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If I have two random variables $X$ and $Y$ which are independent and identically distributed (i.i.d) as Standard normal $\sim N(0,1)$

does $E(X^2) = E(XY)$ since $X$ and $Y$ are i.i.d?

please explain it step by step, I'm a beginner in probability. and if it is true, is it true for any pair of i.i.d random variable no matter what distribution it is (provided that it has a finite mean)?

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    $\begingroup$ By iid, $E(XY) = E(X)E(Y) = E(X)^2$. But in general (and in particular in this case as Clement shows) we do not have $E(X^2) = E(X)^2$. $\endgroup$ – GEdgar Jul 24 '18 at 0:24
  • $\begingroup$ @GEdgar And indeed, whenever this equality holds for a square-integrable r.v., then this by definition is equivalent to having $\operatorname{Var} X =0$, i.e., $X$ is almost surely (a.s.) constant $\endgroup$ – Clement C. Jul 24 '18 at 0:30
  • $\begingroup$ I think the simplest proof would be that x^2 can never be negative where as xy can be. $\endgroup$ – barrycarter Jul 24 '18 at 1:30
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No.

Since $X$ has variance $1$ and expectation $0$ $$\mathbb{E}[X^2] =\mathbb{E}[X^2] - \mathbb{E}[X]^2 = \operatorname{Var}[X] = 1$$ by definition, while, since $X,Y$ are independent,

$$ \mathbb{E}[XY] = \mathbb{E}[X]\mathbb{E}[Y]=0\cdot 0 = 0\,. $$

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