Jackknife standard error from leave-one-out estimates

Question

You estimate a statistic by the jackknife: with $n = 4$ observations, you delete one point at a time and recompute the statistic on the remaining three, giving four leave-one-out estimates:

$\hat{\theta}_{(1)} = 10, \quad \hat{\theta}_{(2)} = 12, \quad \hat{\theta}_{(3)} = 14, \quad \hat{\theta}_{(4)} = 16.$

Compute the jackknife estimate of the standard error.

Accepted Answer

The jackknife standard error takes the spread of the leave-one-out estimates and scales it up by the factor 	frac{n-1}{n}, which corrects for the fact that leave-one-out estimates differ from one another only slightly: 	ext{SE}{	ext{jack}} = \sqrt{\frac{n-1}{n} \sum{i=1}^{n} \left(\hat{	heta}{(i)} - \bar{	heta}{(\cdot)}ight)^2}. Step 1, mean of the leave-one-out estimates. \bar{	heta}{(\cdot)} = \frac{10 + 12 + 14 + 16}{4} = 13. Step 2, sum of squared deviations. (-3)^2 + (-1)^2 + (1)^2 + (3)^2 = 9 + 1 + 1 + 9 = 20. Step 3, apply the 	frac{n-1}{n} factor and take the square root. 	e…

Jackknife standard error from leave-one-out estimates

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