# Kyle Model and Kyle's Lambda: Insider Price Impact

An insider hiding among noise traders, a market maker who prices from total order flow, the price-impact coefficient λ, and why the insider trades gradually.

In "The Glosten–Milgrom model" an informed trader arrives, buys or sells one unit, and leaves; the market maker learns from the direction of each trade. "Adverse selection and informed trading" measured what such customers cost a maker. This lesson gives the informed trader a choice: how much to trade. A single insider knows the asset's value, hides its orders among noise traders, and the market maker prices the total order flow it sees. The model, due to Albert Kyle (1985), produces the price impact coefficient $\lambda$, the most widely used number in market microstructure.

## One auction

There is one asset and one trading round, called an **auction**. Its value $V$ becomes public after the auction. Before trading, everyone agrees that $V$ is normal with mean $p_0$ and variance $\sigma_v^2$:

$$
V \sim N(p_0,\ \sigma_v^2).
$$

Three kinds of participant meet in the auction.

- The **insider** learns $V$ exactly and submits a market order of $x$ shares (positive to buy, negative to sell).
- **Noise traders** submit a total order $u \sim N(0, \sigma_u^2)$ for reasons unrelated to $V$, independent of it. $\sigma_u$ is measured in shares.
- A competitive **market maker** sees only the sum $y = x + u$, the **order flow**, and fills it at one price $p$.

Competition between makers drives their expected profit to zero, so the price equals the expected value given what the maker has seen:

$$
p = E[V \mid y].
$$

This is the conditional expectation of "Conditional expectation and martingales". The maker cannot separate $x$ from $u$: a large buy flow might be an insider who knows the value is high, or just a large noise order.

## The linear equilibrium

An equilibrium is a trading rule for the insider and a pricing rule for the maker such that each is the best response to the other. Guess that both are linear:

$$
x = \beta\,(V - p_0), \qquad p = p_0 + \lambda\,y.
$$

**The insider's best response.** Take the pricing rule as given. With $p = p_0 + \lambda(x + u)$ and $E[u] = 0$, the insider's expected profit for a given $V$ is

$$
E[(V - p)\,x \mid V] = \bigl(V - p_0 - \lambda x\bigr)\,x.
$$

This is a downward parabola in $x$. Setting the derivative $V - p_0 - 2\lambda x$ to zero gives

$$
x = \frac{V - p_0}{2\lambda}, \qquad\text{so}\qquad \beta = \frac{1}{2\lambda}.
$$

Trading more pushes the price against the insider on every share, so it stops halfway: its own order moves the expected price by $\lambda x = (V - p_0)/2$.

**The maker's best response.** Take the trading rule as given. Then $y = \beta(V - p_0) + u$ and $V$ are jointly normal, and for jointly normal variables the conditional expectation is the linear projection:

$$
E[V \mid y] = p_0 + \frac{\operatorname{Cov}(V, y)}{\operatorname{Var}(y)}\,y, \qquad \operatorname{Cov}(V, y) = \beta\sigma_v^2,\quad \operatorname{Var}(y) = \beta^2\sigma_v^2 + \sigma_u^2.
$$

So the maker's rule is linear, as guessed, with

$$
\lambda = \frac{\beta\,\sigma_v^2}{\beta^2\sigma_v^2 + \sigma_u^2}.
$$

**Solving.** Substitute $\beta = 1/(2\lambda)$, that is $\lambda = 1/(2\beta)$: $\ 1/(2\beta) = \beta\sigma_v^2/(\beta^2\sigma_v^2 + \sigma_u^2)$. Cross-multiplying, $\beta^2\sigma_v^2 + \sigma_u^2 = 2\beta^2\sigma_v^2$, hence $\beta^2\sigma_v^2 = \sigma_u^2$ and

$$
\beta = \frac{\sigma_u}{\sigma_v}, \qquad \lambda = \frac{\sigma_v}{2\sigma_u}.
$$

$\lambda$ is in dollars per share: the price move per share of net order flow. It is large when private information is large ($\sigma_v$) and small when there is much noise to hide in ($\sigma_u$).

## Properties of the equilibrium

**Half the information enters the price.** The variance of the price is $\lambda^2\operatorname{Var}(y) = \lambda^2 \cdot 2\sigma_u^2 = \sigma_v^2/2$, and the remaining uncertainty about $V$ after the auction is $\operatorname{Var}(V \mid y) = \sigma_v^2 - \sigma_v^2/2 = \sigma_v^2/2$. The auction reveals half of the insider's information.

**The insider's profit grows with noise.** Its expected profit given $V$ is $(V - p_0)^2/(4\lambda)$. Averaging over $V$, with $E[(V - p_0)^2] = \sigma_v^2$:

$$
E[\pi_I] = \frac{\sigma_v^2}{4\lambda} = \frac{\sigma_u\,\sigma_v}{2}.
$$

More noise does not make the insider trade less carefully; it lowers $\lambda$, so the insider trades more ($\beta$ rises) at the same price impact per unit of information.

**Noise traders pay for it.** The maker breaks even, so the insider's gain is the noise traders' loss. Their expected profit is $E[(V - p)u] = -\lambda\sigma_u^2 = -\sigma_u\sigma_v/2$: they pay $\lambda$ per share on a position of typical size $\sigma_u$.

### Worked example: One auction with σᵥ = 4 and σᵤ = 1,000

1. Take $p_0 = 100$, $\sigma_v = 4$ USD, $\sigma_u = 1{,}000$ shares. The insider's trading intensity, in shares per USD of mispricing, is:
   $$\beta = \sigma_u/\sigma_v = 1{,}000/4$$ → **250**
2. The insider learns $V = 104$ and buys:
   $$x = \beta\,(V - p_0) = 250 \cdot 4$$ → **1000**
3. The maker's price impact, in USD per share:
   $$\lambda = \sigma_v/(2\sigma_u) = 4/2{,}000$$ → **0.002**
4. Noise averages zero, so the expected price closes half the gap:
   $$E[p] = 100 + 0.002 \cdot 1{,}000$$ → **102**
5. The insider's expected profit given this $V$, in USD:
   $$(104 - 102) \cdot 1{,}000$$ → **2000**
6. Averaged over all values of $V$, the expected profit is:
   $$\sigma_u\sigma_v/2 = 1{,}000 \cdot 4/2$$ → **2000**

## Many auctions

Now let the insider trade over $N$ auctions, with fresh independent noise in each, and let the maker use one constant $\lambda$ throughout (the simplification the lab uses). Write $g = V - p$ for the remaining gap between value and price, and $m$ for the number of auctions left, this one included.

With one auction left the problem is the one above: the insider maximises $(g - \lambda x)x$, buys $x = g/(2\lambda)$ and earns $g^2/(4\lambda)$. Suppose that with $m - 1$ auctions left the best expected profit is $c_{m-1}\,g^2/\lambda$ plus a term that does not depend on $g$, with $c_1 = 1/4$. With $m$ left, an order $x$ earns $(g - \lambda x)x$ now and leaves the gap $g - \lambda x - \lambda u$. Noise has mean zero, so it only adds a constant, and the insider maximises

$$
(g - \lambda x)\,x + c_{m-1}\,\frac{(g - \lambda x)^2}{\lambda}.
$$

The derivative is $g - 2\lambda x - 2c_{m-1}(g - \lambda x) = 0$, so $\lambda x = g\,(1 - 2c_{m-1})/(2 - 2c_{m-1})$. With $c_{m-1} = (m-1)/(2m)$ this is $g/(m+1)$, and substituting back gives $c_m = m/(2(m+1))$, which keeps the guess true by induction. With one constant $\lambda$:

$$
\lambda\,x_m = \frac{V - p}{m + 1}.
$$

Each order closes $1/(m+1)$ of the remaining gap: with ten auctions left, $1/11$ of it; with two left, a third; in the last auction, a half. Trading gradually beats trading at once: by spreading the position the insider pays a smaller impact on each share, and the noise in between keeps the maker from inferring too much. Kyle's 1985 equilibrium goes further and lets $\lambda$ change from auction to auction as the maker learns; the insider then trades so that information enters the price at a steady rate. That version is not derived here.

The figure plays ten auctions with $\sigma_v = 4$, noise of 1,000 shares per auction and $V - p_0 = +4$, so the break-even constant $\lambda$ is 0.00106 USD per share. The default plan buys 1,900 shares in the first auction, about $(V - p_0)/(2\lambda)$, and nothing afterwards. On the fixed noise it earns 4,186 USD and closes 47% of the gap, while the optimal insider earns 6,403 USD on the same noise. The light bars show the optimal order given your plan; the first is 343 shares, $4/(11\lambda)$. Drag the bars towards them and watch your profit approach the optimum. Switch to the market maker's view: the maker sees only the total order flow, and with noise of 1,000 shares per auction an order of a few hundred shares is hard to spot.

*[Interactive figure: kyle schedule](https://quantsandbox.dev/guide/microstructure/kyle-model)*

## Estimating lambda

In data, $\lambda$ is estimated by regressing price changes on signed order flow over short intervals:

$$
\Delta p_t = \alpha + \lambda\,y_t + \varepsilon_t,
$$

where $y_t$ is buyer-initiated minus seller-initiated volume in interval $t$. Trades are signed by comparing their price with the mid price, as in "Adverse selection and informed trading". The slope is in price per share. To compare stocks it is often rescaled: price change in basis points per 1% of daily volume, $\lambda \cdot 0.01\,\text{ADV}/p \cdot 10^4$.

> The regression captures all reasons why prices follow order flow, including inventory effects; Kyle's model attributes the whole slope to information.

## Summary

- An insider who knows $V$ trades $x = \beta(V - p_0)$; the maker sees only $y = x + u$ and sets $p = p_0 + \lambda y$.
- In the one-auction equilibrium $\beta = \sigma_u/\sigma_v$ and $\lambda = \sigma_v/(2\sigma_u)$; half the private information enters the price.
- The insider expects $\sigma_u\sigma_v/2$, exactly what the noise traders lose; more noise means more insider profit.
- With one constant $\lambda$ over many auctions, the best order closes $1/(m+1)$ of the remaining gap, so trading gradually beats trading at once.
- $\lambda$ is estimated by regressing price changes on signed order flow, in price per share or basis points per 1% of volume.

[Open the Kyle model lab](https://quantsandbox.dev/kyle)

## Interactive labs

- [Kyle Model: Insider Trading and Price Impact](https://quantsandbox.dev/kyle)

## Practice questions

### 1. In a one-auction Kyle market $\sigma_v = 2$ USD and noise trading has standard deviation $\sigma_u = 500$ shares. What is $\lambda$ in USD per share?

**Answer:** 0.002

$\lambda = \sigma_v/(2\sigma_u) = 2/(2 \cdot 500) = 0.002$ USD per share.

### 2. The standard deviation of noise trading doubles and nothing else changes. The insider's expected profit…

- a) doubles
- b) halves
- c) is unchanged
- d) falls to zero

**Answer:** a) doubles

$E[\pi_I] = \sigma_u\sigma_v/2$ is linear in $\sigma_u$. $\lambda$ halves, the insider's order doubles, and the price still moves halfway to $V$.

### 3. With one constant $\lambda = 0.002$ USD per share, three auctions left and a gap $V - p = 6$ USD, how many shares does the insider buy in this auction?

**Answer:** 750

With $m = 3$ the order closes $1/(m+1) = 1/4$ of the gap: $\lambda x = 6/4 = 1.5$, so $x = 1.5/0.002 = 750$ shares.

---

Source: https://quantsandbox.dev/guide/microstructure/kyle-model

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