The One-Period Kyle (1985) Model Has a Unique Equilibrium: A Monotone Gaussian Bayes inverse-rigidity theorem

The One-Period Kyle (1985) Model Has a Unique Equilibrium: A Monotone Gaussian Bayes inverse-rigidity theorem

Introduction

In the one-period model of insider trading of Kyle85, a risk-neutral informed trader observes an asset value $V\sim N(\mu,\tau^2)$ and, on the realisation $V=v$, submits the market order $\phi(v)$, where the pure strategy $\phi\colon\mathbb R\to\mathbb R$ is a Borel map; liquidity traders submit an independent exogenous demand $U\sim N(0,\sigma^2)$; the risk-neutral market maker observes only the aggregate order flow and, disciplined by competition, prices it at the inverse regression \[ P_\phi(y)=\mathbb E[V\mid Y_\phi=y],\qquad Y_\phi=\phi(V)+U. \] The strategy $\phi$ is an equilibrium if, taking the pricing rule $P_\phi\colon\mathbb R\to\mathbb R$ as given, the order of every type is optimal,

This is the equilibrium concept used throughout the paper. Kyle exhibits the affine equilibrium $\phi(v)=\tfrac{\sigma}{\tau}(v-\mu)$ and $P_\phi(y)=\mu+\lambda y$ with price impact $\lambda=\tfrac{\tau}{2\sigma}$, the workhorse closed form of market microstructure.

The Kyle equilibrium is the only affine equilibrium of the model, and is obtained through a routine calculation. Whether the model admits any other equilibrium has however remained open. Uniqueness is known in neighbouring models. RV94 prove uniqueness, for arbitrary value distributions, in the variant where the insider also observes the liquidity demand. CE00 show that non-Gaussian value distributions can produce genuinely nonlinear single-auction equilibria, so the affine conclusion is a rigidity property of the Gaussian specification. In continuous time, Back92 establishes uniqueness within a class of pricing rules; see Cetin18 for that literature. CCG22 restore uniqueness among virtually all strategies once trading penalties are introduced. The frictionless one-period Gaussian question itself resisted. A direct attack was undertaken by BKL05,BKL13, who introduced complex-analytic techniques to the problem. MMT17 proved that the expected price is real entire for every measurable strategy and established uniqueness when an equilibrium strategy agrees on a nonempty interval with a uniquely continuable analytic function.

This paper resolves the question completely. We prove the following theorem.

Uniqueness removes equilibrium selection from the interpretation of the model's comparative and empirical implications. In the unique equilibrium, the structural parameters satisfy the population identities

The second is the population coefficient of the price-impact regression studied in the empirical market-microstructure literature GH88,Hasbrouck91,BS96,Amihud02.

The proof is probabilistic and real-variable. It uses machinery from probability and standard techniques from analysis, although the way in which the pieces are combined is specific to the Gaussian rigidity problem addressed here. For conditional expectations, kernels, disintegration, and $L^2$ projection, we use the framework of Kallenberg21. The posterior-resampling construction may be viewed as a reversible exchangeable-pair argument in the in the tradition of Stein's method of exchangeable pairs; see Stein86 and Chatterjee07. The value function and its subdifferential rely on standard convex duality; see Rockafellar70. Finally, the bounded-variation input is one-dimensional: the measure derivative of a monotone function, its absolutely continuous--singular decomposition, and Lebesgue--Stieltjes integration by parts; see AFP00. Every specialised statement needed below is proved in the paper; these references provide background for the techniques.

The paper is organised as follows. Section (ref) states the normalised identity theorem and records the preliminary facts needed from the analysis of MMT17: the canonical Bayes version and its linear growth, smoothness of the Gaussian average, location-scale normalisation, monotonicity, and the first-order condition. To keep the argument self-contained, all of those preliminary facts are proved in Appendix (ref). Section (ref) proves the identity theorem. Its main device is the convex value function of the maximisation, which absorbs every jump of $\phi$ without a separate branch analysis. Section (ref) returns to arbitrary Gaussian location and scale proving the uniqueness theorem stated in the introduction.

The identity theorem

Write \[ \gamma(x)=(2\pi)^{-1/2}e^{-x^2/2}, \qquad x\in\mathbb R, \] for the standard normal density. Let $V$ and $U$ be independent random variables with density $\gamma$. For a Borel map $\phi\colon\mathbb R\to\mathbb R$, set $ Y_\phi=\phi(V)+U\,, $ and define

and

For each $x\in\mathbb R$, let

The next lemma collects the real-variable regularity properties needed below. These facts are contained in MMT17, either explicitly or as immediate consequences of their results.

By Lemma (ref), \[ P_\phi(Y_\phi) = \mathbb E[V\mid Y_\phi] \qquad\text{almost surely}. \] Moreover, for every $x\in\mathbb R$, \[ F_\phi(x) = \mathbb E\bigl[P_\phi(x+U)\bigr]. \] Consequently, \[ F_\phi\bigl(\phi(V)\bigr) = \mathbb E\bigl[P_\phi(Y_\phi)\mid V\bigr] = \mathbb E\bigl[\mathbb E[V\mid Y_\phi]\mid V\bigr] \qquad\text{almost surely}. \]

We now state the identity theorem.

Throughout the proof of Theorem (ref), a Borel map $\phi\colon \mathbb R\to\mathbb R$ satisfying its maximisation condition is called a solution. Define the smooth marginal response

The next lemma records consequences of optimality that are contained in MMT17.

The final lemma of the section records the location-scale normalisation.

We conclude the section by verifying the identity solution and proving the backward direction of the identity theorem. This is the familiar Gaussian calculation underlying the affine equilibrium of Kyle85.

Proof of the identity theorem: forward direction

For the forward direction, fix a solution $\phi$ and write \[ X=\phi(V),\qquad \nu=\operatorname{Law}(X),\qquad Y=X+U,\qquad r=F_\phi'. \] Our aim is to prove that $\phi=\operatorname{id}_\mathbb R$.

We begin with a roadmap. The proof is organised around a single nonnegative quantity $\mathcal E$. Convex duality first translates the maximisation problem into identities involving a value function $W$. Posterior resampling then produces two exact balance identities for $\mathcal E$. An $L^2$ projection argument gives the upper bound $\mathcal E\le\tfrac12$, while a signed-root transformation of $W$ gives the matching lower bound $\mathcal E\ge\tfrac12$. Equality is therefore forced at every stage, and the equality cases determine $\phi$ uniquely.

In Section (ref), we introduce the generalised inverse \[ g(x):=\inf\{v\in\mathbb R:\phi(v)\ge x\} \] of the strictly increasing function $\phi$. Since $X=\phi(V)$, we have \[ g(X)=V \qquad\text{almost surely}. \] The first-order condition for the maximisation gives the further identity \[ g(X)=\kappa_\phi(X) \qquad\text{almost surely}. \]

We also introduce \[ H(x):=xF_\phi(x) \] and its Fenchel conjugate \[ W(v):=\sup_{x\in\mathbb R}\{xv-H(x)\}. \] Thus $W(v)$ is the value of the maximisation problem faced at asset value $v$. The strategy $\phi(v)$ is a maximiser, and therefore belongs to the subdifferential of $W$ at $v$. Standard convex duality then gives \[ W'(v)=\phi(v) \qquad\text{for Lebesgue-almost every }v\in\mathbb R; \] see Rockafellar70. Combining the value of the objective at the maximiser with the first-order condition yields \[ W(v)=\phi(v)^2r(\phi(v)) \qquad\text{for every }v\in\mathbb R. \] The advantage of the convex value function is that these identities remain valid even if $\phi$ has jumps. The jumps are encoded by the subdifferential of $W$, so no separate continuity or jump analysis is needed.

In Section (ref), we construct an exchangeable posterior-resampling pair $(X_0,X_1)$. For background on exchangeable pairs and reversible Markov chains, see Stein86,Chatterjee07. Starting from $X_0\sim\nu$, we observe \[ Y=X_0+U. \] Conditional on $Y$, we then draw $X_1$ independently from the conditional law of $X_0$ given $Y$. Thus, conditional on $Y$, the variables $X_0$ and $X_1$ are independent draws from the same posterior distribution. It follows that \[ (X_0,X_1)\overset{d}=(X_1,X_0), \] so the pair is exchangeable.

Let $K$ denote the corresponding Markov operator, \[ (Kf)(x):=\mathbb E[f(X_1)\mid X_0=x]. \] The posterior-mean interpretation of $P_\phi$ gives \[ (Kg)(x)=F_\phi(x) \qquad\text{for $\nu$-almost every }x. \] Since \[ g(X)=\kappa_\phi(X) =F_\phi(X)+Xr(X), \] we obtain \[ (I-K)g(X)=Xr(X) \qquad\text{almost surely}. \]

Define the associated Dirichlet quantity \[ \mathcal E := \frac12\mathbb E\left[ (X_0-X_1)\bigl(g(X_0)-g(X_1)\bigr) \right]. \] Because $g$ is nondecreasing, the integrand is nonnegative and hence $\mathcal E\ge0$. The derivative of the posterior mean gives one representation of this quantity, while reversibility of the pair gives another. After a truncation argument, which avoids imposing any moment condition on $X$, these representations yield the two balance identities \[ \mathbb E[r(X)] = \mathbb E[X^2r(X)] = \mathcal E. \] These identities are the probabilistic core of the proof.

Section (ref) gives the upper bound. Set \[ M:=P_\phi(Y)=\mathbb E[V\mid Y] \] and \[ q:=\mathbb E[M^2]. \] Since conditional expectation is an $L^2$ contraction, \[ 0\le q\le\mathbb E[V^2]=1. \] The tower property gives \[ \mathbb E[VM] = \mathbb E\bigl[\mathbb E[V\mid Y]M\bigr] = \mathbb E[M^2] = q. \] A Gaussian integration-by-parts argument also identifies the coefficient of $M$ in the noise direction: \[ \mathbb E[UM]=\mathcal E. \]

The random variables $V$ and $U$ are orthonormal in $L^2$. Applying Bessel's inequality to $M$ therefore gives \[ \mathbb E[VM]^2+\mathbb E[UM]^2\le\mathbb E[M^2], \] or equivalently, \[ q^2+\mathcal E^2\le q. \] It follows that \[ \mathcal E^2 \le q(1-q) \le\frac14, \] and hence \[ \mathcal E\le\frac12. \]

The key point is that the Dirichlet energy produced by posterior resampling appears as the $L^2$ coefficient of the posterior mean in the noise direction $U$. Since $M$ is automatically square-integrable, the argument does not require any moment assumption on $X=\phi(V)$.

Finally, Section (ref) supplies the matching lower bound and the rigidity statement. Let \[ c:=\sup\{v\in\mathbb R:\phi(v)<0\} \] be the sign-change threshold of $\phi$, and define the signed square root of the value function by \[ h(v):=

\] The monotonicity of $\phi$ and the convexity of $W$ imply that $h$ is nondecreasing. This is the key transformation in the proof.

Indeed, away from the single point $c$, we have \[ h(v)^2=2W(v). \] Since $W'=\phi$ almost everywhere and $W=\phi^2r(\phi)$ everywhere, differentiation gives \[ h'(v)^2 = \frac{W'(v)^2}{2W(v)} = \frac{\phi(v)^2} {2\phi(v)^2r(\phi(v))} = \frac{1}{2r(\phi(v))} \] for almost every $v\in\mathbb R$. Consequently, the two balance identities become \[ \mathbb E[h(V)^2] = 2\mathbb E[X^2r(X)] = 2\mathcal E \] and \[ \mathbb E\left[\frac{1}{h'(V)^2}\right] = 2\mathbb E[r(X)] = 2\mathcal E. \] Thus the same energy controls both the size of $h$ and the reciprocal size of its derivative.

A Gaussian reciprocal-derivative inequality for nondecreasing functions now gives \[ 1 \le \mathbb E[h'(V)]\, \mathbb E\left[\frac{1}{h'(V)}\right] \le \sqrt{ \mathbb E[h(V)^2]\, \mathbb E\left[\frac{1}{h'(V)^2}\right] }. \] Using the two moment identities on the right-hand side yields \[ 1\le2\mathcal E. \] Therefore \[ \mathcal E\ge\frac12. \] Together with the upper bound, this gives \[ \mathcal E=\frac12. \]

Equality must consequently hold throughout the reciprocal-derivative inequality. Its equality case eliminates the singular part of the measure derivative $Dh$, including any jumps, and forces $h$ to be affine: \[ h(v)=av+b \] for some $a>0$ and $b\in\mathbb R$. The two moment identities, together with $\mathcal E=\tfrac12$, then give \[ a^{-2}=1 \qquad\text{and}\qquad a^2+b^2=1. \] Hence $a=1$ and $b=0$, so \[ h=\operatorname{id}_\mathbb R. \]

It follows that \[ W(v)=\frac{v^2}{2} \qquad\text{for every }v\in\mathbb R. \] Since $W'=\phi$ almost everywhere, \[ \phi(v)=v \qquad\text{for Lebesgue-almost every }v. \] The monotonicity of $\phi$ upgrades this almost-everywhere identity to the pointwise identity \[ \phi=\operatorname{id}_\mathbb R. \] The calculation in the backward direction then gives \[ P_\phi(y)=\frac y2, \qquad F_\phi(x)=\frac x2. \] This proves \[ \phi=2P_\phi=2F_\phi=\operatorname{id}_\mathbb R\,. \]

Inverse structure and the value function

Define $H(x):=xF_\phi(x)$, so $H\in C^\infty(\mathbb R)$ and $H'=\kappa_\phi$ by (ref), and let

be the value function of the maximisation, the Fenchel conjugate of $H$.

The convexity of $W$ is the device that absorbs the jumps of $\phi$. Indeed, because $W'=\phi$ almost everywhere and both sides are monotone, $W'_-(s)=\phi(s-)$ and $W'_+(s)=\phi(s+)$ at every $s$. Hence \[ \partial W(s)=[W'_-(s),W'_+(s)] =[\phi(s-),\phi(s+)]. \] No separate continuity, equal-area, or bounded-variation analysis of $\phi$ is required.

Posterior resampling and the balance identities

Because $\phi$ is injective and $V$ has a density, the law $\nu$ is atomless. For every $y\in\mathbb R$, the posterior law of $X$ given $Y=y$ is \[ \pi_y(dx)=\frac{\gamma(y-x)\,\nu(dx)}{p(y)}, \qquad p(y):=\int_{\mathbb R}\gamma(y-t)\,\nu(dt)>0. \] The numerator and the denominator of (ref) are $\mathbb E[V\gamma(y-\phi(V))]$ and $\mathbb E[\gamma(y-\phi(V))]$. Because $X=\phi(V)$ has law $\nu$ and $V=g(X)$ almost surely by (ref), the change-of-variables formula for the push-forward rewrites them as $\int g(x)\gamma(y-x)\,\nu(dx)$ and $p(y)$. Hence

The inverse regression is the posterior mean of the generalised inverse.

Construct a posterior-resampling pair as follows. Let $X_0\sim\nu$, set $Y=X_0+U$, and, conditional on $Y$, let $X_1$ be an independent draw from $\pi_Y$. The joint law of $(X_0,X_1)$ is $\int_{\mathbb R}(\pi_y\otimes\pi_y)\,p(y)\,dy$, which is symmetric: the pair is exchangeable. Let $K$ be its Markov operator, $(Kh)(x)=\mathbb E[h(X_1)\mid X_0=x]$, well posed on $g$ because $g(X_1)$ is distributed as $V\in L^1$. Because $\mathbb E[g(X_1)\mid Y]=P_\phi(Y)$ by (ref),

In particular $F_\phi(X)=Kg(X)\in L^1$, being a conditional expectation of an integrable variable, even though no moment of $X$ is assumed. Combining (ref), (ref), and (ref) gives

Define the nonnegative extended Dirichlet quantity \[ \mathcal E :=\frac12\mathbb E\bigl[(X_0-X_1)(g(X_0)-g(X_1))\bigr]; \] the integrand is nonnegative because $g$ is nondecreasing. Conditioning on $Y$, under which $X_0$ and $X_1$ are independent with common law $\pi_Y$, and using (ref),

Equation (ref) and Tonelli's theorem imply the first balance identity \[ \mathbb E[r(X)] =\int_{\mathbb R}P_\phi'(y)\,p(y)\,dy =\mathcal E. \]

For the second balance, let $a_n(x)=(-n)\vee(x\wedge n)$. Fix $n$. Each of the four expectations $\mathbb E[a_n(X_i)g(X_j)]$, $i,j\in\{0,1\}$, is finite, $a_n$ being bounded and $g(X_j)$ distributed as $V\in L^1$; and exchangeability gives $\mathbb E[a_n(X_0)g(X_1)]=\mathbb E[a_n(X_1)g(X_0)]$. Hence

Hence, the middle step by conditioning on $X_0$ and the last by (ref). Both sides of (ref) have nonnegative integrands that increase pointwise with $n$: on the right, $a_n(x)x=\min(|x|,n)\,|x|$ increases to $x^2$; on the left, for $x>x'$, $a_n(x)-a_n(x')=\int_{x'}^{x}\mathbf 1_{(-n,n)}(t)\,dt$ increases to $x-x'$. Monotone convergence on both sides gives $\mathbb E[X^2r(X)]=\mathcal E$. Thus, as extended nonnegative quantities, \[ \mathbb E[r(X)]=\mathbb E[X^2r(X)]=\mathcal E. \]

The universal upper bound

Let $M=P_\phi(Y)=\mathbb E[V\mid Y]$ and $q=\mathbb E[M^2]$. Conditional expectation is an $L^2$ contraction, so $M\in L^2$ and $q\le\mathbb E[V^2]=1$; note that this cannot be deduced from the growth bound, because no bound on $\mathbb E[Y^2]$ is available.

Let $P_N=(-N)\vee(P_\phi\wedge N)$. Because $P_\phi$ is continuously differentiable and increasing, the set $\{|P_\phi|<N\}$ is an open interval, on whose complement $P_N$ is locally constant; hence $P_N'=P_\phi'\mathbf 1_{\{|P_\phi|<N\}}$ almost everywhere, and $P_N'\uparrow P_\phi'$ pointwise almost everywhere as $N\to\infty$. For fixed $x$, the function $u\mapsto P_N(x+u)$ is bounded, increasing and locally absolutely continuous, so integration by parts with $\gamma'(u)=-u\gamma(u)$ gives \[ \mathbb E[U P_N(x+U)]=\mathbb E[P_N'(x+U)], \] the boundary terms vanishing by boundedness of $P_N$; the left side is finite, hence so is the right, and it is bounded by $N\,\mathbb E|U|$ uniformly in $x$. Taking expectations over $X$, by Fubini on the left and Tonelli on the right, \[ \mathbb E[U P_N(Y)]=\mathbb E[P_N'(Y)]. \] As $N\to\infty$, the left side converges to $\mathbb E[UM]$ by dominated convergence, the dominating function $|U||M|$ being integrable by Cauchy--Schwarz, while the right side increases to (ref) by monotone convergence. Hence \[ \mathcal E=\mathbb E[UM]<\infty. \] Also, by the tower property, $\mathbb E[VM]=\mathbb E[\mathbb E[V\mid Y]\,M]=\mathbb E[M^2]=q$. The centered unit-variance variables $V$ and $U$ are orthonormal in $L^2$, so Bessel's inequality applied to $M$ yields $q^2+\mathcal E^2\le q$. Because $q-q^2\le\tfrac14$, \[ \mathcal E\le\frac12. \] In particular, all quantities in the balance identities are finite. No moment assumption on $X=\phi(V)$ has been imposed.

The reciprocal-derivative inequality and completion

Strict monotonicity and unboundedness give a unique threshold $c:=\sup\{v:\phi(v)<0\}$, with $\phi<0$ strictly to the left of $c$ and $\phi>0$ strictly to the right. The finite convex function $W$ is locally absolutely continuous, so (ref) gives $W(b)-W(a)=\int_a^b\phi$ for all $a<b$; hence $W$ is strictly decreasing on $(-\infty,c)$ and strictly increasing on $(c,\infty)$. Moreover $W>0$ on $\mathbb R\setminus\{c\}$: if $W(v_0)=0$ for some $v_0<c$, strict decrease would give $W(t)<0$ for $v_0<t<c$, contradicting $W\ge0$, and symmetrically to the right. Define

Then $h\in BV_{\mathrm{loc}}(\mathbb R)$ is nondecreasing; on each of $(-\infty,c)$ and $(c,\infty)$ the function $W$ is locally bounded away from zero, so $h$ is locally absolutely continuous there, and the singular part of $Dh$ is at most an upward atom at $c$. For almost every $v\ne c$, by (ref) and (ref),

The value assigned to $h$ at the single point $c$ is immaterial in the following expectations because $V$ has a density. The balance identities, read through (ref) and (ref) with $X=\phi(V)$, give

For a nondecreasing $f\in BV_{\mathrm{loc}}(\mathbb R)$, write $Df=f'\,\mathcal L^1+D^s f$, where $f'$ denotes the Radon--Nikodym density of the absolutely continuous part of $Df$ with respect to Lebesgue measure. By the differentiation theorem for monotone functions, it agrees almost everywhere with the classical pointwise derivative. The measure $D^s f$ is the singular part, including jumps.

The general Gaussian Kyle (1985) model