Introduction to Probability and Statistics for Engineers and Scientists

14.3The Exponential Distribution in Life Testing 587

In general, we have

Y 1 =nX(1)

Y 2 =(n−1)(X(2)−X(1))

..

.

Yj=(n−j+1)(X(j)−X(j−1))

..

.

Yr=(n−r+1)(X(r)−X(r−1))

and

τ=

∑r

j= 1

Yj

The importance of the foregoing representation forτfollows from the fact that the
distributions of theYj’s are easily obtained as follows. SinceX(1), the time of the first
failure, is the minimum ofnindependent exponential lifetimes, each having rate 1/θ,it
follows from Proposition 5.6.1 that it is itself exponentially distributed with raten/θ. That
is,X(1)is exponential with meanθ/n; and sonX(1)is exponential with meanθ. Also, at the
moment when the first failure occurs, the remainingn−1 functioning items are, by the
memoryless property of the exponential, as good as new and so each will have an additional
life that is exponential with meanθ; hence, the additional time until one of them fails is
exponential with rate (n−1)/θ. That is, independent ofX(1),X(2)−X(1)is exponential
with meanθ/(n−1) and soY 2 =(n−1)(X(2)−X(1)) is exponential with meanθ. Indeed,
continuing this argument leads us to the following conclusion:

Y 1 ,...,Yrare independent exponential

random variables each having meanθ (14.3.4)

Hence, since the sum of independent and identically distributed exponential random
variables has a gamma distribution (Corollary 5.7.2), we see that

τ∼gamma(r,1/θ)

That is,τhas a gamma distribution with parametersrand 1/θ. Equivalently, by recalling
that a gamma random variable with parameters (r,1/θ) is equivalent toθ/2 times a
chi-square random variable with 2rdegrees of freedom (see Section 5.8.1), we obtain that

2 τ

θ

∼χ 22 r (14.3.5)