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I need to show that $S_n = \frac{1}{n-1}\sum_{i=1}^{n}(X_i-\bar{X_n})^2$ is an unbiased estimator of $\sigma^2$, given a population of $N$ elements and $X_1, \ldots,X_n$ the, independent, identical distributed random variables, sample size that model the samples.

By using the linearity of expectations and the expression of variances in expectations I can solve this problem for a big part, the only problem is that I can't solve $E(X_i \bar{X_n})$. I suspect that this can be done by using a conditional expectation, but how does that exactly work?

Thanks for your time,

K. Kamal

ArianJ
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  • You missed the square on $S_n$ when you define it. And you did not need to find $E(X_i \bar X_n)$ to prove the result. You can do this by writing $$(n-1)S_n^2=\sum_{i=1}^n\left[(X_i-\mu)-(\bar X_n-\mu)\right]^2$$ and then expanding the r.h.s and taking expectation on both sides. Here $\mu$ is the population mean. – StubbornAtom Jun 28 '18 at 19:59
  • https://math.stackexchange.com/q/1701626/321264 – StubbornAtom May 29 '20 at 18:23

1 Answers1

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$$E(X_i\overline X_n) =E\left(\frac{X_i}n\sum_{j=1}^nX_j\right)=\frac1n \sum_{j=1}^n E(X_iX_j).$$ You need then to find $E(X_iX_j)$. The answers for $j=i$ and $j\ne i$ will differ.

Angina Seng
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