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Chapter 6: Q.6.9 (page 275)

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

Short Answer

Expert verified

The minimum distribution isZ~Expo(nλ)

Step by step solution

01

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.

02

Content Explanation

Define random variable

Z=min(X1,X2,......Xn)

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

P(Z≤z)=1-P(Z>z)=1-P(X1>z,......,Xn>z)=1∏i=1nP((X1>z)=1-P(X1>z)n =1-e-nλz

So, we see thatZ~Expo(nλ)

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