Exponential Random Variable and Conditional Probability

Question

Person 1 enters a queue, Person 1 will eventually abandon this queue, where his impatience is an exponential random variable with rate $θ$. In $s$ minutes later Person 2 will enter the system and will abandon the queue eventually too. Person 2's impatience is also an exponential random variable, is independent of Person 1's patience with rate $μ$. What is the probability that Person 1 will abandon the queue before Person 2 does?

Answer 1

Hints:

• What is the probability that Person 1 is still in the queue at time $s$?
• If they are both in the queue at the same time, what is the probability that Person 1 leaves first?

Answer 2

Let $X_1$ be the time that person 1 spends in the queue and $X_2$ the time that person 2 spends in the queue. The probability that person 1 abandons the queue before person 2 is $$ \mathbb P(X_1\leqslant s) + \mathbb P(s\leqslant X_1\leqslant s+X_2). $$ The first probability follows directly from the distribution function of $X_1$: $$ \mathbb P(X_1\leqslant s) = 1 - e^{-\theta s}. $$ For the second probability, we integrate over the joint density of $X_1$ and $X_2$: \begin{align} \mathbb P(s\leqslant X_1\leqslant s+X_2) &= \iint_{(x,y):s<x<s+y} f_{X_1,X_2}(x,y)\ \mathsf d(x\times y)\\ &= \int_0^\infty \int_s^{s+y}\theta e^{-\theta x}\mu e^{-\mu y}\ \mathsf dx \ \mathsf dy\\ &= \frac{\theta e^{-\theta -s}}{\theta +\mu }. \end{align} Hence the overall probability is $$ 1 - e^{-\theta s} + \frac{\theta e^{-\theta -s}}{\theta +\mu } = 1-\frac{\mu e^{-\theta s}}{\theta +\mu }. $$