Question

Let X1, ... , Xn be independent exponential random variables having a common parameter λ. Determine the distribution of min(X1, ... , Xn)

Short answer

The minimum distribution is$Z~Expo\left(n\lambda \right)$

Step-by-step solution

Content Introduction

In a Poisson point process, when events occur continuously and independently at a constant average rate, the exponential distribution is the probability distribution of the time between occurrences. It's an example of the gamma distribution in action.

Content Explanation

Define random variable

$Z=min(X_{1,}{X}_{2,}......X_n)$

Because of X_i > 0 we have that Z > 0. We have that

$$\begin{gathered} P(Z\le z)=1-P(Z>z) \\ =1-P(X_1>z,......,X_n>z) \\ =1\prod _{i=1}^{n}P\left(\right(X_1>z) \\ =1-P{(X_1>z)}^n=1-e^{-n\lambda z} \end{gathered}$$

So, we see that$Z~Expo\left(n\lambda \right)$