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I would like to confirm with an example that I get what the definition of orthogonality of two random variables means, as defined in this question:

$$\mathbb E[XY^*]=0$$

It's not the first time I ask about this, but in the current post I'd like to ask for an example of the statement

If $Y=X^2$ with symmetric pdf they are dependent yet orthogonal.

Can we then proof that for a normal standard deviation $X \sim N(0,1)$ - perfectly symmetrical - and $Y=X^2$ (which will have a pdf as in here), the $\mathbb E[XY^*=0]$?

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  • $\begingroup$ Aren't $X$ and $Y$ real? What is $Y^*$? $\endgroup$
    – user251257
    Dec 2, 2016 at 18:10
  • $\begingroup$ Yes, they are. It is the general definition as in the referenced post (first link). $\endgroup$ Dec 2, 2016 at 18:13
  • $\begingroup$ I erased the formula. Your comment makes sense, and the formula was just a quick attempt at showing how I would start thinking about the problem. It is not homework. $\endgroup$ Dec 2, 2016 at 18:30
  • $\begingroup$ Note to self: this is a good entry to understand the operations behind this in the discrete case. And for inner product, this is a great one. $\endgroup$ Dec 2, 2016 at 21:52

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This works explicitly for the situation as stated in the question. Since $Y=X^2$ we have $XY = X^3$. Since $X$ is standard normal distributed, it's pdf $$ f_X(x) = \frac1{\sqrt{2\pi}} \exp\left(-\frac{x^2}2 \right) $$ is symmetric around $0$, that is an even function. Thus, we have

$$E(XY) = E(X^3) = \int_{-\infty}^{+\infty} \underbrace{\underbrace{\phantom{f}x^3}_{\text{odd}} \underbrace{f_X(x)}_{\text{even}}}_{\text{odd}} dx = 0.$$

Remember that the integral of any (integrable) odd function $g$ is zero, as: $$ \int_{-\infty}^{+\infty} g(x) dx = \int_{-\infty}^{+\infty} g(-x) \, |-1| dx = \int_{-\infty}^{+\infty} -g(x) dx = -\int_{-\infty}^{+\infty} g(x) dx. $$

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  • $\begingroup$ @AntoniParellada: It has nothing to do with homework or not. What remains unclear to you? I will add more details. $\endgroup$
    – user251257
    Dec 2, 2016 at 18:21
  • $\begingroup$ I read your comment about independence, which makes sense. I don't see then how you set up the equation in your post as the E[XY]. I wrote the equation in my OP to show a bit what I had in mind, but without conviction. Also, I'd like to see the resolution applied to the normal distribution. $\endgroup$ Dec 2, 2016 at 18:28
  • $\begingroup$ Just replace $f_X(x)$ by $\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$, then. $\endgroup$
    – Clement C.
    Dec 2, 2016 at 18:31
  • $\begingroup$ There is no multiplication of pdf's. Do you agree that if $Y=X^2$, then $XY=X^3$? And therefore that $\mathbb{E}[XY]=\mathbb{E}[X^3]$? $\endgroup$
    – Clement C.
    Dec 2, 2016 at 18:36
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    $\begingroup$ LOTUS = Law of the Unconscious Statistician. OK. Now I see it. And I appreciate your help (+1). $\endgroup$ Dec 2, 2016 at 18:43

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