Posterior and Adverse-Selection Decomposition After a Buy

Worked Solution

How to Think About It: Every time someone lifts your ask, you should be nervous -- the buy itself is bad news about $V$. Informed traders only buy when $V = 1$, so a buy skews the posterior upward. The market maker's problem is that the post-trade fair value $m^{+}$ is higher than the pre-trade mid $p$, which means you just sold something that is now worth more than you thought. The gap $m^{+} - p$ is the adverse-selection cost per buy trade, and it grows with $q$ (more informed flow) and with how "surprising" the signal is.

Quick Estimate: Take $p = 0.5$ and $q = 0.3$. Then $P(\text{Buy} \mid V=1) = \frac{1+0.3}{2} = 0.65$ and $P(\text{Buy} \mid V=0) = \frac{1-0.3}{2} = 0.35$. Posterior: $\alpha = \frac{0.65 \times 0.5}{0.65 \times 0.5 + 0.35 \times 0.5} = \frac{0.325}{0.5} = 0.65$. So the mid moves from 0.50 to 0.65 on a buy -- that 0.15 jump is the adverse-selection cost. You need to set $a \geq 0.65$ just to break even on buys.

Formal Derivation:

Part 1 -- Posterior after a buy.

The buy likelihoods are:

$P(\text{Buy} \mid V=1) = q \cdot 1 + (1-q) \cdot \tfrac{1}{2} = \frac{1+q}{2}$

$P(\text{Buy} \mid V=0) = q \cdot 0 + (1-q) \cdot \tfrac{1}{2} = \frac{1-q}{2}$

The total buy probability is:

$P(\text{Buy}) = \frac{1+q}{2} \cdot p + \frac{1-q}{2} \cdot (1-p) = \frac{1 + q(2p - 1)}{2}$

By Bayes' theorem:

$\alpha = P(V=1 \mid \text{Buy}) = \frac{\frac{1+q}{2} \cdot p}{\frac{1 + q(2p-1)}{2}} = \frac{(1+q)\,p}{1 + q(2p-1)}$

Since $V$ is binary, $m^{+} = E[V \mid \text{Buy}] = \alpha$, so:

$m^{+} = \frac{(1+q)\,p}{1 + q(2p-1)}$

Part 2 -- P&L decomposition.

When you sell at the ask $a$ and the asset is worth $V$, your execution P&L is $a - V$. Insert and subtract $m^{+}$:

$a - V = (a - m^{+}) + (m^{+} - V)$

• Spread capture $(a - m^{+})$: the profit you earn because your ask exceeds the post-trade fair value. This is deterministic once you know $a$ and $m^{+}$.
• Adverse selection $(m^{+} - V)$: the loss from mispricing. Because $V > m^{+}$ whenever $V = 1$ and $m^{+} < 1$, the informed buys systematically generate losses here.

Part 3 -- Expected adverse-selection cost.

Conditional on a buy:

$E[m^{+} - V \mid \text{Buy}] = m^{+} - E[V \mid \text{Buy}] = m^{+} - m^{+} = 0$

This is zero by construction -- the conditional decomposition is exact, and $m^{+}$ is the conditional expectation. So the expected P&L conditional on a buy is just the spread capture:

$E[a - V \mid \text{Buy}] = a - m^{+}$

The more informative quantity is the adverse-selection cost measured as the price impact -- the distance $m^{+}$ moved from the prior mid $m = p$:

$m^{+} - p = \frac{(1+q)\,p}{1 + q(2p-1)} - p = \frac{q\,p(1-p)}{\frac{1}{2}[1 + q(2p-1)]} \cdot \frac{1}{2}$

Simplifying:

$m^{+} - p = \frac{q\,p(1-p)}{1 + q(2p-1)}$

This is the per-buy adverse-selection cost. Key properties:

• Increasing in $q$: more informed traders means a bigger upward revision on buys. At $q = 0$, $m^{+} = p$ (no information, no movement). At $q = 1$, $m^{+} = 1$ (a buy guarantees $V = 1$).
• Maximized at $p = \frac{1}{2}$ for fixed $q$: when the prior is most uncertain, each trade carries the most information. At $p = 0$ or $p = 1$, the asset's value is already known and buys carry no news.
• Zero-profit condition: the market maker breaks even on buys when $a = m^{+}$. Any narrower ask is a subsidy to informed traders.

Answer: The posterior is $m^{+} = \frac{(1+q)\,p}{1 + q(2p-1)}$. Execution P&L decomposes as $(a - m^{+}) + (m^{+} - V)$, where the adverse-selection term has conditional expectation zero. The price impact per buy is $m^{+} - p = \frac{q\,p(1-p)}{1 + q(2p-1)}$, increasing in $q$ and maximized at $p = \tfrac{1}{2}$. The break-even ask is $a = m^{+}$.