Zero-Profit Quotes on a Tick Grid in a Binary Glosten-Milgrom Model
A binary asset has payoff $V \in \{0, 1\}$ with prior $P(V = 1) = p$. A single counterparty arrives and is one of two types:
- Informed (with probability - q$): observes $V$ and buys if $V = 1$, sells if $V = 0$.
- Noise (with probability $q$): buys or sells each with probability
/2$, independent of $V$.You are a market maker who must post a bid $b$ and an ask $a$ on a discrete tick grid of size $\delta \in (0, 1)$, so $b = k_b \, \delta$ and $a = k_a \, \delta$ for integers $0 \leq k_b < k_a \leq 1/\delta$.
- Derive the zero-profit ask $a^{*}$, i.e., the smallest tick-grid ask at which the market maker does not lose money in expectation when a buy order arrives.
- Derive the zero-profit bid $b^{*}$, i.e., the largest tick-grid bid at which the market maker does not lose money in expectation when a sell order arrives.
- Give the resulting spread $s^{*} = a^{*} - b^{*}$ and discuss how it depends on $q$, $p$, and $\delta$.
Hints
- Think about what a buy order tells you about $V$. An informed trader only buys when $V = 1$, so seeing a buy is bad news for the market maker.
- Use Bayes' theorem to compute $E[V \mid \text{buy}]$ and $E[V \mid \text{sell}]$. The key likelihood ratio involves the noise fraction $q$.
- On the tick grid, set $a^{*} = \lceil E[V \mid \text{buy}] / \delta \rceil \cdot \delta$ and $b^{*} = \lfloor E[V \mid \text{sell}] / \delta \rfloor \cdot \delta$ to ensure the market maker does not lose money.
Worked Solution
How to Think About It: This is the bread-and-butter adverse selection problem for a market maker. When someone hits your ask, you need to ask yourself: "Given that I just got a buy order, what is $V$ worth?" If the buyer is informed, $V = 1$ for sure. If the buyer is noise, $V$ is still worth $p$ on average. The zero-profit ask is the expected value of $V$ conditional on seeing a buy. The twist here is that quotes live on a discrete grid, so you can't set the ask exactly at the conditional expectation -- you round up to the next tick to avoid losing money.
Quick Estimate: Suppose $p = 0.4$, $q = 0.5$, $\delta = 0.05$. A buy arrives from an informed trader (prob $0.5$) who buys only when $V = 1$, or from noise (prob $0.5$) who buys half the time. We have $P(\text{buy} \mid V=1) = (1-q) + q/2 = 0.75$ and $P(\text{buy} \mid V=0) = q/2 = 0.25$. So $P(V=1 \mid \text{buy}) = (0.75 \cdot 0.4) / (0.75 \cdot 0.4 + 0.25 \cdot 0.6) = 0.30 / 0.45 = 0.667$. The zero-profit ask in continuous space is $0.667$, and rounding up to the nearest tick: $\lceil 0.667 / 0.05 \rceil \cdot 0.05 = 14 \cdot 0.05 = 0.70$. By symmetry for the bid, $P(V=1 \mid \text{sell}) = (0.25 \cdot 0.4)/(0.25 \cdot 0.4 + 0.75 \cdot 0.6) = 0.10/0.55 = 0.182$. The zero-profit bid rounds down to $\lfloor 0.182 / 0.05 \rfloor \cdot 0.05 = 3 \cdot 0.05 = 0.15$. Spread: $0.70 - 0.15 = 0.55$.
Approach: Apply Bayes' theorem to compute the conditional value of $V$ given a buy (or sell), then round to the tick grid so the market maker breaks even.
Formal Solution:
*Step 1: Conditional probabilities of order flow.*
Given $V = 1$, a buy arrives if the counterparty is informed (happens with prob
- q$, always buys) or noise and buys (prob $q/2$):$P(\text{buy} \mid V = 1) = (1 - q) + \frac{q}{2} = 1 - \frac{q}{2}$
Given $V = 0$, a buy arrives only from noise:
$P(\text{buy} \mid V = 0) = \frac{q}{2}$
Similarly for sells:
$P(\text{sell} \mid V = 0) = (1 - q) + \frac{q}{2} = 1 - \frac{q}{2}, \quad P(\text{sell} \mid V = 1) = \frac{q}{2}$
*Step 2: Posterior given a buy.*
By Bayes' theorem:
$P(V = 1 \mid \text{buy}) = \frac{p\bigl(1 - q/2\bigr)}{p\bigl(1 - q/2\bigr) + (1-p)\,q/2}$
Simplify the denominator: $p - pq/2 + q/2 - pq/2 = p + q/2 - pq$. So:
$P(V = 1 \mid \text{buy}) = \frac{p(1 - q/2)}{p + q/2 - pq} = \frac{p(2 - q)}{2p + q(1 - 2p) + q \cdot 0}$
Let us write it cleanly. Multiply numerator and denominator by 2:
$P(V = 1 \mid \text{buy}) = \frac{p(2 - q)}{2p + q - 2pq} = \frac{p(2 - q)}{2p(1 - q) + q}$
Since $V$ is binary, the conditional expected value equals this posterior:
$E[V \mid \text{buy}] = \frac{p(2 - q)}{2p(1 - q) + q}$
*Step 3: Zero-profit ask on the tick grid.*
The market maker sells at the ask when a buy arrives. Expected profit per ask transaction is $a - E[V \mid \text{buy}]$. For zero profit, we need $a \geq E[V \mid \text{buy}]$, and we want the smallest such tick:
$a^{*} = \left\lceil \frac{1}{\delta} \cdot \frac{p(2 - q)}{2p(1 - q) + q} \right\rceil \cdot \delta$
*Step 4: Posterior given a sell.*
By identical logic:
$P(V = 1 \mid \text{sell}) = \frac{p \cdot q/2}{p \cdot q/2 + (1 - p)(1 - q/2)} = \frac{pq}{pq + (1-p)(2-q)}$
Simplify: denominator $= pq + 2 - q - 2p + pq = 2pq + 2 - q - 2p$, so:
$E[V \mid \text{sell}] = \frac{pq}{2 - q - 2p(1 - q)}$
*Step 5: Zero-profit bid on the tick grid.*
The market maker buys at the bid when a sell arrives. Expected profit is $E[V \mid \text{sell}] - b$. For zero profit: $b \leq E[V \mid \text{sell}]$, and we want the largest such tick:
$b^{*} = \left\lfloor \frac{1}{\delta} \cdot \frac{pq}{2 - q - 2p(1 - q)} \right\rfloor \cdot \delta$
*Step 6: Spread.*
$s^{*} = a^{*} - b^{*}$
Key limiting cases:
- $q \to 1$ (all noise): $E[V \mid \text{buy}] \to p$ and $E[V \mid \text{sell}] \to p$. The spread collapses to at most $\delta$ (just the tick size), since there is no adverse selection.
- $q \to 0$ (all informed): $E[V \mid \text{buy}] \to 1$ and $E[V \mid \text{sell}] \to 0$. The spread blows up to 1 (the full range of $V$). The market maker cannot avoid getting picked off.
- $\delta \to 0$ (continuous limit): $a^{*} \to E[V \mid \text{buy}]$ and $b^{*} \to E[V \mid \text{sell}]$, recovering the continuous Glosten-Milgrom result.
- $p = 1/2$ (symmetric prior): The ask and bid are symmetric about /2$.
Answer:
The zero-profit ask and bid are:
$a^{*} = \left\lceil \frac{1}{\delta} \cdot \frac{p(2 - q)}{2p(1-q) + q} \right\rceil \delta, \quad b^{*} = \left\lfloor \frac{1}{\delta} \cdot \frac{pq}{2 - q - 2p(1-q)} \right\rfloor \delta$
The spread $s^{*} = a^{*} - b^{*}$ is driven by adverse selection (controlled by $q$) and discreteness (controlled by $\delta$). More noise ($q \to 1$) narrows the spread; a finer tick ($\delta \to 0$) also narrows it by reducing rounding.
Intuition
This problem is the discrete-tick version of the Glosten-Milgrom (1985) model, which is the foundation of how we think about bid-ask spreads. The core idea is simple: spreads exist because market makers face adverse selection. Some of the orders hitting your quotes come from people who know more than you do. You compensate by shading your prices -- selling a bit above fair value and buying a bit below. The amount of shading is exactly the Bayesian update from observing the direction of the order.
In practice, this shows up everywhere. When you see a spread widen on an illiquid name, it is often because the market maker's model is assigning a higher probability that the next order is informed (lower $q$). The tick grid adds a real-world constraint: even if your theoretical zero-profit ask is 0.667, you might have to post at 0.70 because of the tick size. This means discrete ticks create an extra layer of "spread rent" beyond pure adverse selection -- a key consideration in debates about tick size reform.
- Noise (with probability $q$): buys or sells each with probability