The sample is x=(1,2,4) with n=3

The sample is $x=(1,2,4)$ with $n=3$. The functional of interest is the squared sample mean \[ \hat\theta=\bar x^2. \] Write $\hat\theta_{(i)}$ for the same functional computed after deleting the $i$th observation, and write $\bar\theta_{(\cdot)}=n^{-1}\sum_i\hat\theta_{(i)}$ for the leave-one-out mean. 1. Prove $\bar x=7/3$ and $\hat\theta=49/9$. Compute the three leave-one-out values \[ \hat\theta_{(1)}=9,\qquad \hat\theta_{(2)}=\frac{25}{4},\qquad \hat\theta_{(3)}=\frac94, \] and their mean $\bar\theta_{(\cdot)}=35/6$. 2. The jackknife bias estimate is $(n-1)\bigl(\bar\theta_{(\cdot)}-\hat\theta\bigr)$. Prove that this equals $7/9$, and that the bias-corrected estimator $\hat\theta-\widehat{\mathrm{bias}}$ equals $14/3$. 3. The pseudovalues are $\tilde\theta_i=n\hat\theta-(n-1)\hat\theta_{(i)}$. Prove \[ \bigl(\tilde\theta_1,\tilde\theta_2,\tilde\theta_3\bigr) =\Bigl(-\frac53,\frac{23}{6},\frac{71}{6}\Bigr) \] and that their average is again $14/3$. The jackknife variance is \[ \widehat V =\frac{n-1}{n}\sum_{i=1}^n \bigl(\hat\theta_{(i)}-\bar\theta_{(\cdot)}\bigr)^2 =\frac{1}{n(n-1)} \sum_{i=1}^n\bigl(\tilde\theta_i-\bar{\tilde\theta}\bigr)^2. \] Prove $\widehat V=553/36$. 4. Audit both sentences: (i) “the bias factor is $n$ rather than $n-1$, or the functional is the unsquared mean, giving bias $7/18$ or $0$”; (ii) “the variance uses $(n-1)/n$ times the sum of *pseudovalue* squared deviations, giving $553/9$, or this is a nonparametric bootstrap of the sample maximum.” Stay inside the delete-one jackknife of $\bar x^2$. Do not replace the functional by $\bar x$ or by a bootstrap of the maximum.

$\bar x=7/3$ and $\hat\theta=49/9$. The three leave-one-out means are $3$, $5/2$, and $3/2$.