Jackknife standard error for a proportion-like statistic

Question

You jackknife a statistic that lives between $0$ and $1$ over $n = 4$ observations, giving four leave-one-out estimates:

$0.10, \quad 0.20, \quad 0.30, \quad 0.40.$

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{0.10 + 0.20 + 0.30 + 0.40}{4} = 0.25. Step 2, sum of squared deviations. (-0.15)^2 + (-0.05)^2 + (0.05)^2 + (0.15)^2 = 0.0225 + 0.0025 + 0.0025 + 0.0225 = 0.05. Step 3, apply 	frac{n-1}{n} = 	frac{3}{4} and take the root. 	ext{SE}{	ext{jack}} = \sqrt{\frac{3}{4} 	imes 0.05} = \sqrt{0.0375} \approx 0.194. Takeaway. The jackknife formula is scale-free: whether the estimates are in the hundreds or…

Jackknife standard error for a proportion-like statistic

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