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Tony Tsai

May the force of science be with you

Apr 23, 2015 - 2 minute read - Comments - Statistical Computing

Why Does the Jackknife Estimate of Standard Error Have the Factor (n-1)/n?

Today I read 7.2 The Jackknife in Stastical Computing with R and found the explanation for why the jackknife estimate of standard error have the factor (n1)/n is unclear. I refered to An Introduction to the Bootstrap by Bradley Efron and R. J. Tibshirani, and the slides of jackknife by Rozenn Dahyot to figure out the reason. Here is my understanding for the existence of factor (n1)/n.

The jackknife samples are computed by leaving out one observation xi from a sample x=(x1,x2,,xn) at a time: x(i)=(x1,x2,,xi1,xi+1,,xn) for i=1,2,,n. The ith jackknife replication ˆθ(i) of the statistic ˆθ=s(x) is ˆθ(i)=s(x(i)),i=1,2,,n. Thus, for ˆθ=ˉx, we have s(x(i))=ˉx(i) =1n1jixj =nˉxxin1

The jackknife estimate of standard error is defined by ^sejack=[n1nni=1(ˆθ(i)ˆθ())2]1/2 where ˆθ()=1nni=1ˆθ(i).

The exact form of the factor (n1)/n in the above formula is derived by considering the special case ˆθ=ˉx. For ˆθ=ˉx, it is easy to show that ˉx()=1nni=1ˉx(i) =1nni=1nˉxxin1 =nn1ˉx1n11nni=1xi =nn1ˉx1n1ˉx =ˉx Therefore, ^sejack=[n1nni=1(ˆθ(i)ˆθ())2]1/2 =[n1nni=1(ˉx(i)ˉx())2]1/2 =[n1nni=1(nˉxxin1ˉx)2]1/2 =[1n(n1)ni=1(xiˉx)2]1/2 =[1n1ni=1(xiˉx)2]1/2n =ˆσn =^se(ˉx) where ^se(ˉx) is the unbiased estimate of the standard error of the sample mean, and ˆσ is unbiased estimate of the standard deviation of the population.

The derivation suggests that the factor (n1)/n os exactly what is needed to make ^sejack equal to the unbiased estimate of the standard error of the sample mean ^se(ˉx). It is a somewhat arbitrary convention that ^sejack uses the factor (n1)/n, as in fact, the factor (n1)/n is derived by considering the special case ˆθ=ˉx.

Tags: Jackknife Standard Error Resampling

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