3
$\begingroup$

I understand that, given a set of iid random variables, the variance of the sum is equal to the sum of the variance. Likewise, I know that the variance of the mean is equal to the variance over n.

My question is: if the variances of the respective random variables are not the same, is the variance of the mean still an average of the variances?

$\endgroup$
3
  • $\begingroup$ Do the variables have the same mean, even if they have different variances? $\endgroup$
    – Dave
    Mar 21, 2023 at 18:57
  • $\begingroup$ @Dave - that should not matter if the sample made up of one observation of each of the independent random variables $\endgroup$
    – Henry
    Mar 21, 2023 at 19:21
  • 1
    $\begingroup$ "Is the variance of the mean of a set of independent random variables equal to the average of their respective variances?" No. See en.wikipedia.org/wiki/Variance#Properties $\endgroup$
    – Glen_b
    Mar 21, 2023 at 21:44

1 Answer 1

7
$\begingroup$

Given a set of random variables $X_1,\dots,X_n$, if they are independent, then \begin{align} \text{Var}(\overline X) &= \text{Var}\left(\frac{1}{n} \sum_{i=1}^n X_i\right) \\ &= \frac{1}{n^2}\text{Var}\left(\sum_{i=1}^n X_i\right) \\ &= \frac{1}{n^2}\sum_{i=1}^n \text{Var}\left(X_i\right) \end{align} So the variance of the sample mean is equal to the mean of the variances divided by $n$, regardless of whether the variances are equal or not.

$\endgroup$
8
  • $\begingroup$ +1 though I think you should "is equal to the mean of the variances divided by $n$" without "sample" $\endgroup$
    – Henry
    Mar 21, 2023 at 19:19
  • $\begingroup$ @Henry I used the word "sample" to differentiate it from the "population" mean $E[\cdot]$ $\endgroup$
    – mhdadk
    Mar 21, 2023 at 19:37
  • 1
    $\begingroup$ OK, though I would say $\text{Var}\left(X_i\right)$ is one of the $n$ variances and $\frac1n\text{Var}\left(X_i\right)$ is the mean of the $n$ variances, without any sample or expectation involved $\endgroup$
    – Henry
    Mar 21, 2023 at 20:03
  • $\begingroup$ @Henry I’m not sure what you mean by “$\frac1n\text{Var}\left(X_i\right)$ is the mean of the $n$ variances”. What I mean by “sample mean of the variances” is $$\frac{1}{n}\sum_{i=1}^n \text{Var}\left(X_i\right)$$ Is this what you were asking about? $\endgroup$
    – mhdadk
    Mar 21, 2023 at 20:17
  • 3
    $\begingroup$ @mhdadk Basically (well, this may be a little nitpicking but I agree with Henry's pov), the wording "sample mean of the variances" is not correct -- because "variances" are non-random quantities, not a sample. It is less ambiguous to say "arithmetic mean of variances" or just "mean of variances". $\endgroup$
    – Zhanxiong
    Mar 21, 2023 at 22:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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