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7-2

7-2.5 Bootstrap Estimate of the Standard Error (CD Only)

There are situations in which the standard error of the point estimator is unknown. Usually,
ˆ
these are cases where the form of is complicated, and the standard expectation and variance
operators are difficult to apply. A computer-intensive technique called the bootstrap that was
developed in recent years can be used for this problem.
Suppose that we are sampling from a population that can be modeled by the probability
ˆ

distribution f 1x; 2 . The random sample results in data values x , x , p , x n and we obtain as
2
1

the point estimate of . We would now use a computer to obtain bootstrap samples from the
ˆ
distribution f 1x; 2 , and for each of these samples we calculate the bootstrap estimate ˆ* of .
This results in
Bootstrap Sample Observations Bootstrap Estimate
ˆ *
*
*
1 x 1 , x 2 , p , x n * 1
*
ˆ *
*
2 x 1 , x 2 , p , x n * 2
o o o
*
*
B x 1 , x 2 , p , x n * ˆ *
B
B ˆ
Usually B 100 or 200 of these bootstrap samples are taken. Let * 11 B2 g i 1 * i be the
ˆ
sample mean of the bootstrap estimates. The bootstrap estimate of the standard error of is
ˆ *
just the sample standard deviation of the , or
i
B
a 1 i 2
ˆ *
* 2
i 1 (S7-1)
s ˆ

R B 1
In the bootstrap literature, B 1 in Equation S7-1 is often replaced by B. However, for
the large values usually employed for B, there is little difference in the estimate produced
for .
s ˆ

EXAMPLE S7-1 The time to failure of an electronic module used in an automobile engine controller is tested
at an elevated temperature in order to accelerate the failure mechanism. The time to failure
is exponentially distributed with unknown parameter
. Eight units are selected at random
11.96, x 5.03, x 67.40,
and tested, with the resulting failure times (in hours): x 1 2 3
x 16.07, x 31.50, x 7.73, x 11.10, and x 22.38. Now the mean of an expo-
4
5
8
7
6
nential distribution is 1
, so E(X) 1
, and the expected value of the sample average
ˆ
is E1X2 1 . Therefore, a reasonable way to estimate
is with 1 X . For our sample,
ˆ
x 21.65 , so our estimate of is 1 21.65 0.0462 . To find the bootstrap standard error
we would now obtain B 200 (say) samples of n 8 observations each from an exponential
distribution with parameter
0.0462. The following table shows some of these results:
Bootstrap Sample Observations Bootstrap Estimate
ˆ *
1 8.01, 28.85, 14.14, 59.12, 3.11, 32.19, 5.26, 14.17 1 0.0485
ˆ *
2 33.27, 2.10, 40.17, 32.43, 6.94, 30.66, 18.99, 5.61 2 0.0470
o o o
ˆ *
200 40.26, 39.26, 19.59, 43.53, 9.55, 7.07, 6.03, 8.94 200 0.0459