1
$\begingroup$

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 $\min(X1, X2) \sim \text{Exp}(\lambda_1 + \lambda_2)$.

$\endgroup$
6
  • 1
    $\begingroup$ We want the probability that $X_1>t$ and $X_2>t$. What is the probability that $X_1>t$? $\endgroup$ May 18, 2011 at 2:20
  • 3
    $\begingroup$ Downvote: This is proven in the Wikipedia article about exponential distributions. $\endgroup$ May 18, 2011 at 2:23
  • $\begingroup$ @user6312 So do I simply need to calculate X1>t and X2>t and then that will give me the answer if I multiply the two together? $\endgroup$
    – Cherizzle
    May 18, 2011 at 4:18
  • 1
    $\begingroup$ @Yuval Filmus I don't understand the Wikipedia article.. That is why I thought I'd ask for help.... $\endgroup$
    – Cherizzle
    May 18, 2011 at 4:33
  • $\begingroup$ @Cherizzle: Yes. By integrating, or by quoting your text, you find $\text{P}(X_1>t)=e^{-\lambda_1 t}$, and a similar expression for $\text{P}(X_2>t)$. So $\text{P}(\min(X_1,X_2)>t=e^{-(\lambda_1+\lambda_2)t}$. So the cumulative distibution function of $\min(X_1,X_2)$ is $1$ minus the thing just computed. Differentiate to get the familiar exponential density function with parameter $\lambda_1+\lambda_2$. Or just recognize that the cumulative distribution function is familiar. $\endgroup$ May 18, 2011 at 5:49

1 Answer 1

3
$\begingroup$

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, $\min \{X_1,X_2\} \sim {\rm Exp}(\lambda_1 + \lambda_2)$.

$\endgroup$
1
  • $\begingroup$ Note that this idea can be generalized to show that $\min \{ X_1 , \ldots ,X_n \} \sim {\rm Exp}(\lambda_1 + \cdots + \lambda_n)$, where $X_i$ are independent ${\rm Exp}(\lambda_i)$ variables. $\endgroup$
    – Shai Covo
    May 18, 2011 at 13:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.