One-Period Kyle Model Equilibrium

Market Microstructure · Hard · Free problem

In Kyle's (1985) single-period model, a stock has fundamental value $V \sim N(0, \sigma_V^2)$. An informed trader observes $V$ exactly and submits an order $X_I$. A noise trader independently submits $X_N \sim N(0, \sigma_N^2)$. The market maker only sees the total order flow $Y = X_I + X_N$ and sets a linear price $P = \lambda Y$ to satisfy a zero-profit condition: $E[V \mid Y] = P$.

The informed trader knows that their order moves the price, so they solve: $\max_{X_I} \, E[(V - P)X_I \mid V]$

We look for a linear equilibrium: $X_I = \beta V$ for some constant $\beta$.

  1. Find the equilibrium coefficients $\beta$ and $\lambda$ as functions of $\sigma_V$ and $\sigma_N$. Show that the price impact satisfies $\lambda = \sigma_V / (2\sigma_N)$.
  2. Verify that the market maker's zero-profit condition is satisfied in equilibrium.
  3. Compute the informed trader's expected profit $E[(V - P)X_I]$.

Hints

  1. Conjecture a linear equilibrium $X_I = \beta V$ and $P = \lambda Y$. The market maker's rational $\lambda$ comes from regressing $V$ on $Y$ in a bivariate Gaussian -- and the informed trader's optimal $\beta$ comes from a simple quadratic optimization over order size.
  2. For the informed trader, the problem $\max_{X_I} E[(V - \lambda X_I - \lambda X_N) X_I \mid V]$ simplifies to $\max_{X_I} (V - \lambda X_I) X_I$ since $E[X_N \mid V] = 0$. This gives $X_I^{*} = V / (2\lambda)$, i.e. $\beta = 1 / (2\lambda)$.
  3. Substitute $\beta = 1/(2\lambda)$ into the market maker's formula $\lambda = \beta \sigma_V^2 / (\beta^2 \sigma_V^2 + \sigma_N^2)$ and solve for $\lambda$. You get a quadratic that yields $\lambda = \sigma_V / (2\sigma_N)$.

Worked Solution

How to Think About It: This is a fixed-point problem. The informed trader's optimal order depends on $\lambda$ (how aggressively they can trade before moving the price against themselves). The market maker's rational $\lambda$ depends on $\beta$ (how much information is embedded in the order flow). Equilibrium is where these two are mutually consistent. The market maker is doing a signal extraction problem -- $Y = \beta V + X_N$ is a noisy signal of $V$, and they price at the posterior mean. The informed trader, knowing this, chooses $\beta$ to maximize expected profit given that $\lambda$. The clever part: the two equations $\lambda = f(\beta)$ and $\beta = g(\lambda)$ close into a clean solution.

Quick Sanity Check: Before computing, ask: what should $\lambda$ look like? It should increase in $\sigma_V$ (more fundamental uncertainty means more adverse selection risk, so the market maker prices more aggressively) and decrease in $\sigma_N$ (more noise trading provides cover for the informed trader, compressing $\lambda$). The result $\lambda = \sigma_V / (2\sigma_N)$ has exactly this shape.

Approach: Conjecture a linear equilibrium $X_I = \beta V$. Derive $\lambda$ from the market maker's Bayesian update, then derive $\beta$ from the informed trader's optimization. Solve the resulting

\times 2$ system.

Formal Solution:

Step 1: Market maker's pricing rule.

Given the conjecture $X_I = \beta V$, total order flow is: $Y = \beta V + X_N$

Since $V \sim N(0, \sigma_V^2)$ and $X_N \sim N(0, \sigma_N^2)$ are independent, this is a joint Gaussian system. The conditional expectation (the market maker's rational price) is: $P = E[V \mid Y] = \frac{\text{Cov}(V, Y)}{\text{Var}(Y)} \cdot Y$

Compute each piece: $\text{Cov}(V, Y) = \text{Cov}(V, \beta V + X_N) = \beta \sigma_V^2$ $\text{Var}(Y) = \beta^2 \sigma_V^2 + \sigma_N^2$

So the equilibrium $\lambda$ is: $\lambda = \frac{\beta \sigma_V^2}{\beta^2 \sigma_V^2 + \sigma_N^2} \quad \text{...(1)}$

Step 2: Informed trader's optimization.

Given $P = \lambda Y = \lambda(\beta V + X_N)$ and $X_I = \beta V$, the informed trader's expected profit conditional on $V$ is: $E[(V - P)X_I \mid V] = E[(V - \lambda(\beta V + X_N)) \cdot \beta V \mid V]$ $= \beta V \cdot E[V - \lambda \beta V - \lambda X_N \mid V]$ $= \beta V \cdot (V - \lambda \beta V) \quad \text{(since } E[X_N \mid V] = 0\text{)}$ $= \beta (1 - \lambda \beta) V^2$

But actually, the informed trader optimizes over the choice of $X_I$ directly (not over $\beta$). Given $P = \lambda(X_I + X_N)$, their problem for fixed $V$ is: $\max_{X_I} \, E[(V - \lambda X_I - \lambda X_N) X_I \mid V] = \max_{X_I} \, (V - \lambda X_I) X_I$

(Cross term $E[\lambda X_N X_I] = 0$ since $X_I$ is chosen before $X_N$ is realized -- actually $X_N$ is independent of $V$, so $E[\lambda X_N X_I \mid V] = \lambda X_I \cdot E[X_N] = 0$.) This is a simple quadratic in $X_I$: $\max_{X_I} \, V \cdot X_I - \lambda X_I^2$

First-order condition: $V - 2\lambda X_I = 0 \implies X_I^{*} = \frac{V}{2\lambda}$

So in equilibrium $X_I^{*} = \beta V$ with: $\beta = \frac{1}{2\lambda} \quad \text{...(2)}$

Step 3: Solve the system.

Substitute (2) into (1): $\lambda = \frac{\frac{1}{2\lambda} \sigma_V^2}{\frac{1}{4\lambda^2} \sigma_V^2 + \sigma_N^2}$

Multiply numerator and denominator by $4\lambda^2$: $\lambda = \frac{2\lambda \sigma_V^2}{\sigma_V^2 + 4\lambda^2 \sigma_N^2}$

Divide both sides by $\lambda$ (assuming $\lambda \neq 0$): $1 = \frac{2\sigma_V^2}{\sigma_V^2 + 4\lambda^2 \sigma_N^2}$ $\sigma_V^2 + 4\lambda^2 \sigma_N^2 = 2\sigma_V^2$ $4\lambda^2 \sigma_N^2 = \sigma_V^2$ $\boxed{\lambda = \frac{\sigma_V}{2\sigma_N}}$

Back-substituting into (2): $\boxed{\beta = \frac{1}{2\lambda} = \frac{\sigma_N}{\sigma_V}}$

Step 4: Verify the zero-profit condition.

With $\beta = \sigma_N / \sigma_V$ and $\lambda = \sigma_V / (2\sigma_N)$: $\text{Cov}(V, Y) = \beta \sigma_V^2 = \frac{\sigma_N}{\sigma_V} \cdot \sigma_V^2 = \sigma_N \sigma_V$ $\text{Var}(Y) = \beta^2 \sigma_V^2 + \sigma_N^2 = \frac{\sigma_N^2}{\sigma_V^2} \cdot \sigma_V^2 + \sigma_N^2 = 2\sigma_N^2$ $\frac{\text{Cov}(V,Y)}{\text{Var}(Y)} = \frac{\sigma_N \sigma_V}{2\sigma_N^2} = \frac{\sigma_V}{2\sigma_N} = \lambda \checkmark$

The zero-profit condition is satisfied: $E[V \mid Y] = \lambda Y = P$.

Step 5: Informed trader's expected profit.

The profit per trade is $(V - P) X_I = (V - \lambda Y) \beta V$. Taking expectations: $E[(V - P)X_I] = \beta \cdot E[(V - \lambda(\beta V + X_N)) V]$ $= \beta \cdot E[(1 - \lambda \beta) V^2 - \lambda X_N V]$ $= \beta (1 - \lambda \beta) \sigma_V^2 \quad \text{(since } X_N \perp V\text{)}$

Substitute $\beta = \sigma_N / \sigma_V$ and $\lambda \beta = (\sigma_V / 2\sigma_N)(\sigma_N / \sigma_V) = 1/2$: $E[(V - P)X_I] = \frac{\sigma_N}{\sigma_V} \cdot \left(1 - \frac{1}{2}\right) \cdot \sigma_V^2 = \frac{\sigma_N}{\sigma_V} \cdot \frac{\sigma_V^2}{2} = \frac{\sigma_N \sigma_V}{2}$

$\boxed{E[\text{profit}] = \frac{\sigma_N \sigma_V}{2}}$

Answer: - Price impact: $\lambda = \dfrac{\sigma_V}{2\sigma_N}$ - Informed order size: $\beta = \dfrac{\sigma_N}{\sigma_V}$, so $X_I^{*} = \dfrac{\sigma_N}{\sigma_V} V$ - Informed trader's expected profit: $\dfrac{\sigma_V \sigma_N}{2}$ - In equilibrium, exactly half the fundamental value $V$ is revealed through prices: $P = \lambda Y = V/2$ in expectation (the regression coefficient of $P$ on $V$ is $\lambda \beta = 1/2$).

Intuition

The Kyle model captures the central tension in any informed trading environment: the informed trader wants to trade as aggressively as possible to exploit their edge, but doing so reveals information through the order flow and moves the price against them. In equilibrium, the informed trader holds back -- they trade at scale $\beta = \sigma_N / \sigma_V$, camouflaging their signal behind noise trader volume. The price impact $\lambda = \sigma_V / (2\sigma_N)$ has a beautiful interpretation: when noise trading is thick (large $\sigma_N$), any given order is harder to read as informed, so $\lambda$ is small and the market is liquid. When fundamental uncertainty is high (large $\sigma_V$), the information asymmetry is more severe and the market maker demands a higher price impact to break even.

The result that exactly half the fundamental value is impounded into price ($\lambda \beta = 1/2$) is not a coincidence -- it is a consequence of the monopolist informed trader optimally balancing trade size against market impact, in exactly the same way a monopolist sets quantity at half the competitive level. And the expected profit $\sigma_V \sigma_N / 2$ tells you something practically important: the informed trader's profit scales with both the signal strength and the noise cover. In real markets, this predicts that informed traders will prefer to operate in liquid, noisy markets (large $\sigma_N$) where they can extract more profit while staying hidden. This is why informed order flow is often found in high-volume products -- the noise provides camouflage.

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