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?

  • 2
    $\begingroup$ What is $\bar{X}_n$ here? $\endgroup$ – carmichael561 Jan 3 '17 at 3:14
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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

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 $\sigma^2$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.