Minimum of two exponentially distributed random variables

Question

Let $X_1 ~ \sim \text{Exp}(\lambda_1)$ and $X_2 \sim \text{Exp}(\lambda_2)$ be two independent exponentially distributed random variables.

Show that $\mathbb{P}\{\min(X_1, X_2) > t\} = \exp(-(\lambda_1 + \lambda_2)t)$, and hence that $\min(X1, X2) \sim \text{Exp}(\lambda_1 + \lambda_2)$.

Answer 1

It is instructive to make the following observation, giving intuition for the result.

Let $N_1$ and $N_2$ be independent Poisson processes with rates $\lambda_1$ and $\lambda_2$, respectively. Let $X_i$, $i=1,2$, denote the waiting time until the first event to occur in the process $N_i$. Then $X_i \sim {\rm Exp}(\lambda_i)$, with $X_1$ and $X_2$ independent. Next, let $N = N_1 + N_2$. The process $N$ is a Poisson process with rate $\lambda_1 + \lambda_2$. Finally, let $Z$ be the waiting time until the first event to occur in the process $N$. Thus, on the one hand, $Z \sim {\rm Exp}(\lambda_1 + \lambda_2)$, and on the other hand, $Z = \min \{X_1,X_2\}$. Hence, $\min \{X_1,X_2\} \sim {\rm Exp}(\lambda_1 + \lambda_2)$.