Variance of sample variance (proof explanation) Here is the proof of Variance of sample variance.  Can you please explain me the highlighted places:


*

*Why $(X_i - X_j)$?

*why are there 112 terms, that are equal to 0?

*How do we know, that there are 24,96 and 24 terms of the provided form?

*How do we derive the results (3) result(formulas)?

 A: *

*You can prove more general results.
$$
S^2=\frac{1}{n-1} \sum_{i=1}^n(X_i - \bar{X})^2
=\frac{1}{n^2(n-1)}\sum_{i=1}^n(nX_i - \sum_{j=1}^n X_j)^2 \\=\frac{1}{n^2(n-1)}\sum_{i=1}^n(\sum_{j=1}^n(X_i - X_j))^2 \\=\frac{1}{n^2(n-1)}[ \sum_{i=1}^n\sum_{j \ne i} (X_i-X_j)^2
+\sum_{i=1}^n \sum_{j \ne i}\sum_{k  \ne j, i} (X_i-X_j)(X_i-X_k)]$$


Note that ,
$$\sum_{i=1}^n\sum_{j \ne i} (X_i-X_j)^2 = \frac{1}{n-2} \sum_{i=1}^n\sum_{j \ne i} \sum_{k \ne i,j}(X_i-X_k-(X_j-X_k))^2 \\=
 \frac{1}{n-2} \sum_{k=1}^n\sum_{j \ne k} \sum_{i \ne k,j}(X_k-X_i-(X_k-X_j))^2 \\= \frac{1}{n-2}[2(n-2)\sum_{k=1}^n\sum_{k \ne i} (X_k-X_i)^2-2\sum_{k=1}^n\sum_{j \ne k} \sum_{i \ne k,j}(X_k-X_i)(X_k-X_j)] \\
\Rightarrow \sum_{i=1}^n\sum_{j \ne i} (X_i-X_j)^2 =\frac{2}{n-2} \sum_{i=1}^n \sum_{j \ne i}\sum_{k  \ne j, i} (X_i-X_j)(X_i-X_k),$$
where the first equality is by the fact that $k$ has $n-2$ choices. Thus,
$$
S^2 = \frac{1}{2n(n-1) }\sum_{i=1}^n\sum_{j \ne i} (X_i-X_j)^2
$$
This will be the "straight algebra" in part (ii).


*$(X_i-X_i)^2$ is included in the formula. So essentially there are only $(16-4)(16-4)=144$ nonzero terms, the number of zero terms is $256-144=112$. The reason why $4 \times 16 \times 2 -4^2$ is terms $(X_i-X_i)^2 \times (X_j-X_j)^2$ is counted twice.

*Number of form $E(X_i-X_j)^4$ is $ {{4}\choose {2}} \times 2\times 2 $.
Number of form $E(X_i-X_j)^2(X_i-X_k)^2$ is ${{4}\choose{1}}{{3}\choose{2}} \times 4 \times 2$.


*Use the decomposition $(X_i-X_j)=(X_i-\mu-(X_j-\mu))$.

