Jackknife standard error with three observations

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You jackknife a statistic over n = 3 observations, giving three leave-one-out estimates: 6, \quad 9, \quad 12. 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{6 + 9 + 12}{3} = 9. Step 2, sum of squared deviations. (-3)^2 + 0^2 + (3)^2 = 9 + 0 + 9 = 18. Step 3, apply 	frac{n-1}{n} = 	frac{2}{3} and take the root. 	ext{SE}{	ext{jack}} = \sqrt{\frac{2}{3} 	imes 18} = \sqrt{12} \approx 3.46. With only three points the jackknife is crude, but the mechanics are identical to any larger n. Takeaway. The 	frac{n-1}{n} factor is small when n is small (here 	…

Jackknife standard error with three observations

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