602 Chapter 14*:Life Testing
or, equivalently,
αˆ=
n
∑n
i= 1
x
βˆ
i
n+βˆlog
(n
∏
i= 1
xi
)
=
nβˆ
∑n
i= 1
x
βˆ
i logxi
∑n
i= 1
x
βˆ
i
This latter equation can then be solved numerically forβˆ, which will then also determineαˆ.
However, ratherthanpursuingthisapproachanyfurther, letusconsiderasecondapproach,
which is not only computationally easier but appears, as indicated by a simulation study,
to yield more accurate estimates.
14.5.1 Parameter Estimation by Least Squares
LetX 1 ,...,Xnbe a sample from the distribution
F(x)= 1 −e−αx
β
, x≥ 0
Note that
log(1−F(x))=−αxβ
or
log
(
1
1 −F(x)
)
=αxβ
and so
log log
(
1
1 −F(x)
)
=βlogx+logα (14.5.3)
Now letX(1)<X(2)< ···<X(n)denote the ordered sample values — that is, for
i=1,....n,
X(i)=ith smallest ofX 1 ,...,Xn
and suppose that the data results inX(i)=x(i). If we were able to approximate the quantities
log log(1/[ 1 −F(x(i))]) — say, by the valuesy 1 ,...,yn— then from Equation 14.5.3, we