# Is the variance of the mean of a set of independent random variables equal to the average of their respective variances?

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?

• Do the variables have the same mean, even if they have different variances?
– Dave
Commented Mar 21, 2023 at 18:57
• @Dave - that should not matter if the sample made up of one observation of each of the independent random variables Commented Mar 21, 2023 at 19:21
• "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 Commented Mar 21, 2023 at 21:44

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.
• +1 though I think you should "is equal to the mean of the variances divided by $n$" without "sample" Commented Mar 21, 2023 at 19:19
• @Henry I used the word "sample" to differentiate it from the "population" mean $E[\cdot]$ Commented Mar 21, 2023 at 19:37
• 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 Commented Mar 21, 2023 at 20:03
• @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? Commented Mar 21, 2023 at 20:17