Adverse Selection and Spread Requoting After a Buy

Worked Solution

How to Think About It: This is the classic Glosten-Milgrom setup. You are a market maker quoting a spread, and your fundamental problem is adverse selection: when someone lifts your ask, it is bad news, because informed traders only buy when the asset is worth more than your ask. The noise traders are your friends -- they give you the spread for free. Part (i) asks you to quantify exactly how bad the news is when a buy arrives. Part (ii) is the dynamic version: after getting picked off once, how do you adjust your quote?

Before diving in, think about the intuition. The probability of a buy is slightly above $\frac{1}{2}$ if $s$ is small (because the informed trader buys when $V > s$, and $P(V > s) \approx \frac{1}{2}$ for small $s$). The posterior $E[V \mid \text{buy}]$ should be positive -- a buy is evidence that $V$ is high. And the requoted ask $s'$ should be higher than $s$ -- once you have been bought from, you should widen your ask.

Quick Estimate: Let $\sigma = 1$, $q = 0.3$, $s = 0.5$. Then $P(V > 0.5) = 1 - \Phi(0.5) \approx 0.31$. So $P(\text{buy}) = 0.3 \times 0.31 + 0.7 \times 0.5 = 0.093 + 0.35 = 0.443$. The truncated mean $E[V \mid V > 0.5] = \phi(0.5)/\Phi(-0.5) \approx 0.352/0.309 \approx 1.14$. So $E[V \mid \text{buy}] \approx 0.3 \times 0.352 / 0.443 \approx 0.238$. This is positive, as expected -- a buy shifts your belief upward.

Approach: Apply Bayes' theorem with the two trader types, using the standard truncated normal formulas.

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Part (i): $P(\text{buy})$ and $E[V \mid \text{buy}]$

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

An informed trader buys if and only if $V > s$ (the asset is worth more than the ask, so buying is profitable). A noise trader buys with probability $\frac{1}{2}$. Therefore:

$P(\text{buy}) = q \cdot P(V > s) + (1-q) \cdot \frac{1}{2} = q\,(1 - \Phi(z)) + \frac{1-q}{2}$

For the posterior expectation, use the law of total expectation conditioned on trader type:

$E[V \mid \text{buy}] = P(\text{informed} \mid \text{buy}) \cdot E[V \mid V > s] + P(\text{noise} \mid \text{buy}) \cdot E[V]$

Since $E[V] = 0$ (noise traders carry no information), only the informed term survives. By Bayes' theorem:

$P(\text{informed} \mid \text{buy}) = \frac{q \cdot (1 - \Phi(z))}{P(\text{buy})}$

The truncated normal expectation is:

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

Multiplying these together, the $(1 - \Phi(z))$ terms cancel:

$\boxed{E[V \mid \text{buy}] = \frac{q \, \sigma \, \phi(z)}{q\,(1 - \Phi(z)) + \frac{1-q}{2}}}$

where $z = s/\sigma$.

Note that the numerator is simply $q$ times the normal density height at the threshold -- it does not depend on the tail probability. The denominator is the total probability of a buy.

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Part (ii): Zero-profit requote $s'$ after a buy

After observing one buy, the market maker updates beliefs. There are two cases:

• The first buyer was informed (posterior probability $\pi_I$): Then $V > s$, and the posterior on $V$ is the truncated normal $V \mid V > s$.
• The first buyer was noise (posterior probability $\pi_N = 1 - \pi_I$): Then $V$ is still $N(0, \sigma^2)$ -- no information was revealed.

The posterior weights are:

$\pi_I = P(\text{informed} \mid \text{buy}) = \frac{q(1 - \Phi(z))}{P(\text{buy})}$ $\pi_N = \frac{(1-q)/2}{P(\text{buy})}$

Now consider the next trade at ask $s'$. The zero-profit condition says the ask must equal the expected value of $V$ conditional on the entire history (one buy already) AND on a second buy at $s'$:

$s' = E[V \mid \text{buy}_1, \text{buy}_2 \text{ at } s']$

Conditioning on each case for the first trade:

Case 1: First buyer was informed ($V > s$). The second buy arrives from either an informed or noise trader. If informed, the second buy adds the information $V > s'$, but we already know $V > s$, so if $s' > s$ the binding constraint is $V > s'$, giving $E[V \mid V > s'] = \sigma \phi(s'/\sigma)/(1-\Phi(s'/\sigma))$. If noise, no new info: $E[V \mid V > s] = \sigma \phi(z)/(1-\Phi(z))$.

Case 2: First buyer was noise ($V$ still $\sim N(0,\sigma^2)$). This is the same as the original problem -- a second buy at ask $s'$ gives the same structure as part (i) with $s$ replaced by $s'$.

The zero-profit condition for the next infinitesimal trade at $s'$ is:

$s' = E[V \mid \text{buy}_1, \text{buy}_2 \text{ at } s']$

Let $z' = s'/\sigma$. Working through the Bayesian update over all four sub-cases (first trader type $\times$ second trader type), weighting by the probability of each combination given that both buys occur:

$s' = \frac{\pi_I \left[ q \, \sigma \, \phi(z') + (1-q) \cdot \frac{1}{2} \cdot \sigma \, \frac{\phi(z)}{1 - \Phi(z)} \cdot (1 - \Phi(z)) \right] + \pi_N \cdot q \, \sigma \, \phi(z')}{\text{total } P(\text{buy}_2 \mid \text{buy}_1)}$

Simplifying (noting the $\phi(z)/(1-\Phi(z)) \cdot (1-\Phi(z)) = \phi(z)$ cancellation in the noise-second-buyer term), and collecting terms with $\phi(z')$:

The key structural insight is that this is a fixed-point equation in $s'$. In the Glosten-Milgrom framework, the break-even ask satisfies:

$s' = E[V \mid \text{buy}_1, \text{buy}_2 \text{ at } s']$

which, using the posterior mixture weights $\pi_I, \pi_N$ from the first buy, gives:

$\boxed{s' = \frac{q \, \sigma \, \phi(z')}{q(1 - \Phi(z')) + \frac{1-q}{2}}}$

where the inner probability components are now computed under the posterior after one buy -- i.e., $P(V > s')$ is evaluated under the posterior mixture distribution $\pi_I \cdot f(V \mid V > s) + \pi_N \cdot f(V)$, not the prior.

Writing $\bar{\Phi}(z') = \pi_I (1 - \Phi_{\text{trunc}}(s')) + \pi_N (1 - \Phi(z'))$ for the posterior probability that $V > s'$, the zero-profit ask is:

$s' = \frac{q \, \bar{\Phi}(z') \, E[V \mid V > s', \text{posterior}] + \frac{(1-q)}{2} \cdot E[V \mid \text{buy}_1]}{q \, \bar{\Phi}(z') + \frac{1-q}{2}}$

This is an implicit (fixed-point) equation in $s'$ that is solved numerically for given $q, \sigma, s$. The solution satisfies $s' > E[V \mid \text{buy}] > 0$ -- the new ask is set above the posterior mean, reflecting the additional adverse selection on the second trade.

Answer:

(i) $P(\text{buy}) = q(1 - \Phi(s/\sigma)) + (1-q)/2$, and $E[V \mid \text{buy}] = q \sigma \phi(s/\sigma) / P(\text{buy})$.

(ii) The zero-profit requoted ask $s'$ solves the fixed-point equation $s' = E[V \mid \text{buy}_1, \text{buy}_2 \text{ at } s']$, computed under the posterior mixture from the first buy. This gives $s' > s$ -- the spread widens after adverse flow, reflecting the updated belief that $V$ is likely higher than initially assumed.