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

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    $\begingroup$ We want the probability that $X_1>t$ and $X_2>t$. What is the probability that $X_1>t$? $\endgroup$ – André Nicolas May 18 '11 at 2:20
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    $\begingroup$ Downvote: This is proven in the Wikipedia article about exponential distributions. $\endgroup$ – Yuval Filmus May 18 '11 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 '11 at 4:18
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    $\begingroup$ @Yuval Filmus I don't understand the Wikipedia article.. That is why I thought I'd ask for help.... $\endgroup$ – Cherizzle May 18 '11 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. $\endgroup$ – André Nicolas May 18 '11 at 5:49
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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, $\min \{X_1,X_2\} \sim {\rm Exp}(\lambda_1 + \lambda_2)$.

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  • $\begingroup$ 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 '11 at 13:27

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