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$$\alpha\sum_{i=1}^n \sum_{i\ne j}^n (X_i - X_j)^2 = \hat{\sigma}^2$$

Question: Find the value of $\hat{\sigma}^2$ that makes it an unbiased estimator of $\sigma^2$

The way I have been taught to find unbiased estimators is to use the E($\hat{\sigma}^2$) which is $\sigma^2$ to substitute values that can be converted into $\sigma^2$ and $\mu^2$ from the equation V(X) = E(X^2)-[E(X)^2] So far I understand that there will be n(n-1) way to sum the values because $(X_i - X_i)^2$ cannot exist. However, I'm unsure how to subsitute the expected values into the original equation if I even can. Then I would be able to figure out which value of n could work for the $\alpha$ constant. Sorry if this is poorly explained, I've never seen a problem with two summations before. The answer provided to the problem is $\alpha = \frac{1}{2n(n-1)}$

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  • $\begingroup$ Attempt at explanation. Factor of 2 -every pair is recorded twice. n(n-1) is no. of pairs. $\endgroup$ Feb 14, 2023 at 19:43
  • $\begingroup$ Note $X_i-X_j$ has expectation $0$ and variance $2 \sigma^2$ when $i \not=j$ $\endgroup$
    – Henry
    Feb 15, 2023 at 1:45
  • $\begingroup$ @herbsteinberg Each of the ${n \choose 2}=\frac{n(n-1)}{2}$ pairs is recorded twice, but each of the $n(n-1)$ pairs is recorded once. The $2$ has a different explanation $\endgroup$
    – Henry
    Feb 15, 2023 at 1:53

1 Answer 1

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Assuming $X_i$ are i.i.d. random variables. $E(X_i-X_j)^2=E(X_i)^2)+E(X_j)^2)-2E(X_i)E(X_j)=2\sigma^2$ Therefore $\sum\limits _{i=1}^n\sum\limits _{j=1,j\ne i}^n (X_i-X_j)^2=n(n-1)2\sigma^2$. Therefore $\alpha=\frac{1}{2n(n-1)}$.

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  • $\begingroup$ The question is an inefficient way of estimating $\sigma^2$. Better method - let $m=\frac{1}{n}\sum X_k$ Then an unbiased estimator of $\sigma^2=\frac{1}{n-1}(\sum X_k^2 -nm^2)$. The given answer requires $\frac{n(n-1)}{2}$ adds and squares, while this method requires 2n adds and n squares. Further as more data is added, update is much simpler. $\endgroup$ Feb 17, 2023 at 21:27

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