Adverse Selection Quotes With Three-Point Asset Values

Market Microstructure · Hard · Free problem

An asset has a fundamental value $V \in \{0, 50, 100\}$ with prior probabilities $P(V = 100) = p_2$, $P(V = 50) = p_1$, and $P(V = 0) = 1 - p_1 - p_2$.

You are a market maker quoting a bid $b$ and an ask $a$. Your counterparty is one of two types:

  • With probability
    - q$, the counterparty is perfectly informed: they buy at your ask $a$ if and only if $V > a$, and sell at your bid $b$ if and only if $V < b$.
  • With probability $q$, the counterparty is noise: they buy or sell each with probability $\frac{1}{2}$, regardless of $V$.
  1. Derive the conditional posteriors $P(V = v \mid \text{Buy at } a)$ and $P(V = v \mid \text{Sell at } b)$ for each $v \in \{0, 50, 100\}$.
  1. Impose a zero-expected-profit condition given a fill on each side, and solve for the tightest feasible bid $b$ and ask $a$. State any inequalities on $a$ and $b$ that bind.

Hints

  1. Think about what a buy signal tells you about the counterparty's type and hence about $V$. The informed trader's decision to buy reveals that $V$ exceeds the ask.
  2. Apply Bayes' rule with the likelihood $P(\text{Buy} \mid V = v)$, which mixes the informed trader's deterministic behavior with the noise trader's coin flip. Group terms using $p_0 + p_1 + p_2 = 1$.
  3. Set $E[V \mid \text{Buy}] = a$ and $E[V \mid \text{Sell}] = b$ using your posteriors, then solve the resulting linear equations for $a$ and $b$ in the regime $0 < b < 50 < a < 100$.

Worked Solution

How to Think About It: This is the Glosten-Milgrom adverse selection model with a three-point distribution instead of the usual two-point. The key economic tension: when someone buys from you (lifts your ask), it is bad news -- the informed trader only buys when the asset is worth more than your ask. The noise trader provides cover, but the wider the spread, the less business you do. The zero-profit condition finds the tightest quotes where you break even on each fill, balancing the losses to informed traders against the gains from noise traders.

Before diving in, note that the structure of the problem depends on where $a$ and $b$ sit relative to the values $\{0, 50, 100\}$. The natural case is $0 < b < 50 < a < 100$, which gives the tightest feasible quotes.

Quick Estimate: Consider the symmetric case $p_1 = 0$, $p_2 = 1/2$, so $V$ is equally likely to be $0$ or

00$. Then the model reduces to the classic two-point Glosten-Milgrom. With $0 < b < 50 < a < 100$, the informed trader buys only when $V = 100$ and sells only when $V = 0$. Zero profit on the ask side gives $a = 100 \cdot P(V = 100 \mid \text{Buy})$. The informed trader buys with certainty when $V = 100 > a$, so $P(\text{Buy} \mid V = 100) = (1-q) + q/2$, while $P(\text{Buy} \mid V = 0) = q/2$. With equal priors:

$a = \frac{100 \cdot [(1-q) + q/2]}{[(1-q) + q/2] + [q/2]} = \frac{100(1 - q/2)}{1} = 100(1 - q/2)$

So $a = 100 - 50q$ and by symmetry $b = 50q$. With $q = 0.5$, $a = 75$ and $b = 25$. This makes sense: the spread is $50$ when half the flow is noise, wider than the $0$-to-

00$ midpoint of $50$.

Now let us handle the general three-point case.

Approach: We work in the regime $0 < b < 50 < a < 100$, derive the posteriors using Bayes' rule, then impose $E[V \mid \text{Buy at } a] = a$ and $E[V \mid \text{Sell at } b] = b$.

Formal Solution:

Define $p_0 = 1 - p_1 - p_2$. For $0 < b < 50 < a < 100$:

  • The informed trader ($\text{prob } 1-q$) buys iff $V > a$, i.e., only when $V = 100$.
  • The informed trader sells iff $V < b$, i.e., only when $V = 0$.
  • The noise trader ($\text{prob } q$) buys or sells each with probability
    /2$.

Step 1 -- Buy-side posteriors.

The probability of observing a buy, given $V = v$:

  • $P(\text{Buy} \mid V = 100) = (1 - q) \cdot 1 + q \cdot \frac{1}{2} = 1 - \frac{q}{2}$
  • $P(\text{Buy} \mid V = 50) = (1 - q) \cdot 0 + q \cdot \frac{1}{2} = \frac{q}{2}$
  • $P(\text{Buy} \mid V = 0) = (1 - q) \cdot 0 + q \cdot \frac{1}{2} = \frac{q}{2}$

The total probability of a buy:

$P(\text{Buy}) = p_2\!\left(1 - \frac{q}{2}\right) + p_1 \cdot \frac{q}{2} + p_0 \cdot \frac{q}{2} = p_2(1 - q) + \frac{q}{2}$

where we used $p_0 + p_1 + p_2 = 1$. By Bayes' rule:

$P(V = 100 \mid \text{Buy}) = \frac{p_2\left(1 - \frac{q}{2}\right)}{p_2(1-q) + \frac{q}{2}}$

$P(V = 50 \mid \text{Buy}) = \frac{p_1 \cdot \frac{q}{2}}{p_2(1-q) + \frac{q}{2}}$

$P(V = 0 \mid \text{Buy}) = \frac{p_0 \cdot \frac{q}{2}}{p_2(1-q) + \frac{q}{2}}$

Step 2 -- Sell-side posteriors.

The probability of observing a sell, given $V = v$:

  • $P(\text{Sell} \mid V = 0) = (1 - q) \cdot 1 + q \cdot \frac{1}{2} = 1 - \frac{q}{2}$
  • $P(\text{Sell} \mid V = 50) = (1 - q) \cdot 0 + q \cdot \frac{1}{2} = \frac{q}{2}$
  • $P(\text{Sell} \mid V = 100) = (1 - q) \cdot 0 + q \cdot \frac{1}{2} = \frac{q}{2}$

Total probability of a sell:

$P(\text{Sell}) = p_0(1 - q) + \frac{q}{2}$

By Bayes' rule:

$P(V = 0 \mid \text{Sell}) = \frac{p_0\left(1 - \frac{q}{2}\right)}{p_0(1-q) + \frac{q}{2}}$

$P(V = 50 \mid \text{Sell}) = \frac{p_1 \cdot \frac{q}{2}}{p_0(1-q) + \frac{q}{2}}$

$P(V = 100 \mid \text{Sell}) = \frac{p_2 \cdot \frac{q}{2}}{p_0(1-q) + \frac{q}{2}}$

Step 3 -- Zero-profit conditions.

When you sell at the ask $a$ and the asset is worth $V$, your profit is $a - V$. Zero expected profit given a fill:

$E[V \mid \text{Buy}] = a$

$100 \cdot P(V\!=\!100 \mid \text{Buy}) + 50 \cdot P(V\!=\!50 \mid \text{Buy}) + 0 \cdot P(V\!=\!0 \mid \text{Buy}) = a$

Substituting the posteriors and multiplying through by the denominator $D_a = p_2(1-q) + q/2$:

$100\, p_2\!\left(1 - \frac{q}{2}\right) + 50\, p_1 \cdot \frac{q}{2} = a\left[p_2(1-q) + \frac{q}{2}\right]$

Solving:

$\boxed{a = \frac{100\, p_2\left(1 - \frac{q}{2}\right) + 25\, p_1\, q}{p_2(1-q) + \frac{q}{2}}}$

Similarly, when you buy at the bid $b$ and the asset is worth $V$, your profit is $V - b$. Zero expected profit:

$E[V \mid \text{Sell}] = b$

$0 \cdot P(V\!=\!0 \mid \text{Sell}) + 50 \cdot P(V\!=\!50 \mid \text{Sell}) + 100 \cdot P(V\!=\!100 \mid \text{Sell}) = b$

$50\, p_1 \cdot \frac{q}{2} + 100\, p_2 \cdot \frac{q}{2} = b\left[p_0(1-q) + \frac{q}{2}\right]$

Solving:

$\boxed{b = \frac{25\, p_1\, q + 50\, p_2\, q}{p_0(1-q) + \frac{q}{2}}}$

which simplifies to:

$b = \frac{q(25 p_1 + 50 p_2)}{p_0(1-q) + \frac{q}{2}}$

Step 4 -- Binding inequalities.

For the regime $0 < b < 50 < a < 100$ to hold, we need:

  • $a < 100$: From the ask formula, this requires $q > 0$ (some noise flow). If $q = 0$, only informed traders trade, every buy signals $V = 100$, and no finite ask breaks even.
  • $a > 50$: This requires $p_2(1 - q/2) > (p_2(1-q) + q/2)/2$, which simplifies to the informed flow making buys informative enough to push the conditional mean above $50$.
  • $b > 0$: Requires $p_1 > 0$ or $p_2 > 0$ (at least some probability of mid/high values to give the sell side a non-zero conditional mean).
  • $b < 50$: The symmetric condition on the sell side.

The tightest feasible spread is achieved in this interior regime. If the noise fraction $q$ is too small, the adverse selection is too severe and the market may break down (no quotes satisfy the constraints).

Answer: The zero-profit ask and bid in the regime $0 < b < 50 < a < 100$ are:

$a = \frac{100\, p_2(1 - q/2) + 25\, p_1\, q}{p_2(1-q) + q/2}, \qquad b = \frac{q(25\, p_1 + 50\, p_2)}{p_0(1-q) + q/2}$

where $p_0 = 1 - p_1 - p_2$. The spread $a - b$ is decreasing in $q$: more noise trading tightens the spread. In the limit $q \to 1$ (all noise), $a \to 50 p_1 + 100 p_2 + 25 p_1 = E[V]$ and $b \to E[V]$, collapsing the spread to zero. In the limit $q \to 0$ (all informed), $a \to 100$ and $b \to 0$, the widest possible spread.

Intuition

This is the Glosten-Milgrom model of the bid-ask spread, extended to three values instead of the usual two. The core lesson is that the spread exists because of adverse selection: when someone trades against you, the very fact that they chose to trade is informative. On average, your counterparty knows more than you do, so every fill is bad news. The noise traders are the only reason you can quote at all -- they provide "cover" that dilutes the information content of each trade. The parameter $q$ controls the signal-to-noise ratio of order flow.

In practice, this is exactly why market makers widen spreads when they suspect more informed flow (e.g., ahead of earnings announcements or during unusual volume spikes). The three-point extension matters because real assets do not have binary outcomes -- there is usually a middle state where the informed trader does not trade at all (here, $V = 50$ is between $b$ and $a$, so the informed trader sits out). This "no-trade" region for the informed creates an asymmetry in how buys and sells update your beliefs, and it is the reason the ask and bid formulas are not simply mirror images of each other unless the prior is symmetric around $50$.

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