Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
1  
We want the probability that $X_1>t$ and $X_2>t$. What is the probability that $X_1>t$? –  André Nicolas May 18 '11 at 2:20
3  
Downvote: This is proven in the Wikipedia article about exponential distributions. –  Yuval Filmus May 18 '11 at 2:23
    
@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? –  Cherizzle May 18 '11 at 4:18
1  
@Yuval Filmus I don't understand the Wikipedia article.. That is why I thought I'd ask for help.... –  Cherizzle May 18 '11 at 4:33
    
@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. –  André Nicolas May 18 '11 at 5:49

1 Answer 1

up vote 2 down vote accepted

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

share|improve this answer
    
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. –  Shai Covo May 18 '11 at 13:27

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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