8 Suppose a certain type of component has a lifetime distribution that is exponential with

Example 6.8 Suppose a certain type of component has a lifetime distribution that is exponential with

  parameter l so that expected lifetime is m ⫽ 1l. A sample of n such components is selected, and each is put into operation. If the experiment is continued until all n life-

  times, X 1 ,...,X n , have been observed, then is an unbiased estimator of . X m

  In some experiments, though, the components are left in operation only until

  the time of the rth failure, where r ⬍ n. This procedure is referred to as censoring.

  Let Y 1 denote the time of the first failure (the minimum lifetime among the n com-

  ponents), Y 2 denote the time at which the second failure occurs (the second smallest

  lifetime), and so on. Since the experiment terminates at time Y r , the total accumu- lated lifetime at termination is

  r

  T r 5 g Y i 1 (n 2 r)Y r

  i51

  We now demonstrate that ˆ5T r is an unbiased estimator for . To do so, we need m two properties of exponential variables:

  1. The memoryless property (see Section 4.4), which says that at any time point, remaining lifetime has the same exponential distribution as original lifetime.

  2. When X 1 ,...,X k are independent, each exponentially distributed with parame-

  ter l, min(X 1 ,...,X k ), is exponential with parameter kl.

  Since all n components last until Y 1 ,n ⫺ 1 last an additional Y 2 ⫺Y 1 ,n ⫺ 2 an addi-

  tional Y 3 ⫺Y 2 amount of time, and so on, another expression for T r is T r ⫽ nY 1 ⫹ (n ⫺ 1)(Y 2 ⫺Y 1 ) ⫹ (n ⫺ 2)(Y 3 ⫺Y 2 ) ⫹... ⫹ (n ⫺ r ⫹ 1)(Y r ⫺Y r ⫺1 )

  But Y 1 is the minimum of n exponential variables, so E(Y 1 ) ⫽ 1(nl). Similarly, Y 2 ⫺Y 1

  is the smallest of the n ⫺ 1 remaining lifetimes, each exponential with parameter l (by

  the memoryless property), so E(Y 2 ⫺Y 1 ) ⫽ 1[(n ⫺ 1)l]. Continuing, E(Y i ⫹1 ⫺Y i ) ⫽

  1[(n ⫺ i)l], so

  E(T r ) ⫽ nE(Y 1 ) ⫹ (n ⫺ 1)E(Y 2 ⫺Y 1 ) ⫹ . . . ⫹ (n ⫺ r ⫹ 1)E(Y r ⫺Y r ⫺1 )

  Therefore, E(T r r) ⫽ (1r)E(T r ) ⫽ (1r) ⭈ (rl) ⫽ 1l ⫽ m as claimed.

  As an example, suppose 20 components are tested and r ⫽ 10. Then if the first ten failure times are 11, 15, 29, 33, 35, 40, 47, 55, 58, and 72, the estimate of is m

  The advantage of the experiment with censoring is that it terminates more quickly than the uncensored experiment. However, it can be shown that V(T

  2 r r) ⫽ 1(l r), which is larger than 1(l 2 n), the variance of X in the uncensored experiment.

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  6.1 Some General Concepts of Point Estimation

  Reporting a Point Estimate: The Standard Error

  Besides reporting the value of a point estimate, some indication of its precision should

  be given. The usual measure of precision is the standard error of the estimator used.

  DEFINITION

  The standard error of an estimator is its standard deviation uˆ s u ˆ 5 1V( ˆu) . It is the magnitude of a typical or representative deviation between an estimate and the value of . If the standard error itself involves unknown parameters whose values can be estimated, substitution of these estimates into s u ˆ yields the estimated standard error (estimated standard deviation) of the estimator. The estimated standard error can be denoted either by s ˆ uˆ (the over s empha- ˆ sizes that s uˆ is being estimated) or by . s uˆ

  Example 6.9 Assuming that breakdown voltage is normally distributed, m5X ˆ is the best estima- (Example 6.2

  tor of . If the value of s is known to be 1.5, the standard error of m X is

  continued)

  s 2 X 5 s 1n 5 1.5 120 5 .335 . If, as is usually the case, the value of s is unknown, the estimate s ˆ 5 s 5 1.462 is substituted into s 2 X to obtain the esti-

  mated standard error s ˆ 2 X 5s 2 X 5 s 1n 5 1.462 120 5 .327 .

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