Jackknife bias correction with n = 10

Question

You compute a statistic on your full sample of $n = 10$ observations and get $\hat{\theta} = 8.0$ . The average of the $10$ leave-one-out estimates is $\bar{\theta}_{(\cdot)} = 7.9$ .

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) = 9 	imes (7.9 - 8.0) = 9 	imes (-0.1) = -0.9. Step 2, subtract the bias from the original estimate. \hat{	heta}{	ext{jack}} = \hat{	heta} - \widehat{	ext{bias}} = 8.0 - (-0.9) = 8.9. Check with the one-line form: n\hat{	heta} - (n-1)\bar{	heta}{(\cdot)} = 10(8.0) - 9(7.9) = 80 - 71.1 = 8.9. Takeaway. Whatever the numbers, the corrected estimate is n\hat{	heta} - (n-1)\bar{	heta}{(\cdot)}, which subtracts the amplified leave-one-out drift from the original estimate.

Jackknife bias correction with n = 10

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