ for all $k,j=1,...,J$. \end{enumerate}

Part ((ref)) of the assumption is not restrictive. As before, our goal is to establish identification up to labeling, so we can assume that

equation[equation omitted — 70 chars of source]

without loss of generality: this does not impact the validity of Assumption (ref). Part ((ref)) is an infinitesimal version of the non-parallel regression function conditions used in the previous sections.

Under these assumptions, we first prove identifiability of the slope $\Delta_{ab}m_1$, using the method developed in Section (ref) , for all $x_b$ in a chosen neighborhood of $x_a$. Note that we know $\lambda_1 \neq 0$. By the continuity and differentiability assumptions (Assumption (ref) ((ref)) and ((ref))), there exists $\delta' > 0$, $\delta' < \delta$, such that for all $x_b \in N^1(x_a, \delta')$ and for all $j=1,...,J$, $|m_j(x_b)-m_j(x_a)|<\frac{\epsilon}{2} $, and $D_1 m_j(x_b), j=1,...,J$ take J distinct values. Here we use the fact that twice differentiability of the regression functions implies that they are $\mathcal{C}^1$. Then, as in the proof of Lemma (ref), in the case $ m_1(x_b)-m_1(x_a)>0$, we write $$ \frac 1 t \log \left ( \frac {M(t|x_b)} {M(t|x_a)} \right ) = \frac 1 t \log \left ( \frac { e^{t[m_1(x_b) - m_1(x_a)]} + \sum_{j=2}^{J} \frac{\lambda_j}{\lambda_1}\frac{M_j(t)}{M_1(t)}e^{t[m_j(x_b) - m_1(x_a)]}} {1 + \sum_{j=2}^{J} \frac{\lambda_j}{\lambda_1}\frac{M_j(t)}{M_1(t)}e^{t[m_j(x_a) - m_1(x_a)]}} \right), $$ and in the case $ m_1(x_b)-m_1(x_a)<0$ , we write $$ \frac 1 t \log \left ( \frac {M(t|x_b)} {M(t|x_a)} \right ) = \frac 1 t \log \left ( \frac { 1 + \sum_{j=2}^{J} \frac{\lambda_j}{\lambda_1}\frac{M_j(t)}{M_1(t)}e^{t[m_j(x_b) - m_1(x_b)]}} {e^{t[m_1(x_a) - m_1(x_b)]} + \sum_{j=2}^{J} \frac{\lambda_j}{\lambda_1}\frac{M_j(t)}{M_1(t)}e^{t[m_j(x_a) - m_1(x_b)]}} \right). $$ Similarly, since $m_j(x_b) - m_1(x_a)$, $m_j(x_a) - m_1(x_a)$, $m_j(x_b) - m_1(x_b)$, and $m_j(x_a) - m_1(x_b)$ are less than $\epsilon - \Delta$, this gives in both cases, $$ \forall x_b \in U, \lim_{t \rightarrow \infty} \frac 1 t \log \left ( \frac {M(t|x_b)} {M(t|x_a)} \right ) = \Delta_{ba}m_1. $$ Hence the slope $\Delta_{ab}m_1$ is identifiable for all $x_b \in N^1(x_a, \delta')$. \\

Now we focus on the identifiability of the slopes $\Delta_{ab}m_j$ for all $j=2,...,J$ and $x_b$ in an appropriate neighborhood of $x_a$.

Pick a point $x_b \neq x_a$ in $\mathbb{R}^k$. For notational convenience, define the operator $A(x_a,x_b,t,k)$

equation[equation omitted — 197 chars of source]

where $f: \mathbb{R}^k \rightarrow \mathbb{R}$ is a function that is differentiable in its first argument, and $R_k(t,x)$ is a (rational) function in $t$. Its precise definition will be given shortly. The operator $A(x_a,x_b,t,k)$ generalizes the procedure performed on $D_{x_a}\left [e^{-t{\Delta_{ab}m_1}}M(t|x_a)\right ]$ in Equation ((ref)) to eliminate unknown parameters in ((ref)). Operate $A(x_a,x_b,t,k)$, $k = 2,3,...$ sequentially on $D_{x_a}\left [e^{-t{\Delta_{ab}m_1}}M(t|x_a)\right ]$ to define the expressions

equation[equation omitted — 189 chars of source]

By construction $Q_k(x_a,t)$ satisfies the following recursive formula:

equation[equation omitted — 161 chars of source]

The definition of the operator $A(x_a,x_b,t,k)$, as explained further later, is motivated by two facts: (i) the factor $e^{-t[\Delta_{ab}m_k - \Delta_{ab}m_{k-1}]}$ purges $x_a$ out of the exponent in the leading term of $Q_{k}(x_a,t)$ and (ii) division by the polynomial $R_k(t,x_a)$ then makes the leading term $\lambda_ke^{-tm_k(x_b)}M_k(t)$, which is completely free from $x_a$ and therefore eliminated by $D_{x_a}$. Once this is done, taking the log-derivative with respect to $t$ as in ((ref)) terms and taking the limit $t \rightarrow \infty$ yields $\Delta_{ab}m_{k}$ up to an unknown additive factor $\frac{\frac{\partial}{\partial t}M_k(t)}{M_k(t)}$, which can be differenced out.

Subsequent arguments establish the identifiability of $\Delta_{ab}m_k, k = 2,...,J$ for all $x_b$ in a neighborhood of $x_a$. We proceed in two steps. Step 1 shows that, with an appropriate choice of $R_k(t,x_a)$ in ((ref)), $Q_k(x_a,t), k = 2,3,...,J$ have following representations:

equation[equation omitted — 154 chars of source]

where $R_k^j(t,x_a), k = 2,3...,J, j = k,k+1,...,J$ are polynomials in $t$ with the property that $R_k^k(t,x_a) = R_k(t,x_a)$; a formal definition of these polynomials are provided later. The representations ((ref)) are useful, partly because the unknown functions $M_j(t), j = 1,..,k-1$ do not appear in $Q_{k}(x_a,t)$. Step 2 uses the representations ((ref)) to show that it is possible to identify the slope $\Delta_{ab}m_k, k = 2,...,J$ using the knowledge of $\Delta_{ab}m_1$, $Q_k(x_a,t)$ and $Q_k(x_b,t)$, $k = 2,...,J$ for all $x_b$ in a neighborhood of $x_a$.

The identifiability of the rest of the model (at $x_a$) is then established using the knowledge of $\Delta_{ab}m_k, k = 1,2,...,J$ and conditional moments of $z$ given $x_a$.

Let us start with Step 1, which derives the representation ((ref)) and will be summarized in Lemma (ref). Note that the definitions of the polynomials $R_k(t,x_a), k = 2,...,J$ and $R_k^j(t,x_a), k = 2,...,J, j = k,k+1,...,J$ are given in the course of our derivation.

{\bf Step 1:} \quad Start from $k = 2$. Define $$ R_2^j(t,x_a) = tD_{x_a}(m_j(x_a) - m_1(x_a)), j = 2,...,J, $$ then

align*[align* omitted — 273 chars of source]

yielding the desired representation for the case of $k = 2$. Let $R_2(t,x_a)$ (used in the definition of $A(x_a,x_b,t,2)$) be $R_2^2(t,x_a) = tD_{x_a}[m_2(x_a) - m_1(x_a)]$. With this choice

align*[align* omitted — 483 chars of source]

and the $j=2$ term in the summation drops out. Moreover, this result implies that $R_3(x_a,t)$ should be $$ R_3(x_a,t) = R_3^3(x_a,t) = D_{x_a}\frac{R_2^3(t,x_a)}{R_2(t,x_a)} + t \frac{R_2^3(t,x_a)}{R_2(t,x_a)}D_{x_a}[m_3(x_a) - m_2(x_a)]. $$ Note that the above step requires that $R_2(t,x_a)$ is non-zero: this issue will be discussed shortly.

The fact that the rest of $Q_{k}(x_a,t), k = 4,...,J$ have the representations as in ((ref)) can be shown by induction: suppose ((ref)) holds for $k = h$, that is

equation*[equation* omitted — 116 chars of source]

Define $$ R_{h+1}^j(t,x_a) = D_{x_a^1}\left(\frac{R_h^j(t,x_a)}{R_h^h(t,x_a)}\right) + t\frac{R_h^j(t,x_a)}{R_h^h(t,x_a)}D_{x_a^1}[m_j(x_a) - m_h(x_a)], \qquad j = h+1,...,J. $$ In what follows we sometimes write $$ R_k^j := R_k^j(t,x) $$ and $$ m_{k,l} := m_k(x) - m_l(x). $$ as short hand. Let $R_h(t,x_a) = R_h^h(t,x_a)$, then using this and the definition of the operator $A(x_a,x_b,t,h)$ in ((ref)), obtain

align*[align* omitted — 621 chars of source]

which is the desired result. The next lemma summarizes the foregoing argument. Notice that it relies on the assumption that $R_k(t,x_a) = R_k^k(t,x_a), k = 2,3,...J$ are non-zero, and later we show that the set

equation[equation omitted — 79 chars of source]

is non-empty.

lemDefine $ R_2^j(t,x_a) = tD_{x_a}(m_j(x_a) - m_1(x_a)), j = 2,...,J, $ and $ R_{k+1}^j(t,x_a) = D_{x_a^1}\frac{R_k^j(t,x_a)}{R_k^k(t,x_a)} + t\frac{R_k^j(t,x_a)}{R_k^k(t,x_a)}D_{x_a^1}[m_j(x_a) - m_k(x_a)], k = 3,...,J, j = k+1,...,J. $ Let $R_k(t,x_a) = R_k^k(t,x_a), k = 2,...,J$ in ((ref)). Then $ Q_k(x_a,t) = A(x_a,x_b,t,k-1) A(x_a,x_b,t,k-2) \cdots A(x_a,x_b,t,2) D_{x_a^1}[e^{-t\Delta_{ab}m_1}M(t|x_a)], k = 2,...,J $ have the representations ((ref)) on $S(x_a)$.

{\bf{Step 2:}} This step shows that the knowledge of the function $Q_k(x,t)$ at $x = x_a$ and $x = x_b$ identifies $\Delta_{ab}m_k - \Delta_{ab}m_{k-1}$. The main result is:

lem$\forall x_b \in N^1(x_a, \delta')$, $$ \lim_{t \rightarrow \infty} \frac{\partial}{\partial t} \log \left(\frac {Q_k(x_a,t)} { Q_k(x_b,t)} \right)= \Delta_{ab}m_k - \Delta_{ab}m_{k-1}, k = 2,3,...,J. $$

Lemmas (ref) and (ref) will then be useful to prove the identifiability of $\Delta_{ab}m_k, k = 2,...,J$, for all $x_b$ in a neighborhood of $x_a$, since we already identified $\Delta_{ab}m_1$. The following propositions are useful in proving Lemma (ref). In what follows $\deg_t(f)$ and $\text{lc}_t(f)$ denote the degree and the leading coefficients of a polynomial $f(t)$ with respect to $t$.

propSuppose $x \in N^1(x_a, \delta')$. Then $R_k(t,x)$ is a rational function of $t$ for sufficiently large $t$ and takes the following form: $$ R_k(t,x) = \frac {P_k(t,x)}{P_{k-1}(t,x)^2} $$ where $P_k(t,x), k \geq 3$ are polynomials in $t$ such that $$ \deg_t(P_k(t,x)) = 2^{k-2} -1 $$ and $$ \mathrm{lc}_t(P_k(t,x)) = (\Pi_{g=1}^{k-1}D_{x}(m_k(x) - m_g(x)))\Pi_{j=2}^{k-1}\{(\Pi_{h=1}^{j-1}D_{x}(m_j(x) - m_h(x)))^{2^{k - j - 1}}\}. $$

The proof of the proposition is given in the Appendix.

remThe formula for $R_k(t,x)$ given in Proposition (ref) and the fact that $P_k(t,x)$ is a polynomial in $t$ imply that $R_k \neq 0$ for sufficiently large $t$ for $k = 2,3,...,J$. Consequently $S(x_a)$ in ((ref)) includes (for example) the set $[c,\infty)$ for some constant $c$ and therefore it is not empty. This is important in applying Lemma (ref).
prop$$ \lim_{t \rightarrow \infty} \frac{\partial}{\partial t} \log R_k(x,t) = 0 $$ for all $t \in \mathbb{R}$ and $x \in N^1(x_a,\delta')$.
proof[Proof of Proposition (ref)] By the expression of $ R_k(x,t)$ given in Proposition (ref), \begin{align*} \lim_{t \rightarrow \infty} \frac{\partial}{\partial t} \log R_k(x,t) & = \lim_{t \rightarrow \infty} \frac{\partial}{\partial t} \log \frac {P_k(t,x)}{P_{k-1}(t,x)^2} \\ & = \lim_{t \rightarrow \infty} \frac{\partial}{\partial t} \log P_k(t,x) - 2 \lim_{t \rightarrow \infty} \frac{\partial}{\partial t} \log P_{k-1}(t,x). \end{align*} Since the Proposition shows that $R_k(x,t)$, $P_k(x,t)$ and $P_{k-1}(x,t)$ are well defined for large $t$, so are the above limits. But Proposition (ref) also implies that $P_k(t,x)$ and $P_{k-1}(t,x)$ are polynomials in $t$ with finite degree, therefore the two terms are zero.

Now we are ready to prove the main result in Step 2, that is, Lemma (ref).

proof[Proof of Lemma (ref)] By Lemma (ref) and Proposition (ref), \begin{equation} Q_{k}(x_a,t) = \sum_{j=k}^J \lambda_j R_k^j(t,x_a) e^{t[m_j(x_a) - \Delta_{ab}m_{k-1}]}M_j(t), \quad k = 2,3,...,J, \end{equation} holds for sufficiently large $t$. Then \begin{align*} \frac{\partial}{\partial t} Q_k(x_a,t) &= \sum_{j=k}^J \lambda_j \left( \frac{\partial}{\partial t} R_k^j(t,x_a) + [m_j(x_a) - \Delta_{ab}m_{k-1}] R_k^j(t,x_a) \right) e^{t[m_j(x_a) - \Delta_{ab}m_{k-1}]}M_j(t) \\ &+ \sum_{j=k}^J \lambda_j R_k^j(t,x_a) e^{t[m_j(x_a) - \Delta_{ab}m_{k-1}]}D_tM_j(t). \end{align*} and, for $k \leq J,$ \begin{align*} \frac{\partial}{\partial t} \log(Q_k(x_a,t)) &= \frac{\frac{\partial}{\partial t} Q_k(x_a,t)}{Q_k(x_a,t)} \\ &= \frac{ \frac{\frac{\partial}{\partial t} R_k(t,x_a)}{R_k(t,x_a)} + m_k(x_a) - \Delta_{ab}m_{k-1} + \frac{\frac{\partial}{\partial t} M_k(t)}{M_k(t)} } { 1 + \sum_{j=k+1}^J \frac {\lambda_j}{\lambda_k}\frac{R_k^j(t,x_a)}{R_k(t,x_a)}e^{tm_{j,k}(x_a)}\frac{M_j(t)}{M_k(t)} } \\ &+ \frac{ \sum_{h=k+1}^J \left[ (m_h(x_a) - \Delta_{ab}m_{k-1}) \frac {\lambda_h}{\lambda_k}\frac{R_k^h(t,x_a)}{R_k(t,x_a)}e^{tm_{h,k}(x_a)}\frac{M_h(t)}{M_k(t)} \right] } { 1 + \sum_{j=k+1}^J \frac {\lambda_j}{\lambda_k}\frac{R_k^j(t,x_a)}{R_k(t,x_a)}e^{tm_{j,k}(x_a)}\frac{M_j(t)}{M_k(t)} } \\ &+ \frac{ \sum_{h=k+1}^J \left[ \frac {\lambda_h}{\lambda_k}\frac{\frac{\partial}{\partial t} R_k^h(t,x_a)}{R_k(t,x_a)}e^{tm_{h,k}(x_a)}\frac{M_h(t)}{M_k(t)} + \frac {\lambda_h}{\lambda_k}\frac{R_k^h(t,x_a)}{R_k(t,x_a)}e^{tm_{h,k}(x_a)}\frac{\frac{\partial}{\partial t} M_h(t)}{M_k(t)} \right] } { 1 + \sum_{j=k+1}^J \frac {\lambda_j}{\lambda_k}\frac{R_k^j(t,x_a)}{R_k(t,x_a)}e^{tm_{j,k}(x_a)}\frac{M_j(t)}{M_k(t)} }. \end{align*} Using the notation in the proof of Proposition {(ref)}, for all $h >k$, \begin{align*} \frac{R_k^h(t,x_a)}{R_k(t,x_a)} & = \frac{P_k^h(t,x_a)/(P_{k-1}^{k-1}(t,x_a))^2}{P_k^k(t,x_a)(P_{k-1}^{k-1}(t,x_a))^2} \\ & = \frac{P_k^h(t,x_a)}{P_k^k(t,x_a)}. \end{align*} As noted in the Proof of Proposition (ref), both $P_k^h(t,x_a)$ and $P_k^k(t,x_a)$ are polynomials in $t$, $P_k^k(t,x_a) \neq 0$ for sufficiently large $t$, and their degrees are equal. Hence their ratio goes to a constant as t goes to infinity: $$\lim_{t \rightarrow \infty} \frac{R_k^h(t,x_a)}{R_k(t,x_a)} = c_{h,k,x_a}.$$ For a similar reason, using Proposition (ref), $$\lim_{t \rightarrow \infty} \frac{\frac{\partial}{\partial t} R_k^h(t,x_a)}{R_k(t,x_a)} = 0.$$ Then, using Assumption ((ref)) ((ref)), since $m_{h,k}(x_a) < -\Delta$, we know that the second and third lines of the expression of converge to zero as $t$ goes to $+ \infty$, and we have \begin{equation} \lim_{t \rightarrow \infty} \frac{\partial}{\partial t} \log(Q_k(x_a,t)) = m_k(x_a) - \Delta_{ab}m_{k-1} + \frac{\frac{\partial}{\partial t} M_k(t)}{M_k(t)}. \end{equation} Note that (ref) holds for all $x_b \in \mathbb{R}^{k}$. Let us take $x_b \in N^1(x_a, \delta')$. Note that we can then also write $\frac{\partial}{\partial t} \log(Q_k(x,t))$ taking $x = x_b$: the $\Delta_{ab}m_h$ terms are equal to $0$ and, again since $m_{h,k}(x_b)$ is less than $\epsilon-\Delta$, we have $$\lim_{t \rightarrow \infty} \frac{\partial}{\partial t} \log(Q_k(x_b,t)) = m_k(x_b) + \frac{\frac{\partial}{\partial t} M_k(t)}{M_k(t)},$$ so that, for all $x_b \in N^1(x_a, \delta')$, we have then $$\lim_{t \rightarrow \infty} \frac{\partial}{\partial t} \log(\frac{Q_k(x_a,t)}{Q_k(x_b,t)} ) = \Delta_{ab}m_k - \Delta_{ab}m_{k-1}.$$

To sum up, Lemma (ref) together with the proof of identifiability of $\Delta_{ab}m_1$ allow, by induction, the identifiability of the slopes $\Delta_{ab}m_k$ for all $x_b \in N^1(x_a, \delta')$ and for all $k=1,...,J$:

$$\Delta_{ab}m_1 = \lim_{t \rightarrow \infty} \frac 1 t \log \left ( \frac {M(t|x_a)} {M(t|x_b)} \right ),$$ $$\Delta_{ab}m_k = \sum_{j=2}^k \lim_{t \rightarrow \infty} \frac{\partial}{\partial t} \log(\frac{Q_k(x_a,t)}{Q_k(x_b,t)} ) + \Delta_{ab}m_1. $$

We now state the complete identification result. For the sake of clarity, we name the point of identification $x_0$ instead of $x_a$.

assThere exists $ X=(x_1,...,x_{J-1}) \in N^1(x_0, \delta')^{J-1}$ such that $$ A(x_0,X) = \begin{pmatrix} \Delta_{0,1} m_1 - \Delta_{0,1} m_J & & \ldots & & \Delta_{0,1} m_{J-1} - \Delta_{0,1} m_J\\ \vdots & & \ddots & & \vdots\\ \Delta_{0,J-1} m_1 - \Delta_{0,J-1} m_J & & \ldots & & \Delta_{0,J-1} m_{J-1} - \Delta_{0,J-1} m_J \end{pmatrix} $$ is invertible.

In the above assumption, the notation $\Delta_{0,i} m_j$ denotes $m_j(x_0) - m_j(x_i)$.

lemSuppose Assumptions (ref), (ref) and (ref) hold. Then $F(\cdot|x), x \in B(x_0,\delta')$ uniquely determines $((\lambda_j)_{j=1..J-1},(F_j(\cdot))_{j=1..J},(m_j(x_0))_{j=1..J})$ in the set $(0,1)^{J-1} \times \bar{\mathcal F}(\mathbb{R})^J \times \mathbb{R}^J$ up to labeling.
proof[Proof of Lemma (ref)] Reproducing what was done in the Proof of Lemma (ref), since $$ \dot M(0|x_0) - \dot M(0|x) = \sum_{i=1}^J \lambda_i [(m_i(x_0) - m_i(x)) - (m_J(x_0) - m_J(x))] + (m_J(x_0) - m_J(x)) , $$ we can write $$ \begin{pmatrix} \dot M(0|x_0) - \dot M(0|x_1)\\ \vdots \\ \dot M(0|x_0) - \dot M(0|x_{J-1}) \end{pmatrix} = A(x_0,X) . \begin{pmatrix} \lambda_1\\ \vdots \\ \lambda_{J-1} \end{pmatrix} + \begin{pmatrix} \Delta_{0,1}m_J\\ \vdots \\ \Delta_{0,J-1}m_J \end{pmatrix}. $$ As Assumption (ref) guarantees the invertibility of $A(x_0,X)$ , and since the slopes of the $(m_j)_{j=1..J}$ were all previously identified, the $(\lambda_j)_{j=1..J-1}$ are identified with the formula $$ \begin{pmatrix} \lambda_1\\ \vdots \\ \lambda_{J-1} \end{pmatrix} = A(x_0,X)^{-1} \left[ \begin{pmatrix} \dot M(0|x_0) - \dot M(0|x_1)\\ \vdots \\ \dot M(0|x_0) - \dot M(0|x_{J-1}) \end{pmatrix} - \begin{pmatrix} \Delta_{0,1}m_J\\ \vdots \\ \Delta_{0,J-1}m_J \end{pmatrix} \right]. $$ To identify $(m_j(x_0))_{j=1..J})$, we use the function $$C(x) = \left \{\ddot M(0|x_0) - \ddot M(0|x) + \lambda [m_1(x_0) - m_1(x)]^2 +(1 - \lambda) [m_2(x_0) - m_2(x)]^2 \right \}/2$$ used in the Proof of Lemma (ref), where we can show that $$ C(x_k) = \sum_{i=1}^J \lambda_i \; m_i(x_0) \; \Delta_{0,k}m_i, $$ which gives $$ \begin{pmatrix} C(x_1) \\ \vdots \\ C(x_{J-1})\\ \dot M(0|x_0) \end{pmatrix} = B(x_0, X). diag(\lambda_1,...,\lambda_J) \begin{pmatrix} m_1(x_0)\\ \vdots \\ m_{J}(x_0) \end{pmatrix}, $$ where $$ B(x_0, X) = \begin{pmatrix} \Delta_{0,1} m_1 & & \ldots & & \Delta_{0,1} m_{J} \\ \vdots & & \ddots & & \vdots\\ \Delta_{0,J-1} m_1 & & \ldots & & \Delta_{0,J-1} m_{J} \\ 1 & & \ldots & & 1 \end{pmatrix} \text{ is observable}. $$ $diag(\lambda_1,...,\lambda_J)$ is invertible as $\lambda_j, \, j=1..J$ are assumed to be nonzero. Since $\det B(x_0, X) = \det A(x_0,X)$, $B(x_0, X)$ is invertible. Therefore, we obtain the following identification result: $$ \begin{pmatrix} m_1(x_0)\\ \vdots \\ m_{J}(x_0) \end{pmatrix} = diag(\lambda_1^{-1},...,\lambda_J^{-1}) B(x_0,X)^{-1} \begin{pmatrix} C(x_1) \\ \vdots \\ C(x_{J-1})\\ \dot M(0|x_0) \end{pmatrix}. $$ What now remain to be identified are the $(F_j(\cdot))_{j=1..J}$: we will again use a technique similar to what was done in the proof of Lemma (ref), but using Assumption (ref). As $ M(t|x)= \sum_{i=1}^J \lambda_i e^{tm_i(x)}M_i(t)$, considering $J$ generic points $(c_i)_{i=1..J} \in B(x_0, \delta')^J$, we have $$ \begin{pmatrix} M(t|c_1) \\ \vdots \\ M(t|c_{J}) \end{pmatrix} = D(t, c_1,...,c_J) \; diag(\lambda_1,...,\lambda_J) \; \begin{pmatrix} M_1(t) \\ \vdots \\ M_J(t) \end{pmatrix}, $$ $$ \text{where } D(t, c_1,...,c_J)=(e^{tm_j(c_i)})_{1 \leq i,j \leq J}. $$ We prove in the appendix (Proposition (ref)) that there is a vector of $(J-1)$ points $X^{(J)}=(x_1^{(J)},...,x_{J-1}^{(J)}) \in B(x_0, \delta')^{J-1}$, such that $\mathcal{Z}= \left\lbrace t \in \mathbb{R} | \det D(t, x_0, x_1^{(J)},...,x_{J-1}^{(J)})=0 \right\rbrace$ is finite. Hence, we can invert $ D(t, x_0, x_1^{(J)},...,x_{J-1}^{(J)})$ for all $t \in \mathbb{R} \backslash \mathcal{Z}$. Note that we can write $$ D(t, x_0, x_1^{(J)},...,x_{J-1}^{(J)})= e^{t\sum_{i=1}^J m_i(x_0)} \begin{pmatrix} 1 & & \ldots & & 1 \\ e^{-t \left( \Delta_{0,1} m_1 + \sum_{i=2}^J m_i(x_0)\right) } & & \ldots & & e^{-t \left( \Delta_{0,1} m_J + \sum_{i=1}^{J-1} m_i(x_0)\right)} \\ \vdots & & \ddots & & \vdots\\ e^{-t \left( \Delta_{0,J-1} m_1 + \sum_{i=2}^J m_i(x_0)\right)} & & \ldots & & e^{-t \left( \Delta_{0,J-1} m_J + \sum_{i=1}^{J-1} m_i(x_0)\right)} \\ \end{pmatrix}, $$ and since $(x_1^{(J)},...,x_{J-1}^{(J)}) \in B(x_0, \delta')^{J-1}$, by the above result and Lemma (ref), $D(t, x_0, x_1^{(J)},...,x_{J-1}^{(J)})$ is identified. Therefore $(M_i(t))_{i=1..J}$ are identified for all $t \in \mathbb{R} \backslash \mathcal{Z}$ and since the $(M_i(t))_{i=1..J}$ have domain $(-\infty,+\infty)$, we know that they are continuous (see, e.g, \citeasnoun{gut2013probability} Theorem 8.3 p190) on $\mathbb{R}$. As for each $M_i$, there is a unique continuous extension on $\mathbb{R}$ of its restriction to $\mathbb{R} \backslash \mathcal{Z}$, the $J$ functions are identified. By the same argument of uniqueness of the Laplace transform for a distribution function, this leads to the identification of the $F_i$.

Having showed identification of our model assuming knowledge of $J$, we now consider the case where $J$ is unknown, and show it is identified, using the observable sequence of functions $(Q_k)_{k=1,...}$. As we see below, the number of mixture components $J$ is equal to the largest $j$ for which the function $Q_j$ not identically $0$ in $t$. Therefore one can sequentially compute the $\Delta_{ab}m_j$ using $Q_j$, for increasing $j$. Once there exists $j_0$ such that $Q_{j_0} =0$, then $J=j_0-1$.

prop$$J = \max \left\lbrace j \geq 1 | \exists t_0 \in \mathbb{R}, Q_j(x_a,t_0) \neq 0 \right\rbrace . $$

}

proof[Proof of Proposition (ref)] $$Q_J = \lambda_J R_J(t,x_a) e^{t(m_J(x_a)-\Delta_{ab}m_{J-1})}M_J(t),$$ therefore $$Q_{J+1}(x_a,t)= \lambda_J \frac{\partial}{\partial x_a^1}\left[\frac{R_{J}(t,x_a)}{R_{J}(t,x_a)} e^{-t m_J(x_b)} M_J(t)\right] = 0, \text{ for all } t \in \mathbb{R}. $$ We actually see that we cannot calculate any $\Delta_{ab} m_{J+1}$ with the method of Lemma (ref) because of the logarithm: the identification process must be stopped here. Reciprocally, if $j_0 \leq J$, then for some $t_0 \in \mathbb{R}, Q_{j_0}(x_a, t_0) \neq 0$. Indeed, $j_0 \leq J \Rightarrow \forall j_0 \leq k \leq J,\, \lambda_k > 0 $ and we can write $$ Q_{j_0}(x_a,t) = \lambda_{j_0} R_{j_0}(t,x_a)M_{j_0}(t)e^{tm_{j_0}(x_a)-\Delta_{ab}m_{j_0-1} } \left( 1 + \sum_{j=k_0+1}^J \frac {\lambda_j}{\lambda_{j_0}}\frac{R_{j_0}^j(t,x_a)}{R_{j_0}(t,x_a)}e^{tm_{j,j_0}(x_a)}\frac{M_j(t)}{M_{j_0}(t)} \right). $$ By proposition (ref), we know that $\deg_tR_{j_0}^j = 1$, so there is a constant $b_{x_a, x_b, j, j_0} > 0$ such that $$\frac{R_{j_0}^j(t,x_a)}{R_{j_0}(t,x_a)} \xrightarrow[t\to\infty]{} b_{x_a, x_b, j, j_0}.$$ Using Assumption (ref) ((ref)), since $m_{h,k}(x_a) < -\Delta$, each term in the sum on the right hand side goes to 0 as $t$ goes to $\infty$, implying that for large enough $t$, the term in parenthesis is strictly positive, that is, nonzero.
remNote that it is not essential for our identification strategy to assume to impose Assumption (ref) (ref) $ m$ is $J$-times differentiable in one argument, as stated right after the assumption. Note that the use of the differentiation operator $\frac{\partial}{\partial x^1}$ in the linear operator $A$ is motivated by the fact that it eliminates terms that do not involve $x_a$, therefore with respect to which argument we differentiate is unimportant. The same identification argument applies if at each application of the operator $A$ in the recursive formula (ref) time we use $\frac{\partial}{\partial x^\ell}$ with a different $\ell \{1,...,k\}$ instead of keeping on using the same differential operator $\frac{\partial}{\partial x^1}$ as in the current proof. What we need is, as noted before, that $m$ can be differentiated up to a $J$-th order multi-index. This is less stringent than Assumption (ref) (ref), though we chose to state the result in the current form for notational simplicity.

Application to Identifiability of Auction Models with Unobserved Heterogeneity

It is of great interest to demonstrate that the preceding identification results potentially apply to nonparametric analysis of auction models with unobserved heterogeneity. As recognized in the recent literature, failing to properly taking account for unobserved heterogeneity in empirical auction models can lead to grossly misleading policy implications and counterfactual analyses. The reader is referred to \citeasnoun{haile2018unobserved} for various approaches to nonparametric identifiabilty in auction models when unobserved heterogeneity is present. Here we focus on application of the preceding mixture identification results to models with auction-specific unobserved heterogeneity. In particular, we focus on a symmetric affiliated auction model as considered in \citeasnoun{milgrom1982theory}. Suppose that valuations have the following multiplicative form, with $J$ unknown types of auctions

equation[equation omitted — 117 chars of source]

where $V^k$ is the valuation of bidder $k$, $1 \leq k \leq I$, who knows the number of bidders $I$, observed characteristics $x$, unobserved heterogeneity (i.e. unobserved type of auction) $j$, and a signal $S^k$. The function $\Gamma_j(x)$ depends on the two characteristics $x$ and $j$. The term $U_j^k$ can be interpreted as the “homogenized valuation" for bidder $k$, as used in \citeasnoun{haile2003nonparametric}. Let $B^k$ denote the bid of bidder $k$. The observables in this application is $(I,B^1,...,B^I,x)$. The rest remain unobserved.

We maintain that there are finite number of types in terms of auction heterogeneity. It is then possible to establish identification under quite weak assumptions. In the following result note that (i) valuations can be affiliated, and (ii) unobserved heterogeneity is treated flexibly, as not only it can affect valuations through the index function $\Gamma_j$ in an unrestricted way, the distribution of the homogenized valuation $U_j^k$ is allowed to depend on $j$ freely. Property (i) is important, as many preceding nonparametric identification results for auction with unobserved heterogeneity focus on the independent private value (IPV) model, as they tend to impose independence assumptions across valuations, with the exception of \citeasnoun{CHS}. For example, Property (i) implies that the result in this section applies to the common values model. Property (ii) about the flexible treatment of homogenized valuations is apparently new.

Assume

equation[equation omitted — 115 chars of source]

for every $j \in \{1,...,J\}$. Note that standard approaches to deal with unobserved heterogeneity do so through the index function $\Gamma_j$, and would not allow $(U^1,...,U^I)$ to depend on $j$ . Define $$ w(S,I,x,j) := \mathrm E\left[V^k | S^k = \max_{i \neq k, 1 \leq i \leq I}S^i = S, I, x, j\right] $$ which corresponds to the expected value of a bidder's valuation conditional on $I$, $x$, $j$, and the event that her equilibrium bid is pivotal. This is a quantity sometimes simply called “pivotal expected value". Let $w^k := w(S^k,I,x,j), 1 \leq k \leq I$ denote the pivotal expected value of the $k$-th bidder (whose signal is $S^k$) in an auction with characteristics $(x,j)$ and $I$ bidders. The goal here is to identify the joint distribution of $(w^1,...,w^I)$ in an auction with $(x,I,j)$, along with the distribution $(\lambda_1,...,\lambda_J)$ of the unobserved heterogeneity. Note that such knowledge is sufficient to address important questions often asked in practice: see, for example, footnote 9 of \citeasnoun{haile2018unobserved} for further discussions.

The above setting implies an expression of $w$ of the following form

equation[equation omitted — 96 chars of source]

where $\omega\left(S;I,j\right) = E[V^k|S^k = \max_{i \neq k}S^i = S, I,j]$. Like the homogenized valuation $\{\{U_j^k\}_{k=1}^I\}_{j=1}^J$, $\omega_j^k := \omega\left(S^k,I,j\right)$ is interpreted as a homogenized pivotal expected value of bidder $k$ in an auction of unobserved type $j$. It is well-known that the equilibrium bidding function preserves multiplicative separability in (ref), hence (ref), for each bidder $k$. Thus we obtain $$ B^k = \Gamma_j(x)R_j^k, $$ where $R_j^k$ is the homogenized valuation of bidder $k$ in type $j$ auction. Note that the unobserved auction type can affect equilibrium bids through two channels, that is, the index function $\Gamma_j$ and the homogenized bid $R_j^k$. Define $b^k = \log B^k$, $\gamma_j(x) := \log \Gamma_j(x)$ and $r_j^k := \log R_j^k$, then we have

equation[equation omitted — 105 chars of source]

Note that (ref) implies

equation[equation omitted — 126 chars of source]

conditional on $I$.

We now invoke Lemma (ref) to establish identification of this model. One of the main objects to be identified is the $I$-dimensional joint distribution of the pivotal expected values $w^1,...,w^I$ conditional on $(x,j,I)$, and our identification strategy works for each value of $I$. Thus in the rest of this section we treat $I$ as being fixed at a value, and suppress the index $I$ unless necessary. Let $c = (c_1,...,c_I)' \in {\mathbb R}^I$, and define $b(c) := \sum_{k=1}^{I}c_k b_{k}$, $C(c) := \sum_{k=1}^{n}c_k$ and $r_j(c) := \sum_{k=1}^{I}c_i r_j^k$. By (ref) and the finite mixture structure of the evaluation in (ref) we have $$ b(c) = C(c)\gamma_j(x) + r_j(c) \quad \text{ with probability } \lambda_j, 1 \leq j \leq J $$ where $r(c) {\bot\negthickspace\negthickspace\bot} x$ by (ref). Let $\left(b(c),\{C(c)\gamma_j(\cdot)\}_{j=1}^J,\{r_j(c)\}_{j=1}^J\right)$ play the role of $\left(z,\{m_j(\cdot)\}_{j=1}^J,\{\epsilon_j\}_{j=1}^J\right)$ in Lemma (ref), then $\left(C(c) \gamma_j(\cdot), \lambda_j\right)$ and the distribution of $r_j(c)$ are all identified for every $c \in {\mathbb R}^n$ and each $j \in \{1,...,J\}$. Moreover, we now know $\gamma_j(\cdot)$, $j \in \{1,...,J\}$ since $C(c)$ is known. Note that for each $j$, the marginal distribution of every linear combination $r_j(c)$ of the $I$-vector $(r_j^1,...,r_j^I)$ is identified as $c \in \mathbb{R}^I$ can be chosen arbitrarily. Then by Cram\'er-Wold the joint distribution of $(r_j^1,...,r_j^I)$ is obtained for each $j$. Apply this and the knowledge of $\gamma_j$ to equation (ref) to determine the joint distribution $(b^i,...,b^I)|x,j,I$. Using the first order condition for equilibrium bidding (see, e.g. \citeasnoun{haile2003nonparametric}, \citeasnoun{athey2007nonparametric} and Equation (2.4) in \citeasnoun{haile2018unobserved}) we can now back out the joint distribution of $(w^1,...,w^I)|x,j,I$ as desired. Note that the number of (unobserved) auction types $J$ is also identified by Proposition (ref).

Nonparametric estimation for $J=2$

This section develops a fully nonparametric estimation procedure based on our third identification result in Section (ref) where the number of mixture components is two. We first estimate the slopes of $m_1$ and $m_2$ nonparametrically. Define $\Delta = m_1(x_1) - m_1(x_0)$ and $\nabla = m_2(x_1) - m_2(x_0).$ Let us reintroduce notations. We write, for $j=1,2,$

align*[align* omitted — 501 chars of source]

where $F_i$ is the cumulative distribution function of $\epsilon_i,$ $h_n$ and $b_n$ are carefully chosen bandwidths for kernel density estimation. $\hat{M}^j(t)$ and $\hat{\phi}^j(s)$ are the Nadaraya-Watson regression estimators of respectively the conditional moment generating function and conditional characteristic function of $Z,$ when $X = x_j.$ $X$ being a vector, the kernel function $K$ can have a product form such as $K(X)=\Pi_{l=1}^k k(X^{(l)}).$

Our estimators are

align*[align* omitted — 284 chars of source]

where $(a_n)_n, (s_n)_n$ and $(t_n)_n$ are tuning parameters such that $a_n \to 0,$ $s_n \to \infty$ and $t_n \to \infty$. The notation Log$(\cdot)$ as before corresponds to the principal value of the logarithm of $\cdot$.

We enumerate here the assumptions on the kernel function needed to compute the rates of our estimators.

assThe kernel function $K(.)$ must satisfy the following conditions, $\int |K(U)| \, \mathrm{d} U < \infty $ , $\int K(U)\, \mathrm{d}U = 1,$ $\lim_{||U|| \to \infty} UK(U) \to 0,$ $\int K(U)^{2} \, \mathrm{d}U < \infty,$ $\int |K(U)| \, U'U \, \mathrm{d}U < \infty,$ $\int K(U) \, U \, \mathrm{d}U = 0,$ $\exists \alpha_0, \alpha \leq \alpha_0\ \Rightarrow \int e^{\alpha ||U||} |K(U)| \, U'U \, \mathrm{d}U < \infty, \int e^{\alpha ||U||} K(U)^2 \, \mathrm{d}U < \infty.$

We need the following assumptions on the model parameters.

ass\begin{enumerate}[(i)] • $f_X,$ the density of the random variable $X$, has continuous second order partial derivatives. $f_X$ and all its first and second order partial derivatives are bounded on $\mathbb{R}^k.$ $f_X(x_i)>0$, for $i=0,1.$$m_i,$ $i=1,2$ have continuous second order partial derivatives, and all their first and second order partial derivatives are bounded on $\mathbb{R}^k.$$h_n \underset{n \rightarrow \infty }{\to} 0,$ $nh_n^k \underset{n \rightarrow \infty }{\to} \infty$, and $b_n \underset{n \rightarrow \infty }{\to} 0,$ $nb_n^k \underset{n \rightarrow \infty }{\to} \infty,$$t_n \underset{n \rightarrow \infty }{\to} \infty,$ $t_n h_n \underset{n \rightarrow \infty }{\to} 0,$ and $s_n \underset{n \rightarrow \infty }{\to} \infty,$ $s_n b_n \underset{n \rightarrow \infty }{\to} 0.$ \end{enumerate}
ass\begin{enumerate}[(i)] • $\epsilon_1|x \sim F_1$ and $\epsilon_2|x \sim F_2$ at all $x \in \mathbb{R}^k$ where $F_1$ and $F_2$ do not depend on the value of $x$, • The domains of $M_1(t)$ and $M_2(t)$ are $[0,\infty)$, • $\forall \epsilon >0 \mbox{, } e^{\epsilon t} \frac{M_2(t)}{M_1(t)} \underset{t \rightarrow \infty }{=} O(\mu(t)),$ holds for some $\mu (\cdot),$ where $\mu(t) \xrightarrow[t \rightarrow \infty] {} 0,$$\frac{\phi_1(s)}{\phi_2(s)} \underset{s \rightarrow \infty }{=} O(f(s)),$ holds for some $f(\cdot),$ where $f(t) \xrightarrow [t \rightarrow \infty] {} 0.$ \end{enumerate}
propSuppose Assumptions (ref), (ref) and (ref) hold. Then \begin{enumerate}[(i)] • $\hat{\Delta} - \Delta = O_{\mathbb{P}} \left[ \frac{\mu (t_n)}{t_n} + \frac{1}{t_n} \left( (t_n h_n)^4 + \frac{1}{n h_n^k} \frac{M_1(2t_n)}{M_1(t_n)^2} \right)^{\frac{1}{2}} \right],$ where we assume $\frac{1}{n h_n^k} \frac{M_1(2t_n)}{M_1(t_n)^2} \underset{n \rightarrow \infty }{\to} 0$$\hat{\nabla} - \nabla = \frac{1}{a_n} O_{\mathbb{P}} \left[ f(s_n+a_n) +f(s_n) + \left((b_n s_n)^4 + \frac{1}{n b_n^{k} |\phi_2(s_n + a_n)|^2 } \right)^{1/2} + \left((b_n s_n))^4 + \frac{1}{n b_n^{k} |\phi_2(s_n)|^2 } \right)^{1/2} \right]$ \end{enumerate}

Proof 1 part 1

proof[Proof of Proposition (ref) (ref)] The estimator can be decomposed as $$ \hat{\Delta} = \frac{1}{t_n} \log \left( \frac{\hat{M}^1(t_n)}{\hat{M}^0(t_n)} \right)=\frac{1}{t_n} \log \left( \frac{M^1(t_n)}{M^0(t_n)} \right) + \frac{1}{t_n} \log \left( \frac{\hat{M^1}(t_n)}{M^1(t_n)} \right) - \frac{1}{t_n} \log \left( \frac{\hat{M^0}(t_n)}{M^0(t_n)} \right). $$ The first term in the decomposition is deterministic. Using the proof of Lemma (ref), this approximation error can be written \begin{align*} \frac{M^1(t_n)}{M^0(t_n)} = \, e^{t_n \Delta} \, \frac { 1+ \frac {1 - \lambda}{\lambda} e^{t_n [m_2(x_1) - m_1(x_1)] } \frac {M_2(t_n)}{M_1(t_n)}} { 1 + \frac{1 - \lambda}{\lambda} e^{t_n[m_2(x_0) - m_1(x_0)]}\frac {M_2(t_n)}{M_1(t_n)}} \, = \, e^{t_n \Delta} \left[ 1 + O(\mu (t_n)) \right], \end{align*} where the last equality holds using Assumption (ref) ((ref)). This gives $$ \frac{1}{t_n} \log \left( \frac{M^1(t_n)}{M^0(t_n)} \right) = \Delta + O(\frac{\mu (t_n)}{t_n} ). $$ Let us now focus on the terms $\frac{1}{t_n} \log \left( \frac{\hat{M}^j(t_n)}{M^j(t_n)} \right),$ the two estimation errors. We write $$ \hat{M}^j(t_n) = \frac{\frac{1}{nh_n^k} \sum_{p=1}^n e^{t_n Z_p} K(\frac{X_p - x_j}{h_n})} {\frac{1}{nh_n^k} \sum_{p=1}^n K(\frac{X_p - x_j}{h_n})} = \frac{\hat{N}^j(t_n)}{\hat{D}^j}, $$ and have \begin{equation} \frac{\hat{M}^j(t_n)}{M^j(t_n)} = \frac{\hat{N}^j(t_n)}{\hat{D}^j M^j(t_n)} = \frac{f_X(x_j)}{\hat{D}^j} \frac{\hat{N}^j(t_n)}{ f_X(x_j) M^j(t_n)}. \end{equation} In what follows, we treat separately the two ratios appearing in the last equality in ((ref)), showing that they both converge to 1. Part of the reasoning will be different from usual kernel regression. Indeed,for the second ratio, we need to keep the denominator to compute the convergence rate to counterbalance the numerator going to infinity, as the parameter $t_n$ goes to infinity. Under Assumptions (ref) and (ref), we know from usual results on kernel density estimation that when computing the Mean Square Error of the term $\frac{\hat{D}^j}{f_X(x_j)},$ the bias is of order $h_n^2$ and the variance of order $\frac{1}{n h_n^k}$, so that \begin{equation} \frac{\hat{D}^j}{f_X(x_j)} = 1 + O_{\mathbb{P}} \left( h_n^4 + \frac{1}{n h_n^k} \right) ^{\frac{1}{2}}. \end{equation} As for the second ratio in the decomposition of ((ref)) , the dependence in $t_n$ requires new assumptions when computing bias and variance. For the bias term we denote $G_n(x)= f_X(x) \mathbb{E}(e^{t_n Z}|X=x) $, then by definition of the estimator, $$\mathbb{E}(\hat{N}^j(t_n)) = \mathbb{E} \left(\frac{1}{n h_n^k} \sum_{p=1}^n e^{t_n Z_p} K(\frac{X_p - x_j}{h_n}) \right) = \int_{U \in \mathbb{R}^k} G_n(x_j + h_n U) K(U) \mathrm{d}U.$$ By Assumption (ref), $G_n$ is twice continuously differentiable. Since the kernel is of order 2 (Assumption (ref)), by virtue of the Mean Value Theorem, we have $$\mathbb{E} \left( \frac{\hat{N}^j(t_n)}{ f_X(x_j) M^j(t_n)} \right) - 1 = \frac{1}{G_n(x_j)} \int \frac{h_n^2}{2} U'. \nabla^2 G_n [x_j + h_n \tau_n(U) U]. U K(U) \mathrm{d}U$$ where $\tau_n(u) \in [0;1]$ and $\nabla^2 G_n(x)$ is the hessian matrix of the function $G_n$ evaluated at $x$. We know that $G_n(x) = f_X(x) [\lambda e^{t_n m_1(x)} M_1(t_n) + (1-\lambda) e^{t_n m_2(x)} M_2(t_n)].$ Twice differentiation gives \begin{align*} \nabla^2 G_n(x) = & \, \lambda e^{t_n m_1(x)} M_1(t_n)\ \mathrm{\ t_n^2 f_x(x) \nabla m_1(x) \nabla m_1(x)' \\ & + t_n (\nabla m_1(x) \nabla f_X(x)' + \nabla f_X(x) \nabla m_1(x)' + f_X(x) \nabla^2 m_1(x) ) \\ & + \nabla^2 f_X(x) \mathrm{\}} \\ & + (1-\lambda) e^{t_n m_2(x)} M_2(t_n)\ \mathrm{\ t_n^2 f_x(x) \nabla m_2(x) \nabla m_2(x)' \\ & + t_n (\nabla m_2(x) \nabla f_X(x)' + \nabla f_X(x) \nabla m_2(x)' + f_X(x) \nabla^2 m_2(x) ) \\ & + \nabla^2 f_X(x) \mathrm{\}}. \\ = & \, \lambda \, e^{t_n m_1(x)} M_1(t_n) \{ t_n^2 a_1(x) + t_n b_1(x) + c_1(x)\} \\ & + (1-\lambda) \, e^{t_n m_2(x)} M_2(t_n) \{ t_n^2 a_2(x) + t_n b_2(x) + c_2(x)\}. \end{align*} By boundedness of the first order partial derivatives of $m_i,$ $i=1,2,$ $$\exists \delta, \forall (x,U) \in \mathbb{R}^k \times \mathbb{R}^k, |m_i(x+h_n \tau_n(U) U) - m_i(x)| \leq \delta h_n ||U||,$$ implying that $e^{t_n m_i(x+h_n \tau_n(U) U) - m_i(x)} \leq e^{\delta t_n h_n ||U||}.$ Therefore, as $G_n(x) \geq f_X(x) \lambda e^{t_n m_1(x)} M_1(t_n) ,$ $$\frac{\lambda e^{t_n m_1(x_j + h_n \tau_n(U) U)} M_1(t_n)}{G_n(x_j)} \leq \frac{e^{\delta h_n t_n ||U||}}{f_X(x_j)} \leq \frac{e^{C ||U||}}{f_X(x_j)} ,$$ for some $C \leq \alpha_0,$ for $n$ large enough, under Assumption (ref) ((ref)). The same holds for the $(1-\lambda)$ term. By Assumption (ref), $a_1(x + h_n \tau_n(U) U)$ is bounded by a constant as well as the other coefficients of the $t_n$ polynomial in the expression of $\nabla^2 G_n [x_j + h_n \tau_n(U) U]$. This, together with the previous argument, implies that $\frac{1}{G_n(x_j)} \int U'. \nabla^2 G_n [x_j + h_n \tau_n(U) U]. U K(U) \mathrm{d}U = O(t_n^2).$ The rate of the bias term can therefore be bounded, $$\mathbb{E} \left( \frac{\hat{N}^j(t_n)}{ f_X(x_j) M^j(t_n)} \right) - 1 = O (t_n h_n)^2. $$ For the variance term, an upper bound is \begin{align*} & \frac{1}{n } \mathbb{E} \left[ \left(\frac{1}{h_n^k M^j(t_n) f_X(x_j)} e^{t_n Z} K(\frac{X - x_j}{h_n})\right)^2 \right] \\ &= \frac{1}{n [ h_n^k M^j(t_n) f_X(x_j)]^2} \int \mathbb{E}(e^{2t_n Z}|X)K(\frac{X - x_j}{h_n})^2 f_X(X) \mathrm{d}X \\ &= \frac{1}{nh_n^k} \int \frac{\mathbb{E}(e^{2t_n Z}|h_n U + x_j)}{\mathbb{E}(e^{t_n Z}|x_j)^2} \frac{f_X(h_n U + x_j)}{f_X(x_j)^2} K(U)^2 \mathrm{d}U \\ &= \frac{1}{nh_n^k} \int \frac{\lambda e^{2t_n m_1(h_n U + x_j)}M_1(2t_n) + (1 - \lambda) e^{2t_n m_2(h_n U + x_j)}M_2(2t_n)}{\left( \lambda e^{t_n m_1(x_j)}M_1(t_n) + (1 - \lambda) e^{t_n m_2(x_j)}M_2(t_n)\right)^2} \ \frac{f_X(h_n U + x_j)}{f_X(x_j)^2} K(U)^2 \mathrm{d}U \\ & \leq \frac{1}{nh_n^k} \frac{M_1(2t_n)}{M_1(t_n)^2}\int e^{2 \delta t_n h_n ||U||} \frac{\lambda + (1 - \lambda) e^{2t_n (m_2(x_j) - m_1(x_j)}\frac{M_2(2t_n)}{M_1(2t_n)}}{\left( \lambda + (1 - \lambda) e^{t_n (m_2(x_j) - m_1(x_j))} \frac{M_2(t_n)}{M_1(t_n)}\right)^2} \ \frac{f_X(h_n U + x_j)}{f_X(x_j)^2} K(U)^2 \mathrm{d}U . \end{align*} Using Assumption (ref) ((ref)) and Assumption ((ref)) for $n$ large enough, the integrand is bounded above by $ C' e^{C ||U||} K(U)^2, \, \forall \, U \in \mathbb{R}^k,$ for some $C$ independent of $n$, $C \leq \alpha_0,$ $C' > 0.$ Assumption ((ref)) and ((ref)) guarantee that the variance is of order $O ( \frac{1}{n h_n^k} \frac{M_1(2t_n)}{M_1(t_n)^2}). $ Therefore, \begin{equation} \frac{\hat{N}^j(t_n)}{ f_X(x_j) M^j(t_n)} = 1 + O_{\mathbb{P}} \left( (t_n h_n)^4 + \frac{1}{n h_n^k} \frac{M_1(2t_n)}{M_1(t_n)^2} \right) ^{\frac{1}{2}}. \end{equation} With ((ref)), ((ref)) and ((ref)), and given that by Jensen's inequality $\frac{M_1(2t_n)}{M_1(t_n)^2} \geq 1,$ the second ratio in ((ref)) dominates. $\frac{\hat{M}^j(t_n)}{M^j(t_n)} - 1 = O_{\mathbb{P}} \left( (t_n h_n)^4 + \frac{1}{n h_n^k} \frac{M_1(2t_n)}{M_1(t_n)^2} \right) ^{\frac{1}{2}}$. This finally gives \begin{align*} \hat{\Delta} - \Delta &= O(\frac{\mu (t_n)}{t_n}) + \frac{1}{t_n} \log( \left( 1 + O_{\mathbb{P}} \left( (t_n h_n)^4 + \frac{1}{n h_n^k} \frac{M_1(2t_n)}{M_1(t_n)^2} \right) ^{\frac{1}{2}} \right)^2 )\\ &= O_{\mathbb{P}} \left[ \frac{\mu (t_n)}{t_n} + \frac{1}{t_n} \left( (t_n h_n)^4 + \frac{1}{n h_n^k} \frac{M_1(2t_n)}{M_1(t_n)^2} \right)^{\frac{1}{2}} \right], \end{align*} since we assumed that $\frac{1}{n h_n^k} \frac{M_1(2t_n)}{M_1(t_n)^2} \underset{n \to \infty}{\rightarrow} 0.$

Proof 1 part 2

proof[Proof of Proposition (ref) (ref)] The estimator is $\hat{\nabla} = \frac{-i}{a} \mbox{ Log} \left( \frac{\hat{\phi}^1(s_n+a)}{\hat{\phi}^0(s_n+a)} \left( \frac{\hat{\phi}^1(s_n)}{\hat{\phi}^0(s_n)} \right)^{-1} \right).$ We first compute the rate of convergence of $\frac{\hat{\phi}^0(s_n)}{\hat{\phi}^1(s_n)}$, in a fashion similar to the proof above. From the identification result in Section (ref), we know $$ \lim_{s \rightarrow \infty} \frac{-i}{a} \mbox{ Log} \left( \frac{\phi(s+a|x_1)}{\phi(s+a|x_0)} \left( \frac{\phi(s|x_1)}{\phi(s|x_0)} \right)^{-1} \right) = \frac{1}{a} \left( a \nabla + 2 \pi \left \lfloor{\frac{1}{2} - \frac{a \nabla}{2 \pi}}\right \rfloor \right) . $$ Because we do not know the interval on which the identifying equation will be a constant of $a$, we plug in a sequence $a_n$ going to zero instead of a fixed $a$. For the approximation error, we have \begin{align*} \frac{\phi^1(s_n)}{\phi^0(s_n)}& = \frac{\lambda e^{i s_n m_1(x_1)}\phi_1(s_n) + (1 - \lambda) e^{s_n m_2(x_1)}\phi_2(s_n)}{\lambda e^{i s_n m_1(x_0)}\phi_1(s_n) + (1 - \lambda) e^{s_n m_2(x_0)}\phi_2(s_n)}\\ & = e^{i s_n \nabla} \frac{\frac{\lambda}{1-\lambda} e^{i s_n (m_1(x_1) - m_2(x_1))} \frac{\phi_1(s_n)}{\phi_2(s_n)} + 1}{\frac{\lambda}{1-\lambda} e^{i s_n (m_1(x_0) - m_2(x_0))} \frac{\phi_1(s_n)}{\phi_2(s_n)} + 1}\\ & = e^{i s_n \nabla} (1 + O(f(s_n))). \end{align*} To compute the estimation error, the scheme is initially similar to the previous proof. We write $$\hat{\phi}^j(s_n) = \frac{\frac{1}{n h_n^k} \sum_{p=1}^n e^{is_n Z_p} K(\frac{X_p - x_j}{b_n})} {\frac{1}{n h_n^k} \sum_{p=1}^n K(\frac{X_p - x_j}{b_n})} = \frac{\hat{num}^j(s_n)}{\hat{denom}^j} $$ and work with an equation similar to ((ref)), here $$ \frac{\hat{\phi}^j(s_n)}{\phi^j(s_n)}= \frac{f_X(x_j)}{\hat{denom}^j} \ \frac{\hat{num}^j(s_n)}{ f_X(x_j) \phi^j(s_n)} $$ We compute the convergence rate of the ratios in the last equality. As in ((ref)), we know $\frac{\hat{denom}^j}{f_X(x_j)} = 1 + O_{\mathbb{P}} \left( b_n^4 + \frac{1}{n b_n^k} \right) ^{\frac{1}{2}}.$ Now, let $A_n = \frac{\hat{num}^j(s_n)}{ f_X(x_j) \phi^j(s_n)}:$ $A_n \in \mathbb{C}.$ Working with complex numbers for this proof, we use $|.|$ to denote a modulus. Let us focus on the bias term of $A_n$. We write $g_n(x) = f_X(x) \mathbb{E}(e^{i s_n Z}|X=x)$ so that $A_n = \frac{\hat{num}^j(s_n)}{g_n(x)}$. Since $g_n(x) = f_X(x) (\lambda e^{i s_n m_1(x)}\phi_1(s_n) + (1 - \lambda) e^{i s_n m_2(x)}\phi_2(s_n)),$ we denote $G_n^{lc}(x)=\cos(s_n m_l(x))f_X(x)$ and $ G_n^{ls}(x)=\sin(s_n m_l(x))f_X(x)$ for $l=1,2.$ Then we have \begin{align*} \mathbb{E}(\hat{num}^j(s_n)) = &\mathbb{E} \left(\frac{1}{n b_n^k} \sum_{p=1}^n e^{i s_n Z_p} K(\frac{X_p - x_j}{b_n}) \right) = \int_{U \in \mathbb{R}^k} g_n(x_j + b_n U) K(U) \mathrm{d}U,\\ = & \lambda \phi_1(s_n) \int [\cos(s_n m_1(x_j + b_n U)) + i \sin(s_n m_1(x_j + b_n U))] f_X(x_j + b_n U) K(U) \mathrm{d}U \\ & + (1 - \lambda) \phi_2(s_n) \int [\cos(s_n m_2(x_j + b_n U)) + i \sin(s_n m_2(x_j + b_n U))] f_X(x_j + b_n U) K(U) \mathrm{d}U \\ = & \lambda \phi_1(s_n) \int [G_n^{1c} (x_j + b_n U) + i G_n^{1s} (x_j + b_n U) ] K(U) \mathrm{d}U \\ & + (1 - \lambda) \phi_2(s_n) \int [G_n^{2c} (x_j + b_n U) + i G_n^{2s} (x_j + b_n U) ] K(U) \mathrm{d}U \end{align*} Using the assumption that the kernel is of order 2 (Assumption (ref)), \begin{equation} \int G_n^{1c} (x_j + b_n U) K(U) \mathrm{d}U - G_n^{1c}(x_j), \nonumber \\ = \int \frac{b_n^2}{2} U' \nabla^2 G_n^{1c} [x_j + b_n \tau_n(U) U] \, U K(U) \, \mathrm{d}U, \end{equation} where $\tau_n(U) \in [0;1]$ and $\nabla^2 G_n^{1c}(x)$ is the hessian matrix of the function $G_n^{1c}$ evaluated at $x$. That is, \begin{align*} \nabla^2 G_n^{1c}(x) = & - s_n^2 f_X(x) \cos(s_n m_1(x)) \nabla m_1(x) \nabla m_1(x)' \\ & - s_n \sin(s_n m_1(x)) [\nabla m_1(x) \nabla f_X(x)' + \nabla f_X(x) \nabla m_1(x)' + f_X(x) \nabla^2 m_1(x)] \\ & + \cos(s_n m_1(x)) \nabla^2 f_X(x). \end{align*} Similarly to what is done in the first part of this proof, Assumption (ref) guarantees that $ \int G_n^{1c} (x_j + b_n U) K(U) \mathrm{d}U - G_n^{1c}(x_j) = O(b_n s_n)^2. $ The same rate applies for $G_n^{1s}, G_n^{2c}$ and $G_n^{2s}$, implying $$ \mathbb{E}(\hat{num}^j(s_n)) = \int g_n(x_j + b_n U) K(U) \mathrm{d}U = g_n(x_j) + O( \, (b_n s_n)^2 \, [\lambda |\phi_1(s_n)| + (1 - \lambda) |\phi_2(s_n)| \, ]\, ), $$ which gives, for the bias term, \begin{align*} \mathbb{E}(A_n) & = \frac{1}{ f_X(x_j) \phi^j(s_n)} \mathbb{E}( \hat{num}^j(s_n)) \\ & = 1 + O \left( (b_n s_n)^2 \frac{1}{f_X(x_j)} \frac{\lambda |\phi_1(s_n)| + (1 - \lambda) |\phi_2(s_n)|}{\lambda e^{i s_n m_1(x_j)}\phi_1(s_n) + (1 - \lambda) e^{i s_n m_2(x_j)}\phi_2(s_n)} \right)\\ & = 1 + O \left( (b_n s_n)^2 \frac{1}{ e^{i s_n m_2(x_j)}} \frac{\lambda \frac{|\phi_1(s_n)|}{|\phi_2(s_n)|} + (1 - \lambda)}{\lambda \frac{\phi_1(s_n)}{|\phi_2(s_n)|} e^{i s_n (m_1(x_j) - m_2(x_j) )} + (1 - \lambda)} \right) \\ & = 1 + O(b_n s_n)^2, \end{align*} where the last equality comes from Assumption (ref) ((ref)). As for the variance term, we write \begin{align*} Var(\frac{\hat{num}^j(s_n)}{ f_X(x_j) \phi^j(s_n)})& = \frac{1}{f_X(x_j)^2 |\phi^j(s_n)|^2} \frac{1}{n b_n^{2k}} Var(e^{i s_n Z} K(\frac{X - x_j}{b_n}))\\ & \leq \frac{1}{f_X(x_j)^2 |\phi^j(s_n)|^2} \frac{1}{n b_n^{2k}} \mathbb{E}(|e^{i s_n Z} K(\frac{X - x_j}{b_n})|^2) \\ & \leq \frac{1}{ |\phi^j(s_n)|^2} \frac{1}{n b_n^{k}} \frac{\int f_X(x_j + b_n U) K^2(U) \mathrm{d}U}{f_X(x_j)^2}, \end{align*} and Assumption (ref) ((ref)) guarantees that in the last equality, the third term in the product converges to $\frac{\int K^2(U) \mathrm{d}U}{f_X(x_j)}.$ Moreover, \begin{align*} |\phi^j(s_n)| & =|\lambda e^{i s_n m_1(x_j)}\phi_1(s_n) + (1 - \lambda) e^{i s_n m_2(x_j)}\phi_2(s_n)|\\ & = | \phi_2(s_n)| \ \left| \lambda e^{i s_n m_1(x)} \frac{\phi_1(s_n)}{\phi_2(s_n)} + (1 - \lambda) e^{i s_n m_2(x)} \right| \sim_{n \to \infty} (1- \lambda)|\phi_2(s_n)|, \end{align*} therefore implying $ Var(A_n) = O(\frac{1}{n b_n^{k} |\phi_2(s_n)|^2 }).$ From those two computations, the following reasoning gives a convergence rate for $A_n$ : $Bias(\Re(A_n)) = \Re(Bias(A_n)) = O(b_n s_n)^2.$ Similarly $Bias(\Im(A_n))= O(b_n s_n)^2$. Plus, by definition for a complex random variable $Var(A_n)=Var(\Re(A_n)) + Var(\Im(A_n))$: both the variances of the real part and the imaginary part are smaller than the variance of $A_n$. An upper bound of the rates of convergence of the Mean Square Error of the real and imaginary parts is therefore obtained, $\Re(A_n) - 1 = O_{\mathbb{P}} ((b_n s_n)^4 + \frac{1}{n b_n^{k} |\phi^j(s_n)|^2 } )^{1/2},$ and $\Im(A_n) = O_{\mathbb{P}} ((b_n s_n)^4 + \frac{1}{n b_n^{k} |\phi^j(s_n)|^2 } )^{1/2}$. This gives, $$|A_n -1| = O_{\mathbb{P}} ((b_n s_n)^4 + \frac{1}{n b_n^{k} |\phi^j(s_n)|^2 } )^{1/2},$$ so that the estimation error is \begin{align*} \frac{\hat{\phi}^j(s_n)}{\phi^j(s_n)}&=1+ O_{\mathbb{P}} \left[ \left( b_n^4 + \frac{1}{n b_n^k} \right) ^{\frac{1}{2}} + \left((b_n s_n)^4 + \frac{1}{n b_n^{k} |\phi^j(s_n)|^2 } \right)^{1/2} \right] \\ & = 1+O_{\mathbb{P}} \left((b_n s_n)^4 + \frac{1}{n b_n^{k} |\phi^j(s_n)|^2 } \right)^{1/2}. \end{align*} Finally we obtain \begin{align*} \frac{\hat{\phi}^1(s_n)}{\hat{\phi}^0(s_n)} & = \frac{\hat{\phi}^1(s_n)}{\phi^1(s_n)} \left( \frac{\hat{\phi}^0(s_n)}{\phi^0(s_n)} \right)^{-1} \frac{\phi^1(s_n)}{\phi^0(s_n)}\\ & = e^{i s_n \nabla} [1 + O_{\mathbb{P}}(f(s_n))] \ \left[ 1 + O_{\mathbb{P}} \left((b_n s_n)^4 + \frac{1}{n b_n^{k} |\phi^j(s_n)|^2 } \right)^{1/2} \right]. \end{align*} Plugging in this expression in the definition of the estimator, we obtain \begin{align*} \hat{\nabla} = & \frac{-i}{a_n} Log \left( \frac{\hat{\phi}^1(s_n+a_n)}{\hat{\phi}^0(s_n+a_n)} \left( \frac{\hat{\phi}^0(s_n)}{\hat{\phi}^1(s_n)} \right) \right) \\ = & \frac{-i}{a_n} Log \mathrm{\ e^{i (s_n +a_n) \nabla} [1 + O_{\mathbb{P}}(f(s_n + a_n))] [1 + O_{\mathbb{P}} \left((b_n (s_n + a_n))^4 + \frac{1}{n b_n^{k} |\phi^j(s_n + a_n)|^2 } \right)^{1/2} ] \\ & e^{-i s_n \nabla} [1 + O_{\mathbb{P}}(f(s_n))] [1 + O_{\mathbb{P}} \left((b_n s_n))^4 + \frac{1}{n b_n^{k} |\phi^j(s_n)|^2 } \right)^{1/2} ] \mathrm{\}}\\ = & \frac{-i}{a_n} Log \mathrm{\ e^{i a_n \nabla} [ 1 + O_{\mathbb{P}}(f(s_n+a_n)) + O_{\mathbb{P}}(f(s_n)) + O_{\mathbb{P}} \left((b_n s_n)^4 + \frac{1}{n b_n^{k} |\phi^j(s_n + a_n)|^2 } \right)^{1/2} \\ & + O_{\mathbb{P}} \left((b_n s_n)^4 + \frac{1}{n b_n^{k} |\phi^j(s_n)|^2 } \right)^{1/2} ] \mathrm{\}}. \end{align*} As the term multiplying $e^{i a_n \nabla}$ in the $\mbox{Log}$ converges to $1$, and eventually $a_n \nabla \in (-\pi; \pi),$ the expression above becomes \begin{align*} \hat{\nabla} = \frac{-i}{a_n} \mathrm{\ i a_n \nabla + Log [ 1 + & O_{\mathbb{P}}(f(s_n+a_n)) + O_{\mathbb{P}}(f(s_n)) \\ + & O_{\mathbb{P}} \left((b_n s_n)^4 + \frac{1}{n b_n^{k} |\phi^j(s_n + a_n)|^2 } \right)^{1/2} & + O_{\mathbb{P}} \left((b_n s_n)^4 + \frac{1}{n b_n^{k} |\phi^j(s_n)|^2 } \right)^{1/2} ] \mathrm{\}}, \end{align*} that is, using the first order approximation of the principal value of the log around 1, $$ \hat{\nabla} = \nabla +\frac{1}{a_n} O_{\mathbb{P}} \left( f(s_n+a_n) +f(s_n) + \left((b_n s_n)^4 + \frac{1}{n b_n^{k} |\phi^j(s_n + a_n)|^2 } \right)^{1/2} + \left((b_n s_n)^4 + \frac{1}{n b_n^{k} |\phi^j(s_n)|^2 } \right)^{1/2} \right ). $$

The only restriction imposed on the tuning parameter $a_n$ is that it converges to $0$.

For the sake of simplicity, we now write $\hat{\Delta} - \Delta = O_{\mathbb{P}} (\alpha_n)$ and $\hat{\nabla} - \nabla = O_{\mathbb{P}} (\beta_n )$. The rates $\alpha_n$ and $\beta_n$ depend on the distributions of the error terms, and we show here that the rates are polynomial in $n$ if these distributions are normal.

Indeed if $\epsilon_1|x \sim \mathcal{N}(0, \sigma_1^2)$ and $\epsilon_2|x \sim \mathcal{N}(0, \sigma_2^2),$ with $\delta = \sigma_1^2 - \sigma_2^2 >0 ,$ then Assumption (ref) is satisfied. For the ratio of the mgf, $\forall \epsilon >0 \mbox{, } e^{\epsilon t} \frac{M_2(t)}{M_1(t)} = e^{\epsilon t - \frac{\delta}{2} t^2} \underset{t \rightarrow \infty }{=} O(\mu(t)),$ with $\mu(t) = e^{-(\frac{\delta}{2}-\nu) t^2} \to 0,$ as $t \rightarrow \infty,$ for some $0 < \nu < \frac{\delta}{2}.$ And as for the ratio of the characteristic functions, $\frac{\phi_1(s)}{\phi_2(s)} = e^{- \frac{1}{2} \delta s^2} \underset{s \rightarrow \infty }{=} O(f(s)),$ with $f(s) = e^{- \frac{1}{2} \delta s^2} \xrightarrow [s \rightarrow \infty] {} 0.$ We take a fixed $a$ in the definition of $\hat{\nabla}$ here to simplify the computations, assuming $a$ is small enough. Applying the results from the estimation proofs, the convergence rates are

enumerate[(i)] • $\hat{\Delta} - \Delta = \frac{1}{t_n} O_{\mathbb{P}} \left[e^{-(\frac{\delta}{2}-\nu) t_n^2} + \left( (t_n h_n)^4 + \frac{1}{n h_n^k} e^{\sigma_1^2 t_n^2} \right)^{\frac{1}{2}} \right],$$\hat{\nabla} - \nabla = O_{\mathbb{P}} \left[ e^{- \frac{1}{2} \delta s_n^2} + \left((b_n s_n)^4 + \frac{1}{n b_n^{k} } e^{\sigma_2^2(s_n+a)^2} \right)^{\frac{1}{2}} \right].$

One can show that with the appropriate choice of the sequences $t_h, h_n, s_n,$ and $b_n$, the rates are polynomial in $n$. For example, it is the case if $k=1$, $h_n= n^{\frac{-1}{5} + \epsilon}$, $t_n=\frac{1}{\sigma_1} (\epsilon \log(n))^{\frac{1}{2}}$, $s_n=\frac{1}{\sigma_2} ( \beta \log(n))^{\frac{1}{2}}$ and $b_n = n^{\frac{-1}{5} + \beta}$ for $\epsilon, \beta < \frac{1}{5}.$

Proof 2

We now focus on the estimation of the remaining objects. We showed that if $\lambda \in (0,1),$ then $\lambda = \frac{\mathbb{E}(Z|X=x_1)- \mathbb{E}(Z|X=x_0) - \nabla}{\Delta - \nabla}.$ A natural estimator is therefore $$\hat{\lambda}=\frac{\hat{E}(Z|X=x_1)- \hat{E}(Z|X=x_0) - \hat{\nabla}}{\hat{\Delta} - \hat{\nabla}},$$ where $\hat{E}(Z|X=.)$ is the usual multivariate kernel regression estimator, $ \hat{E}(Z|X=x) = \frac{\sum_{p=1}^n Z_p K(\frac{X_p - x}{d_n})} {\sum_{p=1}^n K(\frac{X_p - x_j}{d_n})}$ where the kernel does not have to be the one used for the previous estimators but will be written $K$ for the sake of simplicity. Similarly the point estimation of the regression functions $m_1$ and $m_2$ is derived from Equation ((ref)). Writing $\hat{C}=\frac{1}{2} \left (\hat{E}(Z^2|X=x_0) - \hat{E}(Z^2|X=x_1) + \hat{\lambda} \hat{\Delta}^2 + (1 - \hat{\lambda}) \hat{\nabla}^2 \right),$ our estimators of $m_1(x_0)$ and $m_2(x_0)$ are

equation[equation omitted — 302 chars of source]

The convergence rate of these estimators can be computed easily. With the usual assumptions for kernel estimation and the appropriate choice of bandwidths $d_n=n^{\frac{-1}{k+4}},$ it is known that $\hat{E}(Z|X=x) - \mathbb{E}(Z|X=x) = O_{\mathbb{P}} (n^{-\frac{2}{k+4}})$ and $\hat{E}(Z^2|X=x) - \mathbb{E}(Z^2|X=x) = O_{\mathbb{P}} (n^{-\frac{2}{k+4}})$, see, e.g, \citeasnoun{hardle1994applied}. Writing $\epsilon_n = n^{-\frac{2}{k+4}} + \alpha_n + \beta_n$, one obtains $\hat{\lambda} = \lambda + O_{\mathbb{P}} (\epsilon_n).$ The estimators of $m_1(x_0)$ and $m_2(x_0)$, being linearizable functions of $\hat{\lambda} $, $\hat{\Delta}$ and $\hat{\nabla}$, their rates of convergence are similarly bounded. This is summarized in the next proposition.

propUnder Assumptions (ref), (ref), assuming that $\lambda \in (0,1),$ $K$ satisfies Assumption (ref), $d_n \to 0,$ and $n d_n^k \to 0$, \begin{enumerate} • $\hat{\lambda} = \lambda + O_{\mathbb{P}} (\epsilon_n),$$\hat{m_i(x_0)} = m_i(x_0) + O_{\mathbb{P}} (\epsilon_n), \mathrm{ for} \: i=1,2$ \end{enumerate}

Proof 3

To estimate the CDF of $\epsilon_1$ and $\epsilon_2$, we use Equation ((ref)) and propose the following estimator

equation*[equation* omitted — 217 chars of source]

In this formula $p(n) \in \mathbb{N}$ will be specified later, $\hat{g}(x)=\hat{m}_1(x) - \hat{m}_2(x),$ $\hat{\delta} = \hat{\Delta} - \hat{\nabla}$, and $\hat{F}(.|.)$ is the kernel regression estimator of the conditional cumulative distribution function, $$\hat{F}(z|x)= \frac{\sum_{j=1}^n \mathbbm{1} (Z_j \leq z) k(\frac{X_j - x}{c_n})} {\sum_{j=1}^n k(\frac{X_j - x}{c_n})}.$$ The kernel function and the bandwidth may differ from the choices for our previous kernel regression estimators, and will be here written as $k$ and $c_n$ respectively.

ass\ \begin{enumerate} • The probability distribution functions of $\epsilon_1$ and $\epsilon_2$, $f_1$ and $f_2,$ are bounded by a constant $c,$$f_j$ is twice differentiable on $\mathbb{R},$ and $f_j,f_j', f_j''$ are continuous and bounded for $j=1,2.$ \end{enumerate}
assWe assume that $k(.)$ satisfies Assumption (ref), and in addition impose, \begin{enumerate} • $||k||_{\infty} < \infty,$ • The kernel function $k$ has support contained in $[-\frac{1}{2}, \frac{1}{2}]^k,$ • Assumptions (K-iii) and (K-iv) of \citeasnoun{einmahl2005uniform} hold for $k(.)$, • $c_n \geq C' \frac{\log(n)}{n},$ $c_n = O(n^{-\gamma_1}),$ for some $\gamma_1 < 1.$ \end{enumerate}

The assumptions from \citeasnoun{einmahl2005uniform} are conditions on the covering and measurability properties of the class of functions $\left\lbrace k(\frac{x-.}{c}) ;\ c>0, \, x \in \mathbb{R}^k \right\rbrace$.

propUnder (ref), (ref), (ref) and (ref), $\hat{F}_1(z) - F_1(z)= O_{\mathbb{P}}\left( (p(n)+1)n^{ \frac{-2}{k+4} + a}+ p(n)^2 \epsilon_n + e^{-\gamma_0 p(n)} \right),$ and $\hat{F}_2(z) - F_2(z)= O_{\mathbb{P}}\left( (p(n)+1)n^{ \frac{-2}{k+4} + a}+ p(n)^2 \epsilon_n + e^{-\gamma_0 p(n)} \right).$
proofFix $z \in \mathbb{R}.$ Write $\xi_j^0=z + j \delta(x_1,x_0) + m_2(x_0),$ $\hat{\xi}_j^0 = z + j \hat{\delta}(x_1,x_0) + \hat{m}_2(x_0)$, $\xi_j^1=z + j \delta(x_1,x_0) + m_1(x_1) - g(x_0)$ and $\hat{\xi}_j^1=z + j \hat{\delta}(x_1,x_0) + \hat{m}_1(x_1) - \hat{g}(x_0)$. \begin{align} \hat{F}_2(z) - F_2(z) = & \frac{1}{1-\lambda} \sum_{j=0}^{p(n)} \left\lbrace \left[\hat{F}(\hat{\xi}_j^1|x_1) - F(\xi_j^1|x_1) \right] - \left[\hat{F}(\hat{\xi}_j^0|x_0) - F(\xi_j^0|x_0) \right] \right\rbrace \nonumber \\ & - \frac{1}{1-\lambda} \sum_{j=p(n)+1}^\infty \left[ F(\xi_j^1|x_1) - F(\xi_j^0|x_0) \right] \\ & + (\frac{1}{1-\hat{\lambda}} - \frac{1}{1-\lambda}) \sum_{j=0}^{p(n)} \hat{F}(\hat{\xi}_j^1|x_1) - \hat{F}(\hat{\xi}_j^0|x_0). \nonumber \end{align} We write $\hat{F}_2(z) - F_2(z) = I_1 - I_2 + I_3,$ and the convergence rate of each part in the right hand side of ((ref)) will be computed separately. For $I_1$, we write \begin{align*} (1 - \lambda) \, I_1= & \sum_{j=0}^{p(n)} \left[\hat{F}(\hat{\xi}_j^1|x_1) - F(\hat{\xi}_j^1|x_1) \right] + \sum_{j=0}^{p(n)} \left[ F(\hat{\xi}_j^1|x_1) - F(\xi_j^1|x_1) \right] \\ & - \sum_{j=0}^{p(n)} \left[\hat{F}(\hat{\xi}_j^0|x_0) - F(\hat{\xi}_j^0|x_0) \right] - \sum_{j=0}^{p(n)} \left[ F(\hat{\xi}_j^0|x_0) - F(\xi_j^0|x_0) \right] . \end{align*} We know $\frac{\partial F(z|x)}{\partial z} = f(z|x) = \lambda f_1(z-m_1(x)) + (1-\lambda)f_2(z-m_2(x)),$ and Assumption (ref) guarantees that $f(y|x)$ is bounded by $c, \, \forall \, (x,y)$ therefore $ y \mapsto F(y|x)$ is Lipschitz continuous with constant $c.$ That is, for $v=0,1,$ $|F(\hat{\xi}_j^v|x_v) - F(\xi_j^v|x_v)| \leq c \, |\hat{\xi}_j^v - \xi_j^v|$ implying \begin{align*} |\sum_{j=0}^{p(n)} F(\hat{\xi}_j^v|x_v) - F(\xi_j^v|x_v) | = O_{\mathbb{P}}(p(n)^2 \epsilon_n). \end{align*} For the two other terms in $I_1,$ we write $$F_n(.|x_i) = \frac{\mathbb{E}(\mathbbm{1} (Z \leq z) k(\frac{X - x_i}{c_n}))}{\mathbb{E}( k(\frac{X - x_i}{c_n}))}.$$ Then under Assumption (ref), we apply Theorem 3 of \citeasnoun{einmahl2005uniform}, which gives the rate of the supremum of $|| \hat{F}(.|x) - F_n(.|x) ||_{\infty}$ over a certain range of bandwidths and over $x \in I$ where $I$ is a compact subset of $\mathbb{R}^k.$ For the specific bandwidth $b_n$ and taking $I = \left\lbrace x_0,x_1 \right\rbrace$ we then have \begin{align} \limsup_{n \to \infty} (nc_n^k)^{1/2} || \hat{F}(.|x) - F_n(.|x) ||_{\infty} & = O_{\mathrm{a.s}}\left(\max(\log \log n,-\log(c_n))^{1/2}\right) \nonumber \\ & = O_{\mathrm{a.s}}\left( (-\log c_n)^{1/2}\right). \end{align} We now examine $ F_n(.|x) - F(.|x).$ Write $f(.,.)$ the joint density of $(Z,X)$, then we define \begin{align*} F_X(x,z)& = \int_{ z' \leq z} f(z',x)\mathrm{d}z' = F(z|x) f_X(x) = [ \lambda F_1(z - m_1(x)) + (1 - \lambda) F_2(z - m_2(x)) ] f_X(x) \end{align*} and write $$ F_n(z|x_i) - F(z|x_i)= \frac{\frac{1}{c_n^k} \mathbb{E}(\mathbbm{1} (Z \leq z) k(\frac{X - x_i}{c_n})) - F_X(x_i,z)}{\frac{1}{c_n^k} \mathbb{E}( k(\frac{X - x_i}{c_n})) } + F_X(x_i,z) \left( \frac{1}{\frac{1}{c_n^k} \mathbb{E}( k(\frac{X - x_i}{c_n}))} - \frac{1}{f_X(x_i)}\right).$$ Under Assumption (ref), (ref) and (ref), we know that $\frac{1}{c_n^k} \mathbb{E}( k(\frac{X - x_i}{c_n})) - f_X(x_i) = O(c_n^2).$ Similarly, $$\frac{1}{c_n^k} \mathbb{E}\left[ \mathbbm{1} (Z \leq z) \, k\left( \frac{X - x_i}{c_n}\right) \right] - F_X(x_i,z)= \frac{c_n^2}{2} \int_{U \in \mathbb{R}^k} U' \, \nabla_X^2 F_X (x_i + b_n \tau_n(U) U, z) \, U k(U) \mathrm{d}U ,$$ and Assumption (ref), (ref) and (ref) guarantee that $\nabla_X^2 F_X (.,.)$ is uniformly bounded over $\mathbb{R}^{k+1}$. Therefore $$\sup_{z \in \mathbb{R}} \left| \ \frac{1}{c_n^k} \mathbb{E}\left[ \mathbbm{1} (Z \leq z) \, k\left( \frac{X - x_i}{c_n}\right) \right] - F_X(x_i,z)\right| = O(c_n^2).$$ which gives, for $i=0,1,$ \begin{equation} || F_n(.|x_i) - F(.|x_i)||_{\infty} = O(c_n^2). \end{equation} Equations ((ref)) and ((ref)) give $||\hat{F}(.|x) - F(.|x) ||_{\infty} = O_{\mathbb{P}}((-\log(c_n))^{1/2} (nc_n^k)^{-1/2} + c_n^2).$ For the appropriate choice of $\gamma_1$ in Assumption (ref) and for any small $a>0,$ $$\sup_{y \in \mathbb{R}} | \hat{F}(y|x_i) - F(y|x_i) | = O_{\mathbb{P}}( n^{-\frac{2}{k+4} + a}), \; i=0,1.$$ This implies that $$ \sum_{j=0}^{p(n)} \left[\hat{F}(\hat{\xi}_j^1|x_i) - F(\hat{\xi}_j^1|x_i) \right]= O_{\mathbb{P}}((p(n)+1)n^{-\frac{2}{k+4} + a}), \; i=0,1. $$ Therefore, $$I_1 = O_{\mathbb{P}} \left( (p(n)+1)n^{-\frac{2}{k+4} + a} + p(n)^2 \epsilon_n \right).$$ Looking at $I_2,$ by construction the second sum appearing in the right hand side of ((ref)) simplifies to $$\frac{1}{1-\lambda} \sum_{j=p(n)+1}^\infty \left[ F(\xi_j^1|x_1) - F(\xi_j^0|x_0) \right] = 1 - F_2(z + (p(n)+1) \delta(x_1,x_0)).$$ Using the exponential version of the Chebyshev's inequality, we have $1- F_2(C)= \mathbb{P}(\epsilon_2 >C) \leq e^{-tC}M_2(t)$ using the assumption that the moment generating functions are finite. Fixing $t_0 \in \mathbb{R}_{+},$ $1 - F_2[z + (p(n)+1) \delta(x_1,x_0)] \leq e^{t_0 z + (p(n)+1) \delta(x_1,x_0) t_0}$ which guarantees the existence of $\gamma_0 > 0$ such that $I_2 = O(e^{-\gamma_0 \, p(n)}).$ As for $I_3,$ we showed in our computation for $I_1$ that $\sum_{j=0}^{p(n)} \hat{F}(\hat{\xi}_j^1|x_1) - \hat{F}(\hat{\xi}_j^0|x_0) \xrightarrow [n \to \infty]{} F_2(z).$ As $\frac{1}{1-\hat{\lambda}} - \frac{1}{1-\lambda}= O_{\mathbb{P}}(\epsilon_n),$ we have $$I_3 = O_{\mathbb{P}} (\epsilon_n).$$ Adding these three parts, we obtain $$\hat{F}_2(z) - F_2(z) = O_{\mathbb{P}} \left( (p(n)+1)n^{-\frac{2}{k+4} + a}+ p(n)^2 \epsilon_n + e^{-\gamma_0 p(n)} \right). $$ Using the equation $F(z|x)= \lambda F_1(z - m_1(x)) + (1 - \lambda) F_2(z - m_2(x)),$ an estimator of $F_1(z)$ is $$ \hat{F}_1(z) = \frac{1}{\hat{\lambda}} \left[ \hat{F}(z + \hat{m}_1(x)) - (1 - \hat{\lambda}) \hat{F}_2(z + \hat{m}_1(x) - \hat{m}_2(x)) \right], $$ which will converge to $F_1(z)$ at the same rate.

In the case where $\epsilon_n$ is slower than $n^{-2 \: \frac{1 - 2a}{k+4}},$ for some $a,$ which happens when for instance the error terms are normally distributed, then $p_n$ is solution to $\epsilon_n p_n = t_0 e^{-t_0 p_n}.$

Conclusion

New nonparametric identification results for finite mixture models are developed. These open up the possibility of flexibly modeling economic behavior in the presence of unobserved heterogeneity.