Introduction to Probability and Statistics for Engineers and Scientists

596 Chapter 14*:Life Testing

which time they had been on test for respective timesy 1 ,...,ys. The likelihood of this
outcome will be

likelihood=K

1

θ

e−x^1 /θ...

1

θ

e−xr/θe−y^1 /θ...e−ys/θ

=

K

θr

exp











−

(∑r

i= 1

xi+

∑s

i= 1

yi

)

θ











(14.3.9)

whereK, which is a function of the testing scheme and the data, does not depend onθ.
(For instance,Kmay relate to a testing procedure in which the decision as to when to stop
depends not only on the observed data but is allowed to be random.) It follows from the
foregoing that the maximum likelihood estimate ofθwill be

θˆ=

∑r

i= 1

xi+

∑s

i= 1

yi

r

(14.3.10)

But

∑r

i= 1 xi+

∑s
i= 1 yiis just the total-time-on-test statistic and so the maximum like-
lihood estimator ofθis indeed the total time on test divided by the number of observed
failures in that time.
The distribution ofτ/Ris rather complicated for the life testing scheme described in
this section* and thus we will not be able to easily derive a confidence interval estimator for
θ. Indeed, we will not further pursue this problem but rather will consider the Bayesian
approach to estimatingθ.

14.3.4 The Bayesian Approach

Suppose that items having independent and identically distributed exponential lifetimes
with an unknown meanθare put on life test. Then as noted in the remark given in Section
14.3.3, the likelihood of the data can be expressed as

f(data|θ)=

K

θr

e−t/θ

wheretis the total time on test — that is, the sum of the time on test of all items used —
andris the number of observed failures for the given data.
Letλ=1/θdenote the rate of the exponential distribution. In the Bayesian approach,
it is more convenient to work with the rateλrather than its reciprocal. From the

* For instance, for the scheme considered,τandRare not only both random but are also dependent.