Exponential Quote Hit Times Across Venues

Expectation · Medium · Free problem

The time (in seconds) until your quote is hit at a given venue is modeled as an exponential random variable with rate $\lambda = 0.25$ (i.e., mean $4$ seconds). You post identical quotes on $n = 5$ independent venues, so the hit times $T_1, T_2, \ldots, T_5$ are i.i.d. $\text{Exp}(0.25)$.

Compute:

  1. $E[\min_i T_i]$ -- the expected time until your first fill across all venues.
  2. $P(\min_i T_i \leq 3)$ -- the probability you get at least one fill within 3 seconds.
  3. $E[\max_i T_i]$ -- the expected time until every venue has filled.

Hints

  1. The minimum of independent exponentials is itself exponential -- what rate?
  2. For the minimum, the rates add: $\min(T_1, \ldots, T_n) \sim \text{Exp}(n\lambda)$. For the maximum, use the tail-sum formula $E[X] = \int_0^{\infty} P(X > t) \, dt$ and expand with the binomial theorem.
  3. The maximum expectation for $n$ i.i.d. $\text{Exp}(\lambda)$ variables simplifies to $H_n / \lambda$ where $H_n = 1 + 1/2 + \cdots + 1/n$ is the $n$-th harmonic number.

Worked Solution

How to Think About It: This is bread-and-butter order statistics for exponentials, and it shows up constantly in market microstructure. You post on 5 venues and each fill arrives as an independent Poisson process. The minimum of independent exponentials is exponential with the summed rate -- that is the memoryless "race" property. The maximum is harder: there is no closed-form shortcut like the minimum, but for i.i.d. exponentials you can use the classic harmonic-number formula.

Quick Estimate: Each venue fills in 4 seconds on average. With 5 venues racing, the first fill should come about 5 times faster: roughly $4/5 = 0.8$ seconds. For the last fill, a rule of thumb is the harmonic sum: $H_5 = 1 + 1/2 + 1/3 + 1/4 + 1/5 \approx 2.28$, so the max is about $4 \times 2.28 \approx 9.1$ seconds. For part (ii), 3 seconds is nearly 4 mean lifetimes of the minimum (mean $0.8$), so the probability should be very close to 1.

Approach: Use the well-known properties of exponential order statistics.

Formal Solution:

Part (i): $E[\min_i T_i]$

The survival function of the minimum is:

$P(\min_i T_i > t) = \prod_{i=1}^{5} P(T_i > t) = \left(e^{-\lambda t}\right)^5 = e^{-5\lambda t}$

So $\min_i T_i \sim \text{Exp}(5\lambda) = \text{Exp}(1.25)$, and:

$E[\min_i T_i] = \frac{1}{5\lambda} = \frac{1}{5 \times 0.25} = \frac{1}{1.25} = 0.8 \text{ seconds}$

Part (ii): $P(\min_i T_i \leq 3)$

Since $\min_i T_i \sim \text{Exp}(1.25)$:

$P(\min_i T_i \leq 3) = 1 - e^{-5\lambda \cdot 3} = 1 - e^{-3.75} \approx 1 - 0.0235 = 0.9765$

So there is about a 97.7% chance of getting at least one fill within 3 seconds.

Part (iii): $E[\max_i T_i]$

The CDF of the maximum is:

$P(\max_i T_i \leq t) = \prod_{i=1}^{5} P(T_i \leq t) = \left(1 - e^{-\lambda t}\right)^5$

Using the tail-sum formula for the expectation of a non-negative random variable:

$E[\max_i T_i] = \int_0^{\infty} P(\max_i T_i > t) \, dt = \int_0^{\infty} \left[1 - \left(1 - e^{-\lambda t}\right)^5\right] dt$

Expand $\left(1 - e^{-\lambda t}\right)^5$ using the binomial theorem and substitute $u = e^{-\lambda t}$:

$\left(1 - e^{-\lambda t}\right)^5 = \sum_{k=0}^{5} \binom{5}{k}(-1)^k e^{-k\lambda t}$

So:

$1 - \left(1 - e^{-\lambda t}\right)^5 = -\sum_{k=1}^{5} \binom{5}{k}(-1)^k e^{-k\lambda t} = \sum_{k=1}^{5} \binom{5}{k}(-1)^{k+1} e^{-k\lambda t}$

Integrating term by term:

$E[\max_i T_i] = \sum_{k=1}^{5} \binom{5}{k} \frac{(-1)^{k+1}}{k\lambda} = \frac{1}{\lambda}\sum_{k=1}^{5} \frac{(-1)^{k+1}}{k}\binom{5}{k}$

For i.i.d. exponentials, this simplifies to the well-known harmonic formula:

$E[\max_i T_i] = \frac{1}{\lambda} \cdot H_n = \frac{1}{\lambda}\left(1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}\right)$

To verify: expand the binomial sum explicitly:

$\sum_{k=1}^{5} \frac{(-1)^{k+1}}{k}\binom{5}{k} = \frac{5}{1} - \frac{10}{2} + \frac{10}{3} - \frac{5}{4} + \frac{1}{5} = 5 - 5 + \frac{10}{3} - \frac{5}{4} + \frac{1}{5}$

$= \frac{10}{3} - \frac{5}{4} + \frac{1}{5} = \frac{200 - 75 + 12}{60} = \frac{137}{60}$

And indeed $H_5 = 1 + 1/2 + 1/3 + 1/4 + 1/5 = 60/60 + 30/60 + 20/60 + 15/60 + 12/60 = 137/60$. So:

$E[\max_i T_i] = \frac{1}{0.25} \cdot \frac{137}{60} = 4 \cdot \frac{137}{60} = \frac{137}{15} \approx 9.133 \text{ seconds}$

Answer:

  1. $E[\min_i T_i] = \dfrac{1}{n\lambda} = 0.8$ seconds
  2. $P(\min_i T_i \leq 3) = 1 - e^{-3.75} \approx 0.9765$
  3. $E[\max_i T_i] = \dfrac{H_n}{\lambda} = \dfrac{137}{15} \approx 9.133$ seconds

Intuition

This problem captures a fundamental tension in execution strategy: spreading quotes across venues. The minimum fill time scales as

/(n\lambda)$ -- linear speedup, just like parallelizing independent tasks. This is why firms post on multiple venues: you get filled faster, and the probability of at least one fill within any time window grows rapidly with $n$. The maximum, however, only grows logarithmically (via the harmonic number $H_n \approx \ln n + 0.577$). So the last venue to fill takes disproportionately long relative to the first, but not catastrophically so.

The deeper lesson is about the memoryless property and the "race" interpretation of exponentials. Each venue is running an independent Poisson clock. The first to ring is exponential with the pooled rate -- this is why superpositions of Poisson processes are Poisson. The gap between the min and max expectations ($0.8$ vs $9.1$ seconds here) quantifies the difference between "at least one fill" and "all fills," which matters for things like inventory management and sweep-to-fill order routing.

Open the full interactive solver →