Introduction to Probability and Statistics for Engineers and Scientists

14.3The Exponential Distribution in Life Testing 595

Hence,

∂

∂θ

logf(i 1 ,...,ir,x 1 ,...,xr)=−

r

θ

+

∑r

1

xi+(n−r)T

θ^2

Equating to 0 and solving yields thatθˆ, the maximum likelihood estimate, is given by

θˆ=

∑r

i= 1

xi+(n−r)T

r

Hence, if we letRdenote the number of items that fail by timeTand letX(i)be the
ith smallest of the failure times,i=1,...,R, then the maximum likelihood estimator
ofθis

θˆ=

∑R

i= 1

X(i)+(n−R)T

R
Letτdenote the sum of the times that all items are on life test. Then, because theR
items that fail are on test for timesX(1),...,X(R)whereas then−Rnonfailed items are all
on test for timeT, it follows that

τ=

∑R

i= 1

X(i)+(n−R)T

and thus we can write the maximum likelihood estimator as

θˆ=τ

R

In words, the maximum likelihood estimator of the mean life is (as in the life testing
procedures of Sections 14.3.1 and 14.3.2) equal to the total time on test divided by the
number of items observed to fail.

REMARK

As the reader may possibly have surmised, it turns out that for all possible life test-
ing schemes for the exponential distribution, the maximum likelihood estimator of the
unknown meanθwill always be equal to the total time on test divided by the number of
observed failures. To see why this is true, consideranytesting situation and suppose that
the outcome of the data is thatritems are observed to fail after having been on test for
timesx 1 ,...,xr, respectively, and thatsitems have not yet failed when the test ends — at