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I'm trying to prove $$ \mathrm{Var}\left(\sum\limits_{i=1}^n{g(X_i)}\right) = n(Var(g(X_1))) $$ where $X_1...X_n$ are IID variables.

I have been trying to use proof for a similar question - https://stats.stackexchange.com/a/31181 - but then I'm stuck when trying to prove $$ \sum\limits_{i=1}^n{E(g(X_i))^2} = nE(g(X_1))^2$$

How to deal with the last equality? Or is there any easier way to prove an equality from the title?

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  • $\begingroup$ I don't think that the same mean and variance is sufficient to make this claim. You will need $(X_i)$ to be an IID sequence for this to hold. $\endgroup$
    – tchakravarty
    Dec 29, 2013 at 17:41
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    $\begingroup$ $X_i$ are IID sequence, I have just simplified question too much. $\endgroup$
    – Mateusz Kowalski
    Dec 29, 2013 at 17:45
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    $\begingroup$ Then the last line is trivial. As long as the expectation of the function exists, it will be the same for all of the random variables in the sequence. Hence the summation is just $n$ times the expectation of any one of those r.v.s. $\endgroup$
    – tchakravarty
    Dec 29, 2013 at 17:46
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    $\begingroup$ This looks like it should carry the self-study tag $\endgroup$
    – Glen_b
    Dec 29, 2013 at 17:57

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Let $Y_i = g(X_i)$. Show that the $Y_i$s are iid (which I hope is straightforward for you).

Then simply show

$\sum\limits_{i=1}^n{E(Y_i)^2} = nE(Y_1)^2$

which I assume you already know how to do.

If you have trouble with that, you might want to consider $Y_i = g(X_i)^2$ and then the last result reduces to $\sum\limits_{i=1}^n{E(Y_i)} = nE(Y_1)$

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