Limit Order Fill Probability

Expectation · Medium · Free problem

You post a limit order on an exchange. Market orders arrive according to a Poisson process with rate $\lambda = 0.8$ orders per second, and your order rests for $T = 2.5$ seconds before you cancel it. The order fills if at least one market order arrives during $[0, T]$. Let $N$ be the number of market orders that arrive in that window.

Compute:

  1. The fill probability $P(N \geq 1)$.
  2. The probability that exactly one market order arrives, $P(N = 1)$.
  3. The conditional expectation $E[N \mid N \geq 1]$.

Hints

  1. Compute the Poisson rate parameter $\mu = \lambda T$ first -- everything flows from that single number.
  2. The fill probability is the complement of the zero-arrival event: $P(N \geq 1) = 1 - e^{-\mu}$.
  3. For the conditional expectation, note that $N \cdot \mathbf{1}_{N=0} = 0$, so $E[N \mid N \geq 1] = E[N] / P(N \geq 1)$.

Worked Solution

How to Think About It: This is a textbook Poisson exercise dressed up in a trading context. The key parameter is $\lambda T = 0.8 \times 2.5 = 2.0$ -- that is the expected number of market orders hitting your resting limit order. Since the expected count is 2, you should expect to get filled most of the time (the only way you don't is if zero orders arrive, which is the $e^{-\lambda T}$ tail). The conditional expectation in part (iii) asks: given you did get filled, how many orders came through? Since you are conditioning away the $N = 0$ case, the answer must be higher than the unconditional mean of 2.

Quick Estimate: With $\lambda T = 2$, the probability of zero arrivals is $e^{-2} \approx 0.135$, so the fill probability is about $0.865$ -- you get filled roughly 86.5% of the time. For the conditional expectation, you are redistributing the mean of 2 over only the 86.5% of outcomes where $N \geq 1$, so $E[N \mid N \geq 1] \approx 2 / 0.865 \approx 2.31$. The exact one-arrival probability is

e^{-2} \approx 0.271$.

Approach: All three parts follow directly from the Poisson PMF $P(N = k) = e^{-\mu} \mu^k / k!$ with $\mu = \lambda T = 2$, plus the definition of conditional expectation.

Formal Solution:

Set $\mu = \lambda T = 0.8 \times 2.5 = 2.0$, so $N \sim \text{Poisson}(2)$.

(i) Fill probability:

$P(N \geq 1) = 1 - P(N = 0) = 1 - e^{-2} \approx 0.8647$

(ii) Exactly one arrival:

$P(N = 1) = e^{-2} \cdot \frac{2^1}{1!} = 2e^{-2} \approx 0.2707$

(iii) Conditional expectation $E[N \mid N \geq 1]$:

By definition:

$E[N \mid N \geq 1] = \frac{E[N \cdot \mathbf{1}_{N \geq 1}]}{P(N \geq 1)}$

Since $N \cdot \mathbf{1}_{N \geq 1} = N - N \cdot \mathbf{1}_{N = 0} = N$ (because $N \cdot \mathbf{1}_{N=0} = 0$ always), we get:

$E[N \mid N \geq 1] = \frac{E[N]}{P(N \geq 1)} = \frac{\mu}{1 - e^{-\mu}} = \frac{2}{1 - e^{-2}} \approx \frac{2}{0.8647} \approx 2.313$

Answer:

Intuition

This problem illustrates a pattern that comes up constantly in electronic trading: modeling order flow as a Poisson process and asking what happens during a fixed time window. The fill probability

- e^{-\lambda T}$ is the single most important formula for a market maker thinking about how long to rest a limit order. If you double your resting time, you don't double your fill probability -- you push deeper into the diminishing-returns region of
- e^{-x}$. At $\lambda T = 2$ you are already at 86.5%; doubling to $\lambda T = 4$ only gets you to 98.2%.

The conditional expectation $E[N \mid N \geq 1] = \mu / (1 - e^{-\mu})$ is a useful object too: it tells you how much adverse flow you eat conditional on getting filled. Notice it is always greater than 1 (you can't get filled with zero orders), and for large $\mu$ it converges to $\mu$ since the conditioning event becomes near-certain. The key technique -- that $E[N \cdot \mathbf{1}_{N=0}]$ vanishes -- is a small trick worth remembering. It lets you avoid summing an infinite series and instead just divide the unconditional mean by the survival probability.

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