Variance of sample variance (proof explanation)

Question

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

1. Why $(X_i - X_j)$?
2. why are there 112 terms, that are equal to 0?
3. How do we know, that there are 24,96 and 24 terms of the provided form?
4. How do we derive the results (3) result(formulas)?

Answer 1

1. 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).

2. $(X_i-X_i)^2$ 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.

3. 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 ${{4}\choose{1}}{{3}\choose{2}} \times 4 \times 2$.

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