Jackknife of the mean

Topic: Estimating error rates · more questions on this topic
‹ The 0.632 rule Costs of cross-validation ›

For \hat\theta=\bar x show that \hat\theta_{(i)}=\bar x+(\bar x-x_i)/(n-1), that the jackknife bias is 0, and that \widehat{\mathrm{se}}_{\mathrm{jack}}=s/\sqrt{n} with s^2 the unbiased sample variance.

Worked answer (click to show)

Leaving out x_i gives

\hat\theta_{(i)}=\frac{n\bar x-x_i}{n-1} =\bar x+\frac{\bar x-x_i}{n-1},

because n\bar x-x_i=(n-1)\bar x+(\bar x-x_i). Averaging over i, \sum_i(\bar x-x_i)=0, so \hat\theta_{(\cdot)}=\bar x=\hat\theta and

\widehat{\mathrm{bias}}_{\mathrm{jack}}=(n-1)(\hat\theta_{(\cdot)}-\hat\theta)=0.

Next,

\sum_{i=1}^n\bigl(\hat\theta_{(i)}-\hat\theta_{(\cdot)}\bigr)^2 =\sum_{i=1}^n\frac{(\bar x-x_i)^2}{(n-1)^2} =\frac{(n-1)s^2}{(n-1)^2} =\frac{s^2}{n-1},

where \sum_i(\bar x-x_i)^2=(n-1)s^2. The jackknife standard error is

\widehat{\mathrm{se}}_{\mathrm{jack}} =\Biggl(\frac{n-1}{n}\cdot\frac{s^2}{n-1}\Biggr)^{1/2} =\frac{s}{\sqrt{n}}.

The jackknife reproduces the usual standard error of the mean.