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Suppose that $X_1, \ldots, X_n \sim N(\mu, \sigma^2)$ are iid. I am wondering how we can show $Var\left(\frac{1}{n}\sum\limits_{i=1}^{n}X_i\right) = \infty$. Is there an easy way to do this without resorting to integrals or moment generating functions?

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    $\begingroup$ What is $\bar{X}_n$ here? $\endgroup$ – carmichael561 Jan 3 '17 at 3:14
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    $\begingroup$ It's definitely not the arithmetic mean... which is usually what $\bar{X}_n$ is, because if this were the case, the variance of $\dfrac{1}{n}\bar{X}_n$ should be $\dfrac{\sigma^2}{n^3}$... $\endgroup$ – Clarinetist Jan 3 '17 at 3:18
  • $\begingroup$ Sorry, I made a mistake, I changed it! $\endgroup$ – user321627 Jan 3 '17 at 3:25
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    $\begingroup$ This still doesn't make sense. Actually, it is a well-known result that $$\text{Var}\left(\dfrac{1}{n}\sum_{i=1}^{n}X_i\right) = \dfrac{\sigma^2}{n}\text{.}$$ $\endgroup$ – Clarinetist Jan 3 '17 at 3:28
  • $\begingroup$ It seems that you are looking for the variance of the sample mean. This is $Var(\overline X)=\frac{\sigma^2}{n}$, not $\infty$ $\endgroup$ – callculus Jan 3 '17 at 3:29
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Since $X_1, \dots, X_n$ are iid, $$\text{Var}\left(\dfrac{1}{n}\sum_{i=1}^{n}X_i\right) = \dfrac{1}{n^2}\text{Var}\left(\sum_{i=1}^{n}X_i\right)=\dfrac{1}{n^2}\sum_{i=1}^{n}\text{Var}(X_i)$$ due to independence, and $\text{Var}(X_i) = \sigma^2$.

$\sum_{i=1}^{n}\text{Var}(X_i)$ sums $\sigma^2$ $n$ times, so $\sum_{i=1}^{n}\text{Var}(X_i) = n\sigma^2$; hence, $$\text{Var}\left(\dfrac{1}{n}\sum_{i=1}^{n}X_i\right) = \dfrac{n\sigma^2}{n^2} = \dfrac{\sigma^2}{n}\text{.}$$ This does not require that the $X_i$ are normally distributed; it only requires that the $X_i$ are independent, and that they all have finite variance $\sigma^2$.

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