Here is the proof of Variance of sample variance. Can you please explain me the highlighted places:
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$\begingroup$ It tells you why the 112 zero ones are zero right after your highlight of that bit -- because $i=j$. $\endgroup$– Glen_bDec 6, 2017 at 21:27
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$\begingroup$ @Glen_b I understood, but why 112? $\endgroup$– Daniil YefimovDec 6, 2017 at 21:40
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$\begingroup$ Oh, sorry, I misunderstood the issue you wanted clarified on that. I'll delete these comments soon. $\endgroup$– Glen_bDec 7, 2017 at 0:25
1 Answer
- 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 $(X_i-X_j)=(X_i-\mu-(X_j-\mu))$.