Optimal Size Cap Under Adverse Selection
You are a market maker posting an ask price $a$. Buy orders arrive with random size $Q \sim \text{Exp}(\lambda)$. With probability $\alpha$, the buyer is informed -- meaning the true value will drift up by $\delta > 0$ per unit during your holding period. With probability
You can choose to accept any incoming order up to a size cap $C$ -- that is, you fill $\min(Q, C)$ shares. Your per-share edge before adverse selection is $s/2$ (i.e., the half-spread).
- Compute $E[\text{PnL} \mid \text{cap} = C]$ as a function of $C$, $\lambda$, $\alpha$, $\delta$, and $s$.
- Derive the optimal cap $C^{*}$ that maximizes expected P&L.
- Under what conditions is it optimal to set $C^{*} = \infty$ (i.e., accept all sizes)?
Hints
- Think about how your P&L per share differs between informed and uninformed flow -- the cap lets you trade off spread revenue against adverse selection cost.
- Use the truncated expectation $E[\min(Q, C)] = (1 - e^{-\lambda C})/\lambda$ for an exponential random variable, and note that informed and uninformed orders may have different size distributions.
- Differentiate the expected P&L with respect to $C$ and set the marginal gain from uninformed flow equal to the marginal loss from informed flow to find the optimal cap.
Worked Solution
How to Think About It: This is a classic market-making trade-off: you earn the spread on every share you fill, but informed orders cost you $\delta$ per share. Bigger orders are more likely to be informed (an informed trader wants to trade as much as possible), so capping your fill size limits your adverse selection losses at the cost of also capping your spread revenue from large uninformed orders. The question is where to set that cap.
Your P&L per share filled is $s/2$ from uninformed flow and $s/2 - \delta$ from informed flow. If $\delta < s/2$, the spread covers the adverse selection even on informed orders, and you should accept everything. The interesting case is $\delta > s/2$, where large informed orders hurt you.
Quick Estimate: Suppose $\lambda = 1$, $\alpha = 0.3$, $\delta = 0.08$, $s/2 = 0.05$. The per-share blended edge is $(0.7)(0.05) + (0.3)(0.05 - 0.08) = 0.035 - 0.009 = 0.026$. The expected fill (uncapped) is $E[Q] = 1$, so the uncapped expected P&L is about $0.026$. But large informed orders drag this down disproportionately. A cap around $C = 2$ would truncate the worst informed fills while keeping most of the uninformed revenue. We expect the optimal cap to be on the order of
Approach: Compute the expected P&L by conditioning on informed vs. uninformed, using the truncated mean of the exponential distribution, then optimize over $C$.
Formal Solution:
The filled quantity is $\min(Q, C)$. For $Q \sim \text{Exp}(\lambda)$:
$E[\min(Q, C)] = \frac{1}{\lambda}(1 - e^{-\lambda C})$
The expected P&L, conditioning on order type:
$E[\text{PnL}] = (1 - \alpha) \cdot \frac{s}{2} \cdot E[\min(Q, C)] + \alpha \cdot \left(\frac{s}{2} - \delta\right) \cdot E[\min(Q, C)]$
Since $E[\min(Q, C)]$ is the same for both types (both draw $Q$ from $\text{Exp}(\lambda)$), this simplifies to:
$E[\text{PnL}] = \left[\frac{s}{2} - \alpha \delta\right] \cdot \frac{1}{\lambda}(1 - e^{-\lambda C})$
Notice: if the per-share blended edge $s/2 - \alpha \delta$ is positive, then E[PnL] is increasing in $C$, so $C^{*} = \infty$. If $s/2 - \alpha \delta$ is negative, then E[PnL] is negative for any $C > 0$, and you should not trade at all ($C^{*} = 0$).
This is too simple -- it misses the key feature that informed traders tend to send larger orders. Let us instead model size-dependent adverse selection: the probability that an order of size $q$ is informed increases with $q$. A natural model is that informed and uninformed traders draw sizes from different exponentials:
- Uninformed: $Q_U \sim \text{Exp}(\lambda_U)$
- Informed: $Q_I \sim \text{Exp}(\lambda_I)$ with $\lambda_I < \lambda_U$ (informed orders are larger on average)
The expected P&L with cap $C$ is:
$E[\text{PnL}(C)] = (1-\alpha) \cdot \frac{s}{2} \cdot \frac{1}{\lambda_U}(1 - e^{-\lambda_U C}) + \alpha \cdot \left(\frac{s}{2} - \delta\right) \cdot \frac{1}{\lambda_I}(1 - e^{-\lambda_I C})$
Differentiating with respect to $C$:
$\frac{d}{dC} E[\text{PnL}] = (1-\alpha) \cdot \frac{s}{2} \cdot e^{-\lambda_U C} + \alpha \cdot \left(\frac{s}{2} - \delta\right) \cdot e^{-\lambda_I C}$
Setting equal to zero (noting $s/2 - \delta < 0$ in the interesting case):
$(1-\alpha) \cdot \frac{s}{2} \cdot e^{-\lambda_U C} = \alpha \cdot \left(\delta - \frac{s}{2}\right) \cdot e^{-\lambda_I C}$
$e^{(\lambda_U - \lambda_I) C} = \frac{\alpha(\delta - s/2)}{(1-\alpha)(s/2)}$
Solving for $C^{*}$:
$C^{*} = \frac{1}{\lambda_U - \lambda_I} \ln\left(\frac{\alpha(\delta - s/2)}{(1-\alpha)(s/2)}\right)$
This is valid only when: - $\lambda_U > \lambda_I$ (informed orders are genuinely larger) - $\delta > s/2$ (informed flow is toxic enough to matter) - The argument of the log is greater than 1 (otherwise $C^{*} \leq 0$ and you should not trade)
The second-order condition confirms this is a maximum: at $C^{*}$, the marginal uninformed revenue from increasing the cap equals the marginal informed loss.
Part 3: It is optimal to set $C^{*} = \infty$ (accept all sizes) when the marginal P&L from filling the next share is non-negative for all sizes. From the first-order condition, this happens when the right-hand side is $\leq 1$, i.e.:
$\alpha(\delta - s/2) \leq (1-\alpha)(s/2)$
$\alpha \delta \leq \frac{s}{2}$
In words: the probability-weighted adverse selection cost per share is less than the half-spread. When this holds, the spread is wide enough to compensate for adverse selection even on the largest orders.
Answer: The optimal cap is $C^{*} = \frac{1}{\lambda_U - \lambda_I} \ln\!\left(\frac{\alpha(\delta - s/2)}{(1-\alpha)(s/2)}\right)$ when $\delta > s/2$ and informed orders are larger on average. Accept all sizes ($C^{*} = \infty$) whenever $\alpha \delta \leq s/2$.
Intuition
The core idea here is that size is a signal of information. In real markets, informed traders want to trade large because their edge is perishable -- they need to monetize their signal before it decays. Uninformed traders (hedgers, index rebalancers) have no urgency and their sizes are more random. By capping how much you fill, you are effectively saying: "I will provide liquidity up to a point, but if you want to trade huge, go find someone else." This is adverse selection management through quantity rationing rather than price adjustment.
This shows up constantly in practice. Market makers on electronic exchanges use "size tiers" -- different prices or fill rates for different order sizes. The optimal cap balances two forces: too small and you leave money on the table from harmless uninformed flow, too large and you get run over by informed whales. The elegant result that $C^{*} = \infty$ when $\alpha \delta \leq s/2$ has a clean interpretation: if your spread is wide enough relative to the toxicity of informed flow, there is no reason to ration. This is why wide-spread, illiquid names often have more generous size quotes -- the spread itself is the protection.