First Fill Time Across Multiple Venues
You post the same quote on $n = 4$ independent venues. The time-to-first-fill on venue $i$ is exponential with rate $\lambda_i$ (per second), where
$(\lambda_1, \lambda_2, \lambda_3, \lambda_4) = (0.10, 0.15, 0.20, 0.05).$
Compute:
- $E[\min_i T_i]$, the expected time until your first fill anywhere.
- $P(\min_i T_i \leq 5)$, the probability that at least one venue fills within 5 seconds.
- The probability that venue 3 is the first to fill.
Hints
- What happens when you take the minimum of independent exponential random variables? Think about how survival probabilities multiply.
- The minimum of independent exponentials with rates $\lambda_1, \ldots, \lambda_n$ is itself exponential with rate $\Lambda = \sum \lambda_i$. Use this to handle parts (i) and (ii) directly.
- For part (iii), condition on the minimum occurring at time $t$ and ask which venue "wins" the race. The probability that venue $j$ is first equals $\lambda_j / \Lambda$.
Worked Solution
How to Think About It: This is a bread-and-butter order statistics problem using the memoryless property of exponentials. On a trading desk, you post to multiple venues and care about the fastest fill. The key fact: the minimum of independent exponentials is itself exponential, with rate equal to the sum of the individual rates. Once you know that, all three parts are one-liners. The intuition is that each venue is independently "racing" to fill you, and the combined rate of the race is just the total hazard rate.
Quick Estimate: The total rate is $\Lambda = 0.10 + 0.15 + 0.20 + 0.05 = 0.50$ per second. So the expected first fill is
Approach: Apply the standard result: if $T_1, \ldots, T_n$ are independent exponentials with rates $\lambda_1, \ldots, \lambda_n$, then $\min_i T_i \sim \text{Exp}(\lambda_1 + \cdots + \lambda_n)$, and $P(\text{venue } j \text{ is first}) = \lambda_j / (\lambda_1 + \cdots + \lambda_n)$.
Formal Solution:
Let $\Lambda = \sum_{i=1}^{4} \lambda_i = 0.10 + 0.15 + 0.20 + 0.05 = 0.50$.
Since the $T_i$ are independent exponentials, $M = \min_i T_i \sim \text{Exp}(\Lambda)$. This follows because
$P(M > t) = \prod_{i=1}^{4} P(T_i > t) = \prod_{i=1}^{4} e^{-\lambda_i t} = e^{-\Lambda t}.$
Part (i):
$E[\min_i T_i] = \frac{1}{\Lambda} = \frac{1}{0.50} = 2 \text{ seconds}.$
Part (ii):
$P(\min_i T_i \leq 5) = 1 - e^{-\Lambda \cdot 5} = 1 - e^{-2.5} \approx 1 - 0.0821 = 0.9179.$
Part (iii):
The probability that venue $j$ fills first equals $\lambda_j / \Lambda$. This is because
$P(T_j = \min_i T_i) = \int_0^{\infty} \lambda_j e^{-\lambda_j t} \prod_{k \neq j} e^{-\lambda_k t} \, dt = \lambda_j \int_0^{\infty} e^{-\Lambda t} \, dt = \frac{\lambda_j}{\Lambda}.$
So
$P(\text{venue 3 is first}) = \frac{\lambda_3}{\Lambda} = \frac{0.20}{0.50} = 0.40.$
Answer:
- (i) $E[\min_i T_i] = 2$ seconds
- (ii) $P(\min_i T_i \leq 5) = 1 - e^{-2.5} \approx 0.918$
- (iii) $P(\text{venue 3 first}) = 0.40$
Intuition
This problem captures a core pattern in electronic trading: when you blast a quote to multiple venues, the fill dynamics reduce to a race among independent exponential clocks. The minimum-of-exponentials result is one of those facts that shows up constantly -- in queuing theory, reliability engineering, and market microstructure. The total hazard rate is additive, so adding more venues always shrinks your expected fill time, and each venue's probability of being first is simply its share of the total rate.
The deeper lesson is about the memoryless property doing all the heavy lifting. Because exponentials are memoryless, the identity of the winner is independent of when the minimum occurs. This clean separation -- "who wins" versus "when" -- is what makes the problem tractable. In practice, venue fill rates are not truly exponential (they depend on queue position, volatility, etc.), but this model gives you a useful first-order framework for thinking about venue allocation and smart order routing.