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Let $X_1, X_2 \geq 0$ be two non-negative identically distributed random variables. I wonder if the following equation holds.

$$ \mathbb{E}[X_1X_2] = \mathbb{E}[Y^2] $$ where $Y$ is a random variable having the common distribution of $\{X_1, X_2\}$.

My attempt: We know that, in general, $\mathbb{E}[X_1X_2] \neq \mathbb{E}[X_1^2]$ and/or $\mathbb{E}[X_1X_2] \neq \mathbb{E}[X_2^2]$. One can take $X_1 \in \{0,1\}$ with probability $1/2$ and take $X_2 := -X_1$. However, the nonnegativity excludes this case.

Also, if we look at $$ \mathbb{E}[Y^2] = \int y^2\, dF_{Y} = \int y^2 \, dF_{X_1,X_2} $$ where the last equality hold since $Y$ has the common joint distribution of $\{X_1,X_2\}$. But I get stuck to see if this is equal to $ \mathbb{E}[X_1X_2] = \int x_1x_2 \,dF_{X_1,X_2}$. Any comment is appreciated.

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  • $\begingroup$ What if $X_1$ and $X_2$ are independent? $\endgroup$
    – user140541
    Jul 26, 2021 at 8:35

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It is easy to find such examples. If $EX_1X_2=EX_1^{2} (=EX_2^{2})$ then $E(X_1-X_2)^{2}=EX_1^{2}+EX_2^{2}-2EX_1X_2=0$ so $X_1=X_2$ a.s.. Can you come up with two non-negative r.v.'s with the same distribution (say $exp(1)$) which are not a.s. equal?

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