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It is known that the sample variance is an unbiased estimator:

$$s^2 = \frac 1{n-1} \sum_{i=1}^n (X_i - \bar X)^2$$

I would like show that $\sigma '^2 = (X_1 - X_2)^2 $ is a biased estimator.

My work:

$$E((X_1 - X_2)^2)= E(X_1^2) - 2E(X_1 X_2) + E(X_2^2)$$

I wasn't taught of how to specifically simplify these kinds of expression, but I suspect that $E(X_1^2)=E(X_2^2)$ since it's symmetrical.

I don't have any further ideas about how to show that the expected value is not the population variance. Please give me some hints to work on it. Thanks.

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It is unbiased if $\mathsf{E}(\hat{\sigma}^2)=\sigma^2$.

$$ \begin{align} \mathsf{E}(\hat{\sigma}^2)=\mathsf{E}((X_1-X_2)^2)&=\mathsf{E}(X_1^2)+\mathsf{E}(X_2^2)-2\mathsf{E}(X_1X_2)\\ &=2(\sigma^2+\mu^2)-2\mu^2\\ &=2\sigma^2\neq\sigma^2 \end{align}$$

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  • $\begingroup$ Thanks for the answer. I would like to ask why $E(X_1^2)= \sigma^2 + \mu^2$. Since $X_1$ is not simply taken from the population, it is choosen from a sample of sample size at least $2$. I may be completely misunderstanding though. $\endgroup$ – lEm Dec 19 '16 at 6:28
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    $\begingroup$ $E(X_1^2)= E((X_1-\mu)+\mu)^2=E(X_1-\mu)^2+\mu^2+2E(X_1-\mu)\mu=E(X_1-\mu)^2+\mu^2$, which is $\sigma^2 + \mu^2$. Note that the cross-product term term vanishes and Expectation of a constant is the same constant. $\endgroup$ – Roronoa Dec 19 '16 at 6:31
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    $\begingroup$ @Bubububu: When we say $X_1, X_2, \dots, X_n$ is a random sample from a population with mean $\mu$ and variance $\sigma^2$, we mean that each $X_i$ has $E(X_i) = \mu$ and $Var(X_i) = \sigma^2.$ Also, that the $X_i$ are all indep. $\endgroup$ – BruceET Dec 19 '16 at 8:24
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Hint: Your approach works fine. Use the fact that $X_1, X_2, \ldots , X_n$ are independently and identically distributed.

Alternatively, let $\mu=E(X_1)=E(X_2)$. The population mean, assuming they exist, will be same due to identical distribution. Now we write $E(X_1-X_2)^2$ as $E\big\{(X_1-\mu)-(X_2-\mu)\big\}^2$.

Note that $E\big\{(X_1-\mu)-(X_2-\mu)\big\}^2=E(X_1-\mu)^2+E(X_2-\mu)^2$ (The cross-product term vanishes!), which equals twice the value of the population variance.

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    $\begingroup$ Alternatively: $E(X_1^2) - 2E(X_1)E(X_2) + E(X_2^2) \\= E(X_1^2) - 2\mu^2 + E(X_2^2) = [E(X_1^2) - \mu^2] + [E(X_2^2 - \mu^2] = 2\sigma^2.\\$ $E(X_1 X_2) = E(X_1)E(X_2) = \mu^2$ by independence. $\endgroup$ – BruceET Dec 19 '16 at 6:19
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    $\begingroup$ I was giving the same hint @BruceET $\endgroup$ – msm Dec 19 '16 at 6:23
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    $\begingroup$ Happens frequently. I see your contemporaneous Answ. Great minds... (+1) $\endgroup$ – BruceET Dec 19 '16 at 6:28

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