Confusion about the jack-knife Suppose that I am trying to use the jack-knife to estimate the variance of some estimator $E$. If I have $n$ data points, I begin by computing $n$ estimates (call them $B_1$, ..., $B_n$), each obtained from leaving one of the $n$ data points out of my sample. If I understand it correctly, the Wikipedia page then suggests that my estimate of the variance should be
$$\hat{Var(E)} = \frac{\sum_{i = 1}^n (B_i - \bar{B_i})^2}{n-1}$$
where $\bar{B_i}$ is the average of my $n$ estimates. The standard error is then $\sqrt(\hat{Var(E)})$.
On the other hand, the notes here seem to suggest
$$\hat{Var(E)} = \sum_{i = 1}^n (B_i - \bar{B_i})^2\left(\frac{n-1}{n}\right)$$
What am I missing here? And which formula should I use? Many thanks in advance!
Extra question: if the answer is indeed the second formula, will standard errors indeed get smaller (as one would expect) as $n$ gets large? I ask because $\sum_{i = 1}^n (B_i - \bar{B_i})^2/n \approx Var(B_i)$ for large $n$; so it appears that the second estimate is roughly $Var(B_i)(n-1)$. It is not clear to me that this is decreasing in $n$ (even if $Var(B_i)$ goes to zero as $n$ gets large).
I'm not a statistician... and this is the economics forum, so forgive me for the possible mistakes. I got most of this information from these slides.
Consider a statistic of interest  $\theta$ with a consistent but biased estimate $\hat \theta$. The Jacknife estimates $\hat \theta_{(i)}$ with is the same statistic, based on the sample that is obtained from removing observation $i$. Doing this for all $i$ and taking the mean over the values $\hat \theta_{(i)}$ gives:
$$
\hat \theta_{(.)} = \frac{1}{n} \sum_{i = 1}^n \hat \theta_{(i)}
$$
As stated above, the estimator $\hat \theta$ is biased, so let:
$$
\mathbb{E}(\hat \theta) = \theta + \frac{b}{n} + O\left(\frac{1}{n^2}\right)
$$
Where $b$ is the first order bias of the estimator $\hat \theta$. Then for the Jacknife estimates, we have a similar expression:
$$
\mathbb{E}(\hat \theta_{(i)}) = \theta + \frac{b}{n-1} + O\left(\frac{1}{n^2}\right).
$$
So, ignoring the higher order terms, we get:
$$
\mathbb{E}(\hat \theta_{(i)}) - \frac{b}{n-1} =  \mathbb{E}(\hat \theta) - \frac{b}{n},\\
\to (n-1)\mathbb{E}(\hat \theta_{(i)} - \hat \theta) = \frac{b}{n}
$$
This shows that: $(n-1) (\hat \theta_{(i)} - \hat \theta)$ is an `unbiased' estimator for $\dfrac{b}{n}$ (at least when we are ignoring all other higher order bias terms). Then using this correction, we have the following 'bias-corrected estimate':
$$
pv_{(i)} = \hat \theta + (n-1)(\hat \theta - \hat \theta_{(i)}) = n \hat \theta + (n-1) \hat \theta_{(i)}.
$$
This is called the pseudovalue.
Now, of course it would be stupid to only use $pv_{(i)}$ for one particular value of $i$ . So a better bias-corrected estimate is the one that averages over all these, the jacknife estimator:
$$
\hat \theta_{jack} = \frac{1}{n}\sum_{i = }^n pv_i =  n \hat \theta + (n-1) \hat \theta_{(.)}.
$$
The pseudovalues $pv_{(i)}$ are not necessarily i.i.d. but assume they are anyway. Then, their variance is given by:
$$
\begin{align*}
{\rm var}(pv_i) &amp;= \frac{1}{n-1} \sum_{i = 1}^n (pv_i - \hat \theta_{jack})^2,\\
&amp;= \frac{1}{n-1} \sum_{i = 1}^n (n \hat \theta - (n-1)\hat \theta_{(.)})^2,\\
&amp;= \frac{1}{n-1} \sum_{i = 1}^n (n \hat \theta + (n-1) \hat \theta_{(i)} - n \hat \theta - (n-1) \hat \theta_{(.)} )^2,\\
&amp;= \frac{(n-1)^2}{n-1} \sum_{i = 1}^n (\hat \theta_{(i)}- \hat \theta_{(.)})^2,\\
&amp;= (n-1) \sum_{i = 1}^n (\hat \theta_{(i)} - \hat \theta_{(.)})^2
\end{align*}
$$
Assuming a central limit theorem holds for $\hat \theta_{jack}$, then:
$$
\frac{\theta_{jack} - \theta}{s_n} \to^d N(0,1)
$$
where $s_n$ is the variance of $pv_i$ divided by $n$.
$$
s_n^2 = \frac{n-1}{n} \sum_{i = 1}^n \left(\hat \theta_{(i)} - \hat \theta_{(.)}\right)^2.
$$
This 'variance', $s^2_n$, is the estimator that you give in your question.