let $x=(x_1,\cdots,x_n) $ are n independent samples from unknown distribution. The Jackknife Samples are selected by taking the original data vector and deleting one observation from the set. Thus, there are n unique Jackknife samples, and the ith Jackknife sample vector is defined as:
$$x_{-i}=(x_1,\cdots,x_{i-1},x_{i+1},\cdots,x_n) $$
The ith Jackknife Replicate is defined as the value of the estimator $\hat{\theta}= s(x)$ evaluated at the ith Jackknife sample. $$\hat{\theta}_{-i}= s(x_{-i})$$ The Jackknife Standard Error is defined as $$\hat{se}=\sqrt{\frac{n-1}{n}\sum_{i=1}^{n} (\hat{\theta}_{-i}-\hat{\theta}_{.})^2}$$ where $$\hat{\theta}_{.}=\frac{1}{n}\sum_{i=1}^{n}\hat{\theta}_{-i}$$ The Jackknife Bias is defined as $$\hat{bias}_{jack}=(n-1)(\hat{\theta}_{.}- \hat{\theta})$$ where $\hat{\theta}=s(x)$ is the estimator taking the entire sample as argument.
The question is where come from the factor "n-1"? or prove
$$(n-1)\left(\frac{1}{n} \sum_{i=1}^{n}\hat{\sigma^2_{-i}}-\hat{\sigma^2}\right)=-\frac{S^2}{n}$$ where
$$\hat{\sigma^2}=\frac{1}{n}\sum_{i=1}^{n}(x_i-\bar{x})^2$$ $$\hat{\sigma_{-i}^2}=\frac{1}{n-1}\sum_{j=1\neq i}^{n}(x_j-\bar{x}_{-i})^2$$ $$S^2=\frac{1}{n-1}\sum_{i=1}^{n}(x_i-\bar{x})^2$$
I read this http://people.bu.edu/aimcinto/jackknife.pdf and it say:
I consider as an estimator the uncorrected variance of the sample: $$s(x)=\frac{1}{n}\sum_{i=1}^{n}(x_i-\bar{x})^2$$ If I take as my estimator $\hat{\theta}=s(x)$ the (biased) sample variance, I have that its bias, that is, $E[\hat{\theta}-\theta]$, is equal to $-\frac{\sigma^2}{n}$. If I use the Jackknife bias as an estimate for the bias of my estimator, and I have that my estimator $\hat{\theta})$ is equal to the uncorrected sample variance, then the Jackknife bias formula reduces to $-\frac{S^2}{n}$, where $S^2$ is now the regular, corrected, unbiased estimator of sample variance. Thus, the bias here is constructed from a heuristic notion to emulate the bias of the uncorrected sample variance.
My try
$$(n-1)(\hat{\sigma^2_{.}}-\hat{\sigma^2})=(n-1)\left(\frac{1}{n}\sum_{i=1}^{n} \hat{\sigma^2_{-i}} - \frac{1}{n}\sum_{i=1}^{n}(x_i-\bar{x})^2 \right)$$
$$=\frac{n-1}{n}\sum_{i=1}^{n}\left( \hat{\sigma^2_{-i}}- (x_i-\bar{x})^2\right)$$
$$=\frac{n-1}{n}\sum_{i=1}^{n}\left( \frac{1}{n-1}\sum_{j=1,j\neq i}(x_j-\bar{x}_{-i})^2- (x_i-\bar{x})^2\right)$$
$$=\frac{n-1}{n}\sum_{i=1}^{n}\left(\frac{1}{n-1} \sum_{j=1}^{n} (x_j-\bar{x}_{-i})^2-\frac{1}{n-1}(x_i-\bar{x}_{-i})^2- (x_i-\bar{x})^2\right)$$ since $$\bar{x}_{-i}=\frac{n \bar{x}-x_i}{n-1}$$
$$=\frac{n-1}{n}\sum_{i=1}^{n}\left( \frac{1}{n-1}\sum_{j=1}^n ( x_j-\frac{n \bar{x}-x_i}{n-1})^2-\frac{1}{n-1}(x_i-\frac{n \bar{x}-x_i}{n-1})^2- (x_i-\bar{x})^2\right)$$
$$=\frac{1}{n}\sum_{i=1}^{n} \sum_{j=1}^n ( x_j-\frac{n \bar{x}-x_i}{n-1})^2-\frac{1}{n}\sum_{i=1}^{n}(x_i-\frac{n \bar{x}-x_i}{n-1})^2-\frac{n-1}{n}\sum_{i=1}^{n} (x_i-\bar{x})^2$$
$$=\frac{1}{n}\sum_{i=1}^{n} \sum_{j=1}^n ( x_j-\frac{n \bar{x}-x_i}{n-1})^2-\frac{1}{n}\sum_{i=1}^{n}(\frac{nx_i-n \bar{x}}{n-1})^2-\frac{n-1}{n}(n-1)S^2$$
$$=\frac{1}{n}\sum_{i=1}^{n} \sum_{j=1}^n ( x_j-\frac{n \bar{x}-x_i}{n-1})^2-\frac{1}{n}\sum_{i=1}^{n}\frac{n^2}{(n-1)^2}(x_i- \bar{x})^2-\frac{n-1}{n}(n-1)S^2$$
$$=\frac{1}{n}\sum_{i=1}^{n} \sum_{j=1}^n ( x_j-\frac{n \bar{x}-x_i}{n-1})^2-\frac{n}{(n-1)^2}\sum_{i=1}^{n}(x_i- \bar{x})^2-\frac{n-1}{n}(n-1)S^2$$
$$=\frac{1}{n}\sum_{i=1}^{n} \sum_{j=1}^n ( x_j-\frac{n \bar{x}-x_i}{n-1})^2-\frac{n}{(n-1)^2}(n-1)S^2-\frac{n-1}{n}(n-1)S^2$$
$$=\frac{1}{n}\sum_{i=1}^{n} \sum_{j=1}^n ( x_j-\frac{n \bar{x}-x_i}{n-1})^2-\frac{n}{(n-1)}S^2-\frac{n-1}{n}(n-1)S^2$$
$$=\frac{1}{n}\sum_{i=1}^{n} \sum_{j=1}^n ( x_j-\frac{n \bar{x}-x_i}{n-1})^2-S^2 \left(\frac{n}{n-1}+\frac{(n-1)^2}{n}\right)$$
$$=\frac{1}{n}\sum_{i=1}^{n} \sum_{j=1}^n ( x_j-\frac{n \bar{x}-x_i}{n-1})^2-S^2 \left(\frac{n^2+(n-1)^3}{n(n-1)}\right)$$
$$=\frac{1}{n}\sum_{i=1}^{n} \sum_{j=1}^n ( x_j-\frac{n \bar{x}-x_i}{n-1})^2-S^2 \left(\frac{n^2+n^3-3n^2+3n-1}{n(n-1)}\right)$$
$$=\frac{1}{n}\sum_{i=1}^{n} \sum_{j=1}^n ( x_j-\frac{n \bar{x}-x_i}{n-1})^2-S^2 \left(\frac{n^3-2n^2+3n-1}{n(n-1)}\right)$$
if I calculate $$\frac{1}{n}\sum_{i=1}^{n} \sum_{j=1}^n ( x_j-\frac{n \bar{x}-x_i}{n-1})^2$$ it is done! $$=\frac{1}{n}\sum_{i=1}^{n} \sum_{j=1}^n (\frac{(n-1)(x_j -\bar{x})+(x_i-\bar{x})}{n-1})^2$$
$$=\frac{1}{n}\sum_{i=1}^{n} \sum_{j=1}^n ((x_j -\bar{x})+\frac{(x_i-\bar{x})}{n-1})^2$$
$$=\frac{1}{n}\sum_{i=1}^{n} \sum_{j=1}^n ((x_j -\bar{x})^2+\frac{(x_i-\bar{x})^2}{(n-1)^2})$$
$$=\frac{1}{n}\sum_{i=1}^{n} n (x_j -\bar{x})^2+\frac{1}{n}\sum_{j=1}^{n} n\frac{(x_i-\bar{x})^2}{(n-1)^2}+0$$
$$=(n-1)S^2+\frac{S^2}{n-1}=S^2\frac{(n-1)^2+1}{n-1}=S^2\frac{n^2-2n+2}{n-1}$$
so
$$(n-1)(\hat{\sigma^2_{.}}-\hat{\sigma^2})=S^2\frac{n^2-2n+2}{n-1}-S^2\frac{n^3-2n^2+3n-1}{n(n-1)}$$
$$(n-1)(\hat{\sigma^2_{.}}-\hat{\sigma^2})=S^2\left(\frac{n^3-2n^2+2n-n^3+2n^2-3n+1}{n(n-1)}\right)$$
$$(n-1)(\hat{\sigma^2_{.}}-\hat{\sigma^2})=S^2\left(\frac{-n+1}{n(n-1)}\right)=-\frac{S^2}{n}$$
It is done. is these calculations correct?(it seems)