Today I read **7.2 The Jackknife** in Stastical Computing with R and found the explanation for why the jackknife estimate of standard error have the factor $(n-1)/n$ is unclear. I refered to An Introduction to the Bootstrap by Bradley Efron and R. J. Tibshirani, and the slides of jackknife by Rozenn Dahyot to figure out the reason. Here is my understanding for the existence of factor $(n-1)/n$.

The jackknife samples are computed by leaving out one observation $x_i$ from a sample $\mathbf{x} = (x_1, x_2, \cdots, x_n)$ at a time:
$$
\mathbf{x}_{(i)} = (x_1, x_2, \cdots, x_{i-1}, x_{i+1}, \cdots, x_n)
$$
for $i = 1, 2, \cdots, n$. The $i$th jackknife replication $\hat{\theta}_{(i)}$ of the statistic $\hat{\theta} = s(\mathbf{x})$ is
$$
\hat{\theta}_{(i)} = s(\mathbf{x}_{(i)}), \forall i = 1, 2, \cdots, n.
$$
Thus, for $\hat{\theta} = \bar{x}$, we have
$$
\begin{equation} \nonumber
\begin{aligned}
s(\mathbf{x}_{(i)}) &= \bar{x}_{(i)} \\

&= \frac{1}{n-1}\sum_{j \ne i}{x_j} \\

&= \frac{n\bar{x} - x_i}{n-1}
\end{aligned}
\end{equation}
$$

The jackknife estimate of standard error is defined by $$ \hat{se}_{jack} = [\frac{n-1}{n}\sum_{i=1}^n{(\hat{\theta}_{(i)} - \hat{\theta}_{(\cdot)})^2}]^{½} $$ where $\hat{\theta}_{(\cdot)} = \frac{1}{n}\sum_{i=1}^n{\hat{\theta}_{(i)}}$.

The exact form of the factor $(n-1)/n$ in the above formula is derived by considering the special case $\hat{\theta} = \bar{x}$. For $\hat{\theta} = \bar{x}$, it is easy to show that
$$
\begin{equation} \nonumber
\begin{aligned}
\bar{x}_{(\cdot)} &= \frac{1}{n}\sum_{i=1}^n{\bar{x}_{(i)}} \\

&= \frac{1}{n}\sum_{i=1}^n{\frac{n\bar{x} - x_i}{n-1}} \\

&= \frac{n}{n-1}\bar{x} - \frac{1}{n-1}\frac{1}{n}\sum_{i=1}^n{x_i} \\

&= \frac{n}{n-1}\bar{x} - \frac{1}{n-1}\bar{x} \\

&= \bar{x}
\end{aligned}
\end{equation}
$$
Therefore,
$$
\begin{equation} \nonumber
\begin{aligned}
\hat{se}_{jack} &= [\frac{n-1}{n}\sum_{i=1}^n{(\hat{\theta}_{(i)} - \hat{\theta}_{(\cdot)})^2}]^{½} \\

&= [\frac{n-1}{n}\sum_{i=1}^n{(\bar{x}_{(i)} - \bar{x}_{(\cdot)})^2}]^{½} \\

&= [\frac{n-1}{n}\sum_{i=1}^n{(\frac{n\bar{x} - x_i}{n-1} - \bar{x})^2}]^{½} \\

&= [\frac{1}{n(n-1)}\sum_{i=1}^n{(x_i - \bar{x})^2}]^{½} \\

&= \frac{[\frac{1}{n-1}\sum_{i=1}^n{(x_i - \bar{x})^2}]^{½}}{\sqrt{n}} \\

&= \frac{\hat{\sigma}}{\sqrt{n}} \\

&= \hat{se}(\bar{x})
\end{aligned}
\end{equation}
$$
where $\hat{se}(\bar{x})$ is the unbiased estimate of the standard error of the sample mean, and $\hat{\sigma}$ is unbiased estimate of the standard deviation of the population.

The derivation suggests that the factor $(n-1)/n$ os exactly what is needed to make $\hat{se}_{jack}$ equal to the unbiased estimate of the standard error of the sample mean $\hat{se}(\bar{x})$. It is a somewhat arbitrary convention that $\hat{se}_{jack}$ uses the factor $(n-1)/n$, as in fact, the factor $(n-1)/n$ is derived by considering the special case $\hat{\theta} = \bar{x}$.