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If we have $n$ i.i.d random variables $X_1,\ldots,X_n$, and some real-valued function $g: \mathbb{R} \to \mathbb{R}$, is it true that $\mathbb{E}(g(X_1)) = \cdots = \mathbb{E}(g(X_n))$? I think this is true since all the random variables follow the same distribution, when it comes down to it, calculating the expectation for each random variable will follow the same procedure, resulting in the same answer. Can anybody provide a more rigorous justification/proof? Or is this a fine proof?

let $i,j \in \{1,\ldots,n\}$ with $i \neq j$.

Discrete case:

Since $X_1,\ldots,X_n$ are i.i.d, they all have the same associated probability function, so $$\mathbb{E}(g(X_i)) = \sum_{\text{all }x} g(x)\mathbb{P}(X_i = x) = \sum_{\text{all} \ x} g(x)\mathbb{P}(X_j = x) = \mathbb{E}(X_j)$$

Continuous case:

Since $X_1,\ldots,X_n$ are i.i.d, they all have the same associated pdf $f$, so $$\mathbb{E}(g(X_i)) = \int_{\text{all } x} g(x) f(x) \, dx = \mathbb{E}(g(X_j))$$

Hence, the result.

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    $\begingroup$ Your argument is correct if density exists but the result is true for arbitrary distributions.. You don't even have to distinguish between discrete and continuous cases. $\endgroup$ Jun 15, 2018 at 5:37

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Hints: It suffices to show that if $X$ and $Y$ are equal in distribution,then also $g(X)$ and $g(Y)$ for any Borel measurable function $g: \mathbb{R} \to \mathbb{R}$, because expectation is determined by the distribution of a variable (and not the variable itself). You can then use induction.

This easily follows using the definitions.

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