Optimal Accept-Reject Threshold for Fills
A venue sends you fill opportunities one at a time. Each opportunity reveals a scalar feature $X \sim N(0,1)$. Conditional on $X = x$, the expected mark-to-mid P&L of accepting the fill is
$E[\pi \mid X = x] = 0.06x - 0.01 \quad \text{(ticks)}$
and $\text{Var}(\pi \mid X = x) = 1$ (ticks$^2$). You may accept or reject each opportunity independently; rejecting yields zero P&L.
- Find the threshold policy $\mathbf{1}\{X \geq t\}$ that maximizes $E[\pi \cdot \mathbf{1}\{X \geq t\}]$.
- Compute the resulting acceptance rate $P(X \geq t)$ and the unconditional expected P&L per opportunity in terms of $\phi$ and $\Phi$ (the standard normal PDF and CDF).
- If you must keep the acceptance rate at exactly 20%, what threshold do you use and what is the expected P&L per opportunity?
Hints
- You should accept a fill whenever its expected P&L is positive. What value of $X$ makes the conditional expectation exactly zero?
- The unconditional expected P&L involves $E[X \cdot \mathbf{1}\{X \geq t\}]$ for standard normal $X$. Use the identity $E[X \cdot \mathbf{1}\{X \geq t\}] = \phi(t)$ where $\phi$ is the standard normal PDF.
- For the constrained problem, set - \Phi(t_c) = 0.20$ to find $t_c$, then plug into the same P&L formula from part (b).
Worked Solution
How to Think About It: This is a classic fill-or-kill scoring problem. Every opportunity has a signal $X$ that tells you something about whether the fill will be profitable. You want to accept only when the expected P&L is positive -- anything else is giving away edge. The threshold is just the break-even point of the conditional expectation. Once you have that, the unconditional P&L calculation is a truncated-normal integral, which is bread and butter for anyone working in execution or market-making.
The twist in part (c) is a capacity constraint: you can only accept 20% of flow, so you raise the bar. You trade off some marginal fills for higher average quality.
Quick Estimate: The conditional P&L is $0.06x - 0.01$, which is zero at $x = 1/6 \approx 0.167$. Since $X \sim N(0,1)$, roughly 43% of flow exceeds that threshold -- so unconstrained, you accept a bit less than half. Among those accepted fills, the average signal is roughly $E[X \mid X \geq 0.167] \approx \phi(0.167)/(1 - \Phi(0.167)) \approx 0.395/0.434 \approx 0.91$, giving average P&L around $0.06 \times 0.91 - 0.01 \approx 0.045$ ticks per accepted fill. Per opportunity (including rejects), that is $0.045 \times 0.434 \approx 0.019$ ticks.
Approach: We maximize the unconditional expected P&L by choosing the optimal threshold, then evaluate using standard truncated-normal identities.
Formal Solution:
*Part (a): Optimal threshold*
The expected P&L from a threshold policy is
$V(t) = E[\pi \cdot \mathbf{1}\{X \geq t\}] = E[(0.06X - 0.01) \cdot \mathbf{1}\{X \geq t\}]$
Differentiate with respect to $t$ using Leibniz's rule:
$V'(t) = -(0.06t - 0.01) \cdot \phi(t)$
Setting $V'(t) = 0$ gives $0.06t - 0.01 = 0$, so
$t^{*} = \frac{0.01}{0.06} = \frac{1}{6}$
The second derivative is negative at $t^{*}$ (since $V''(t^{*}) = -0.06 \cdot \phi(1/6) < 0$), confirming this is a maximum. The economic logic is clear: accept whenever the conditional expected P&L is non-negative.
*Part (b): Acceptance rate and expected P&L*
The acceptance rate is
$P(X \geq t^{*}) = 1 - \Phi\!\left(\frac{1}{6}\right) \approx 1 - 0.5662 = 0.4338$
For the unconditional expected P&L, we need $E[X \cdot \mathbf{1}\{X \geq t\}]$ when $X \sim N(0,1)$. The standard identity is
$E[X \cdot \mathbf{1}\{X \geq t\}] = \phi(t)$
Therefore
$V(t^{*}) = 0.06 \, \phi\!\left(\frac{1}{6}\right) - 0.01 \left[1 - \Phi\!\left(\frac{1}{6}\right)\right]$
Plugging in $\phi(1/6) \approx 0.3945$ and
- \Phi(1/6) \approx 0.4338$:$V(t^{*}) \approx 0.06 \times 0.3945 - 0.01 \times 0.4338 = 0.02367 - 0.00434 \approx 0.0193 \text{ ticks per opportunity}$
*Part (c): Constrained acceptance rate of 20%*
We need $P(X \geq t_c) = 0.20$, so $\Phi(t_c) = 0.80$, giving
$t_c = \Phi^{-1}(0.80) \approx 0.8416$
The expected P&L per opportunity is
$V(t_c) = 0.06 \, \phi(0.8416) - 0.01 \times 0.20$
With $\phi(0.8416) \approx 0.2803$:
$V(t_c) \approx 0.06 \times 0.2803 - 0.002 = 0.01682 - 0.002 = 0.0148 \text{ ticks per opportunity}$
Note the per-accepted-fill P&L is $0.0148/0.20 = 0.074$ ticks, which is higher than the unconstrained per-fill average of $0.0193/0.434 \approx 0.045$ ticks. You accept fewer but better fills.
Answer:
- (a) Optimal threshold: $t^{*} = 1/6 \approx 0.167$. Accept when the conditional expected P&L is non-negative.
- (b) In closed form: acceptance rate $= 1 - \Phi(1/6) \approx 43.4\%$; unconditional expected P&L $= 0.06\,\phi(1/6) - 0.01[1 - \Phi(1/6)] \approx 0.0193$ ticks per opportunity.
- (c) Constrained threshold: $t_c = \Phi^{-1}(0.80) \approx 0.842$; expected P&L $\approx 0.0148$ ticks per opportunity -- about 23% less total P&L, but 65% higher P&L per accepted fill.
Intuition
This problem is the atomic unit of execution optimization: given a noisy signal about fill quality, where do you draw the line? The unconstrained answer is trivially "accept when expected P&L is positive," but the interesting economics emerge when you add constraints. A 20% acceptance-rate cap forces you to raise the bar, and you see the fundamental trade-off: total P&L drops (you reject some marginally profitable fills) but average quality per fill rises sharply. In practice, this trade-off is everywhere -- market makers face it when managing their queue position, stat-arb desks face it when sizing into crowded signals, and execution algos face it when deciding which passive fills to pursue.
The mathematical backbone is truncated-normal expectations, specifically the identity $E[X \cdot \mathbf{1}\{X \geq t\}] = \phi(t)$ for standard normal $X$. This identity (and its generalizations) shows up constantly in quant finance -- from computing conditional VaR to pricing barrier options to evaluating selection-corrected returns. Getting fluent with truncated-normal calculations is one of the highest-leverage technical skills for anyone working in execution or market microstructure.