The 0.632 rule

Topic: Estimating error rates · more questions on this topic
‹ Test-set size Jackknife of the mean ›

Compute 1-(1-1/n)^n for n=5 and n=20, and show that it tends to 1-1/e. If \mathrm{err}=0.05 and \widehat{\mathrm{Err}}^{(\mathrm{BOOT})}=0.20, compute the .632 estimate.

Worked answer (click to show)

A point is absent from a bootstrap sample only if every draw misses it. For n=5,

1-\Bigl(1-\frac15\Bigr)^5=1-(0.8)^5=1-0.32768=0.672.

For n=20,

1-\Bigl(1-\frac{1}{20}\Bigr)^{20}=1-(0.95)^{20}=0.642.

Since (1-1/n)^n\to e^{-1}, the inclusion probability tends to 1-e^{-1}=0.632. The .632 estimate is

0.368(0.05)+0.632(0.20)=0.0184+0.1264=0.145.