Adverse Selection with Gaussian Fundamentals

Worked Solution

How to Think About It: This is the Glosten-Milgrom adverse selection model with continuous values. The market maker faces a classic problem: noise traders are profitable (they buy and sell randomly, paying the spread), but informed traders are toxic (they buy only when the asset is worth more than the ask, so the market maker systematically sells cheap and buys dear). The break-even spread is the one where noise trader profits exactly offset informed trader losses. The wider the spread, the more you make on noise flow and the less you lose on informed flow (because informed traders trade less often when the spread is wide).

Key Insight: The market maker's expected loss on an informed trade is the expected distance between $V$ and the quote, conditional on the informed trader choosing to trade. This is a conditional expectation involving truncated Gaussian distributions.

The Method:

Let $\Phi$ and $\phi$ denote the standard normal CDF and PDF. Define $z = s/\sigma$.

Step 1: Informed trader behavior.

The informed trader buys when $V \geq a = \mu + s$, which happens with probability $P(V \geq \mu + s) = 1 - \Phi(z)$. She sells when $V \leq b = \mu - s$, with the same probability by symmetry.

Step 2: Market maker's loss on informed buy.

When the informed trader buys at the ask $a = \mu + s$, the market maker sells at $\mu + s$ but the asset is worth $V \geq \mu + s$. The expected loss is:

$E[V - (\mu + s) \mid V \geq \mu + s] \cdot P(V \geq \mu + s)$

Using the truncated normal formula:

$E[V \mid V \geq \mu + s] = \mu + \sigma \cdot \frac{\phi(z)}{1 - \Phi(z)}$

So the expected loss per informed buy (unconditional on whether a buy happens) is:

$E[(V - \mu - s) \cdot \mathbf{1}_{V \geq \mu + s}] = \sigma \phi(z) - s(1 - \Phi(z))$

By symmetry, the expected loss per informed sell is the same.

Step 3: Market maker's profit on noise trade.

A noise trader buys or sells with equal probability, paying the half-spread $s$ each time. The market maker earns $s$ per noise trade.

Step 4: Expected profit per trade.

Note that an informed trader only trades if $V \geq \mu + s$ (buys) or $V \leq \mu - s$ (sells), and does nothing if $\mu - s < V < \mu + s$. So conditioned on an informed trader arriving, there may be no trade. On each arrival:

• With probability $q$: noise trader, always trades, market maker earns $s$.
• With probability
  - q$: informed trader. Trades only if $V \geq \mu + s$ (buys) or $V \leq \mu - s$ (sells). Otherwise no trade.

Expected profit per arrival:

$\Pi = q \cdot s - (1-q) \cdot 2[\sigma\phi(z) - s(1 - \Phi(z))]$

The factor of 2 accounts for both buy and sell sides. The informed loss term $\sigma\phi(z) - s(1 - \Phi(z))$ represents the expected loss on each side (buy or sell).

Simplifying:

$\Pi = q \cdot s - 2(1-q)[\sigma\phi(z) - s(1-\Phi(z))]$

$= s[q + 2(1-q)(1-\Phi(z))] - 2(1-q)\sigma\phi(z)$

Step 5: Break-even condition $\Pi = 0$:

$s^{*}[q + 2(1-q)(1-\Phi(z^{*}))] = 2(1-q)\sigma\phi(z^{*})$

where $z^{*} = s^{*}/\sigma$. This is a nonlinear equation in $z^{*}$ that must be solved numerically in general.

Step 6: Comparative statics.

• More informed traders (higher
  - q$): The right side grows and the $q$ term on the left shrinks, so $s^{*}$ increases. More adverse selection requires wider spreads.
• Higher volatility ($\sigma$): The informed trader's informational advantage is proportional to $\sigma$, so $s^{*}$ increases roughly linearly in $\sigma$. In fact $z^{*} = s^{*}/\sigma$ is determined purely by $q$, so $s^{*} = z^{*}(q) \cdot \sigma$ -- the break-even spread scales linearly with volatility.
• Limiting cases: If $q = 1$ (all noise), $s^{*} = 0$ -- no adverse selection, zero spread. If $q = 0$ (all informed), there is no finite spread that breaks even -- the market shuts down (no noise flow to offset losses).

Practical Interpretation: This is why market makers widen spreads around earnings announcements (higher $\sigma$) and in illiquid names with high insider-trading risk (lower $q$). The break-even spread is the minimum viable bid-ask spread for a competitive market maker.

Answer: The break-even half-spread satisfies $s^{*} = z^{*} \cdot \sigma$ where $z^{*}$ solves $z[q + 2(1-q)(1 - \Phi(z))] = 2(1-q)\phi(z)$. The spread scales linearly with volatility $\sigma$ and increases with the fraction of informed traders