Jackknife correction when leave-one-out drifts down

Question

You compute a statistic on your full sample of $n = 5$ observations and get $\hat{\theta} = 2.00$ . The average of the $5$ leave-one-out estimates is $\bar{\theta}_{(\cdot)} = 1.96$ .

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) = 4 	imes (1.96 - 2.00) = 4 	imes (-0.04) = -0.16. Step 2, subtract the bias from the original estimate. \hat{	heta}{	ext{jack}} = \hat{	heta} - \widehat{	ext{bias}} = 2.00 - (-0.16) = 2.16. Check with the one-line form: n\hat{	heta} - (n-1)\bar{	heta}{(\cdot)} = 5(2.00) - 4(1.96) = 10 - 7.84 = 2.16. Here the leave-one-out average dipped below the full estimate, so the estimated bias is negative and the correction nudges the estimate upward. Takeaway. The direction of the correc…

Jackknife correction when leave-one-out drifts down

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