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)?

  • 2
    $\begingroup$ 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. $\endgroup$ – user2469 Jul 24 '18 at 1:30


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,

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.