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Is the variance of the mean of a set of possibly dependent random variables less than or equal to the average of their respective variances?

Mathematically, given random variables $X_1, X_2, ..., X_n$ that may be dependent:

Let $\bar{X} = \frac{1}{n}\sum_{i=1}^n X_i$ be the mean of these random variables.

Is it true that:

$$\text{Var}(\bar{X}) \leq \frac{1}{n}\sum_{i=1}^n \text{Var}(X_i)$$

I know that for independent random variables, we have the following equality:

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

Which clearly satisfies the inequality. However, I'm unsure if this holds for dependent variables.

If this inequality is true, is there a proof or intuitive explanation?

If it's not always true, are there conditions under which it holds? What about the following inequality? $$\text{Var}(\bar{X}) \leq \text{Max}_{i=1}^n \text{Var}(X_i)$$

Any insights, proofs, or counterexamples would be greatly appreciated. Thank you!

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    $\begingroup$ Cross post of math.stackexchange.com/questions/4942864/… which has four upvoted answers $\endgroup$
    – Henry
    Commented Jul 7 at 21:49
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    $\begingroup$ I’m voting to close this question because it has already been cross-posted and received answers there. $\endgroup$ Commented Jul 8 at 1:43

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If the $X_i$ are positively correlated, then the variance of the empirical mean is going to be bigger than in the independent case.

If the $X_i$ are negatively correlated, then the variance of the empirical mean is going to be smaller than in the independent case.

And if they are dependent with 0 correlation? Then we have equality.

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  • $\begingroup$ That looks as if it is addressing $\text{Var}(\bar{X}) = \frac{1}{n^2}\sum_{i=1}^n \text{Var}(X_i)$ rather than $\text{Var}(\bar{X}) \leq \frac{1}{n}\sum_{i=1}^n \text{Var}(X_i)$ (which is true, with equality only when $P(X_i-E[X_i]=X_j-E[X_j])=1$ for all $i,j$ i.e. when there is perfect positive correlation and identical variances). $\endgroup$
    – Henry
    Commented Jul 7 at 21:54

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