A large jackknife correction with n = 20

Question

You compute a statistic on your full sample of $n = 20$ observations and get $\hat{\theta} = 100$ . The average of the $20$ leave-one-out estimates is $\bar{\theta}_{(\cdot)} = 101$ .

Compute the jackknife bias-corrected estimate.

Accepted Answer

Step 1, estimate the bias. \widehat{	ext{bias}} = (n - 1)\left(\bar{	heta}{(\cdot)} - \hat{	heta}ight) = 19 	imes (101 - 100) = 19 	imes 1 = 19. Step 2, subtract the bias from the original estimate. \hat{	heta}{	ext{jack}} = \hat{	heta} - \widehat{	ext{bias}} = 100 - 19 = 81. Notice how a modest one-unit gap between the leave-one-out average and the full-sample estimate becomes a 19-unit correction, because the factor (n-1) = 19 amplifies it. When the leave-one-out estimates barely move, that is a cue to double-check: a large correction from a tiny gap can signal an unstable estima…

A large jackknife correction with n = 20

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