Jackknife bias correction of an estimator

Question

You compute a statistic on your full sample of $n = 8$ observations and get $\hat{\theta} = 5.0$ . Deleting one point at a time and averaging the $8$ leave-one-out estimates gives $\bar{\theta}_{(\cdot)} = 5.1$ .

Compute the jackknife bias-corrected estimate.

Accepted Answer

Many estimators are slightly biased, and the bias typically shrinks like 1/n. The jackknife exploits this: comparing the full-sample estimate with the average leave-one-out estimate reveals the bias, amplified by the factor (n-1). Step 1, estimate the bias. \widehat{	ext{bias}} = (n - 1)\left(\bar{	heta}{(\cdot)} - \hat{	heta}ight) = 7 	imes (5.1 - 5.0) = 7 	imes 0.1 = 0.70. Step 2, subtract the bias from the original estimate. \hat{	heta}{	ext{jack}} = \hat{	heta} - \widehat{	ext{bias}} = 5.0 - 0.70 = 4.3. Equivalently, \hat{	heta}{	ext{jack}} = n\hat{	heta} - (n-1)\bar{	heta}…

Jackknife bias correction of an estimator

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