How to derive jackknife bias for variance and mean I am having a hard time to understand how one derives the jackknife bias for the variance and mean. 
1) Why do we need an inflation factor of $(n-1)$ when calculating the jackknife bias of the mean?
2) How does one derive the jackknife bias for the variance? 
 A: Let $\sigma^{2}$ be the variance of your random variable, and $\hat{\sigma}_{n} = \frac{1}{n}\sum_{i=1}^{n}(X_{i}-\bar{X})$ your estimate of the variance. Then it is well known fact that the bias of this estimator is $-\sigma^{2}/n$.
How does the Jackknife works?. For the sample $\left\{ X_{1},..., X_{n}\right\}$ it calculates the variances, $\hat{\sigma}_{n}^{(-i)}$, where $(-i)$ denotes that the ith measurement has been omitted. This estimator has thus a bias of $-\sigma^{2}/(n-1)$.
Then calculates its estimate of the bias as,
$$
\mathbf{bias}_{jack} = (n-1)\left(\frac{1}{n}\sum_{i}\hat{\sigma}_{n}^{(-i)}-\hat{\sigma}_{n}\right)
$$
If we calculate its expected value,
$$
E[\mathbf{bias}_{jack}] = (n-1)\left(\frac{1}{n}\sum_{i}E[\hat{\sigma}_{n}^{(-i)}]-E[\hat{\sigma}_{n}]\right) = \frac{\sigma^{2}}{n} + O\left(\frac{1}{n^{2}}\right)
$$
that is, it gives us an estimate of the actual bias up to order $O\left(\frac{1}{n^{2}}\right)$.
If your estimate is unbiased, so it is also the estimate of the Jackknife.
A: The previous answer was almost right, except it is wrong by a minus sign. 
\begin{eqnarray}
 \theta_{(\cdot )}   &=&  \frac{1}{n}  \sum_i  \theta_{ (i)}  = \frac{1  }{ n(n-1)} \sum_i \sum_{j, j \ne i} ( x_j - \bar{x}_{(i)} )^2  \\
&=& \frac{1}{ n(n-1)^3 }  \sum_{ij} [ (n-1)^2 (x_j -\bar{x} )^2  + 2 (n-1) (x_j - \bar{x}) ( x_i - \bar{x} ) + ( x_i - \bar{x} )^2   ]\\
 &-& \frac{n}{(n-1)^3 }\sum_i ( x_i -\bar{x} )^2  \\
&=& \frac{n(n-1)}{ (n-1)^2  } \hat{\sigma}^2 
\end{eqnarray} with $\bar{x} $ being the sample mean.  
Note that 
\begin{eqnarray}
  \sum_{i,j} (x_j - \bar{x}) ( x_i - \bar{x} )  &=&  \frac{1}{n^2} \sum_{i,j,k,l} (x_i - x_k) (x_j - x_l )  \\
&=&   \frac{1}{n^2}  \sum_{j,l} (x_j - x_l) \sum_{i,k}   (x_i - x_k )   = 0 . 
\end{eqnarray}
 One would get the previous wrong answer if mistakenly used  $ \langle \sum_{ij}  (x_i - \bar{x} )(  x_j -\bar{x} )  \rangle = n \langle \sum_{i}  (x_i - \bar{x}  )^2  \rangle  = n^2 \hat{\sigma}^2   $. 
Therefore we have
\begin{equation}
\langle (n-1) ( \hat{\theta}_{(\cdot)} - \hat{\theta} ) \rangle = - \frac{ 1 }{ n-1 }  \hat{\sigma}^2 = - \frac{ \sigma^2 }{ n } .
\end{equation}
This is the same as the expected bias from the sample variance estimator for $\hat{ \sigma}^2  $.  
