Adverse Selection and the Break-Even Spread
You are a market maker on a binary contract that pays $\
Order flow is a mixture of two trader types: - With probability $q \in (0,1)$: a noise trader, who buys or sells with equal probability $\frac{1}{2}$ each, regardless of the outcome. - With probability
- Compute your expected P\&L per trade as a function of $s$ and $q$.
- Find the minimum spread $s^{*}$ such that $E[\text{P\&L}] \geq 0$.
- Briefly explain how $s^{*}$ scales with $q$ and what this tells you about the cost of adverse selection.
Hints
- Condition on the trader type (noise vs. informed) and compute your expected P\&L in each case separately.
- Against a noise trader, you earn half-spread $s$ on both sides (their expected value equals the mid). Against a perfectly informed trader, your loss is $50 - s$ regardless of the outcome.
- Set $q \cdot s - (1-q)(50-s) = 0$ and solve for $s^{*}$.
Worked Solution
How to Think About It: You make money from noise traders (who trade randomly) and lose money to informed traders (who always trade against you). The informed trader buys when the contract is worth $\
Quick Estimate: If $q = 0.5$ (half noise, half informed), you lose roughly $50/2 = 25$ on average to informed traders per trade, offset by profit of $s$ from noise traders. Break-even: $s \approx 12.5$. We will see the exact formula gives $s^{*} = 25(1-q)$, which for $q=0.5$ gives $s^{*} = 12.5$. Intuition confirmed.
Formal Solution:
Step 1: Compute expected P\&L.
Condition on who arrives. Consider a single trade.
*Noise trader (probability $q$):* Buys with probability $\frac{1}{2}$, sells with probability $\frac{1}{2}$. - If noise trader buys: you sell at ask $= 50 + s$. The contract is worth $\
*Informed trader (probability
Combining:
$E[\text{P\&L}] = q \cdot s + (1-q) \cdot (-(50-s))$
$= qs - (1-q)(50-s)$
$= qs - 50(1-q) + (1-q)s$
$= s[q + (1-q)] - 50(1-q)$
$= s - 50(1-q)$
Step 2: Break-even spread.
Set $E[\text{P\&L}] = 0$:
$s^{*} - 50(1-q) = 0 \implies s^{*} = 50(1-q)$
Step 3: Scaling with $q$.
As $q \to 1$ (all noise traders): $s^{*} \to 0$. You can quote a vanishingly tight spread -- no informed flow means no adverse selection cost.
As $q \to 0$ (all informed traders): $s^{*} \to 50$. You would need to quote a spread of $\
The break-even spread is linear in the fraction of informed flow $(1-q)$: every percentage point of informed flow costs you 50 cents per trade in adverse selection.
Answer:
$E[\text{P\&L}] = s - 50(1-q)$
$s^{*} = 50(1-q)$
The break-even spread is proportional to the fraction of informed traders. More toxic flow requires a wider spread to survive.
Intuition
This is the classic Glosten-Milgrom model in miniature. The core insight is that market makers price the spread to break even against the mix of informed and uninformed flow -- not to make money on every trade, but to make enough from noise traders to offset the losses to informed traders. The break-even spread $s^{*} = 50(1-q)$ is entirely driven by adverse selection: the more informed the order flow, the wider you need to quote.
This framework is the foundation for how practitioners think about market quality. A tight bid-ask spread is only sustainable if the flow is mostly uninformed. When a market maker sees their spreads getting wider (or their inventory drifting systematically), the natural interpretation is that the fraction of informed flow has increased -- perhaps because news is imminent, or because toxic algorithmic flow has found their venue. In practice, estimating $q$ (the noise-to-informed ratio) from order flow data is a central problem in market microstructure research.