Just the jackknife bias estimate

Question

You compute a statistic on your full sample of $n = 25$ observations and get $\hat{\theta} = 0.50$ . The average of the $25$ leave-one-out estimates is $\bar{\theta}_{(\cdot)} = 0.52$ .

Compute the jackknife estimate of the bias (not the corrected estimate, just the bias).

Accepted Answer

The jackknife bias estimate is the gap between the leave-one-out average and the full-sample estimate, amplified by (n-1): \widehat{	ext{bias}} = (n - 1)\left(\bar{	heta}{(\cdot)} - \hat{	heta}ight). Plug in. \widehat{	ext{bias}} = 24 	imes (0.52 - 0.50) = 24 	imes 0.02 = 0.48. So the estimator appears biased upward by about 0.48. (Had the question asked for the corrected value, it would be \hat{	heta} - \widehat{	ext{bias}} = 0.50 - 0.48 = 0.02, a large adjustment relative to the original 0.50.) Takeaway. The jackknife bias estimate on its own is (n-1)\left(\bar{	heta}{(\cdot)} - …

Just the jackknife bias estimate

Your answer

Discussion

Learn the concepts

Related questions