Jackknife standard error with five observations

Question

You jackknife a statistic over n = 5 observations, deleting one at a time and recomputing, which gives five leave-one-out estimates: 2.0, \quad 2.5, \quad 3.0, \quad 3.5, \quad 4.0. Compute the jackknife estimate of the standard error.

Accepted Answer

Use 	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{2.0 + 2.5 + 3.0 + 3.5 + 4.0}{5} = 3.0. Step 2, sum of squared deviations. (-1)^2 + (-0.5)^2 + 0^2 + (0.5)^2 + (1)^2 = 1 + 0.25 + 0 + 0.25 + 1 = 2.5. Step 3, apply 	frac{n-1}{n} = 	frac{4}{5} and take the root. 	ext{SE}{	ext{jack}} = \sqrt{\frac{4}{5} 	imes 2.5} = \sqrt{2.0} \approx 1.41. Takeaway. Same recipe at any n: spread of the leave-one-out estimates times 	frac{n-1}{n}, then square root.

Jackknife standard error with five observations

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