Introduction to Probability and Statistics for Engineers and Scientists

14.3The Exponential Distribution in Life Testing 591

From Equation 14.3.8, we obtain that the likelihood of the datar,x 1 ,...,xris as
follows:

f(r,x 1 ,...,xr|θ)

=fX 1 ,...,Xr(x 1 ,...,xr)P

{

Xr+ 1 >T−

∑r

i= 1

xi

}

,

∑r

i= 1

xi<T

=

1

θr

e−

ir= 1 xi/θ

e−(T−

ri= 1 xi)/θ

=

1

θr

e−T/θ

Therefore,

logf(r,x 1 ,...,xr|θ)=−rlogθ−

T

θ

and so

∂

∂θ

logf(r,x 1 ,...,xr|θ)=−

r

θ

+

T

θ^2

On equating to 0 and solving, we obtain that the maximum likelihood estimate forθis

θˆ=T

r

SinceTis the total time on test of all items, it follows once again that the maximum
likelihood estimate of the unknown exponential mean is equal to the total time on test
divided by the number of observed failures in this time.
If we letN(T) denote the number of failures by timeT, then the maximum likelihood
estimator ofθisT/N(T). Suppose now that the observed value ofN(T)isN(T)=r.To
determine a 100(1−α) percent confidence interval estimate forθ, we will first determine
the valuesθLandθU, which are such that

PθU{N(T)≥r}=

α

2

, PθL{N(T)≤r}=

α

2

where byPθ(A) we mean that we are computing the probability of the eventAunder the
supposition thatθis the true mean. The 100(1−α) percent confidence interval estimate
forθis

θ∈(θL,θU)

To understand why those values ofθfor which eitherθ<θLorθ>θU are not
included in the confidence interval, note thatPθ{N(T)≥r}decreases andPθ{N(T)≤r}