Competing Poisson Processes
Two independent Poisson processes $N_1$ and $N_2$ have rates $\lambda_1 > 0$ and $\lambda_2 > 0$. Let $T_1$ and $T_2$ be their respective first arrival times, and define $T = \min(T_1, T_2)$ as the time of the first event across both processes.
- Show that $T \sim \text{Exp}(\lambda_1 + \lambda_2)$ and that $P(T = T_1) = \frac{\lambda_1}{\lambda_1 + \lambda_2}$.
- Conditioned on $T$, identify which process produced the first event. Specifically, show that the identity of the winner is independent of $T$.
- Let $K$ be the number of $N_1$ arrivals that occur strictly before the first arrival of $N_2$. Find $P(K = k)$ for $k = 0, 1, 2, \ldots$ and compute $E[K]$.
Hints
- The survival function of the minimum of independent random variables factors into a product of individual survival functions.
- After computing $P(T_1 < T_2)$ via integration, check whether the joint density of $(T, I)$ factors into the product of the marginal of $T$ and the marginal of $I$.
- Use the memoryless property: each time an event fires, the remaining waiting times reset. This turns the count $K$ into a sequence of independent coin flips with probability $\lambda_1/(\lambda_1 + \lambda_2)$.
Worked Solution
How to Think About It: This is one of the most important results in applied probability -- the "racing exponentials" property. The punchline is simple: when independent exponentials compete, the minimum is exponential with the sum of rates, and the winner is chosen with probability proportional to its rate. This shows up constantly in queuing theory, order flow modeling, and any system where independent events race against each other. Before any math, you should know the answers: the first event arrives at rate $\lambda_1 + \lambda_2$, process 1 wins with probability $\lambda_1 / (\lambda_1 + \lambda_2)$, and -- crucially -- who wins is independent of when the race ends.
Quick Estimate: Suppose $\lambda_1 = 3$ and $\lambda_2 = 7$. The first event should arrive at rate
Approach: Use the memoryless property and the CDF/survival function of exponentials for part (a), then a direct conditioning argument for parts (b) and (c).
Formal Solution:
*Part (a): Distribution of $T$ and the winning probability*
Since $T_1 \sim \text{Exp}(\lambda_1)$ and $T_2 \sim \text{Exp}(\lambda_2)$ are independent, the survival function of $T = \min(T_1, T_2)$ is:
$P(T > t) = P(T_1 > t)\,P(T_2 > t) = e^{-\lambda_1 t} \cdot e^{-\lambda_2 t} = e^{-(\lambda_1 + \lambda_2)t}$
This is the survival function of $\text{Exp}(\lambda_1 + \lambda_2)$, so $T \sim \text{Exp}(\lambda_1 + \lambda_2)$.
For the winning probability:
$P(T_1 < T_2) = \int_0^{\infty} P(T_2 > t)\,\lambda_1 e^{-\lambda_1 t}\,dt = \int_0^{\infty} e^{-\lambda_2 t}\,\lambda_1 e^{-\lambda_1 t}\,dt = \lambda_1 \int_0^{\infty} e^{-(\lambda_1 + \lambda_2)t}\,dt = \frac{\lambda_1}{\lambda_1 + \lambda_2}$
*Part (b): Independence of the winner and the arrival time*
Let $I$ be the indicator that process 1 jumped first ($I = 1$ if $T = T_1$, $I = 0$ if $T = T_2$). We want to show $I$ is independent of $T$. Compute the joint density by conditioning on which process wins:
$f_{T, I}(t, 1) = P(T_1 \in dt,\, T_1 < T_2)/dt = \lambda_1 e^{-\lambda_1 t} \cdot e^{-\lambda_2 t} = \lambda_1 e^{-(\lambda_1 + \lambda_2)t}$
This factors as:
$f_{T, I}(t, 1) = \frac{\lambda_1}{\lambda_1 + \lambda_2} \cdot (\lambda_1 + \lambda_2)e^{-(\lambda_1 + \lambda_2)t} = P(I = 1) \cdot f_T(t)$
Similarly for $I = 0$. Since the joint factors into marginals, $I$ and $T$ are independent. In words: knowing when the first event happened tells you nothing about which process produced it.
*Part (c): Distribution of $K$ -- arrivals of $N_1$ before first $N_2$ arrival*
Think of it as a sequence of independent races. Each time an event occurs (from either process), it came from $N_1$ with probability $p = \lambda_1/(\lambda_1 + \lambda_2)$ and from $N_2$ with probability $q = \lambda_2/(\lambda_1 + \lambda_2)$, independently of all previous outcomes (by the memoryless property and the independence result from part (b)).
$K$ counts the number of $N_1$ arrivals before the first $N_2$ arrival. This is the number of successes before the first failure in independent Bernoulli trials with success probability $p$. So $K$ has a Geometric distribution (counting failures):
$P(K = k) = p^k \cdot q = \left(\frac{\lambda_1}{\lambda_1 + \lambda_2}\right)^k \cdot \frac{\lambda_2}{\lambda_1 + \lambda_2}, \quad k = 0, 1, 2, \ldots$
The expected value is:
$E[K] = \frac{p}{q} = \frac{\lambda_1}{\lambda_2}$
This makes intuitive sense: process 1 fires $\lambda_1 / \lambda_2$ times as fast as process 2, so on average it gets that many arrivals in before process 2 fires.
Answer:
- (a) $T \sim \text{Exp}(\lambda_1 + \lambda_2)$ and $P(T = T_1) = \lambda_1 / (\lambda_1 + \lambda_2)$.
- (b) The identity of the winning process is independent of the arrival time $T$.
- (c) $P(K = k) = \left(\frac{\lambda_1}{\lambda_1 + \lambda_2}\right)^k \frac{\lambda_2}{\lambda_1 + \lambda_2}$ and $E[K] = \lambda_1 / \lambda_2$.
Intuition
The racing exponentials result is one of the workhorses of stochastic modeling. It says that when independent memoryless processes compete, the combined process is also memoryless with the pooled rate, and the identity of each event is an independent coin flip weighted by rates. This is the foundation of continuous-time Markov chains: transitions out of any state happen at the total rate, and you pick the destination proportionally to the individual rates. In finance, this shows up whenever you model competing order arrivals, default events in a credit portfolio (first-to-default baskets), or competing limit orders in an order book.
The result in part (c) -- that $K$ is geometric with mean $\lambda_1/\lambda_2$ -- is a direct consequence of the memoryless property. Every time process 1 fires, the race "resets" completely. So you are just flipping a biased coin repeatedly until process 2 finally wins. A common mistake is trying to condition on the value of $T_2$ and integrate over the number of Poisson arrivals in $(0, T_2)$ -- this works but is far more painful than the coin-flip argument. The elegant approach leverages the independence result from part (b) to avoid any integration at all.