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

  • $\begingroup$ 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

1 Answer 1


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: $$ \int_{-\infty}^{+\infty} g(x) dx = \int_{-\infty}^{+\infty} g(-x) \, |-1| dx = \int_{-\infty}^{+\infty} -g(x) dx = -\int_{-\infty}^{+\infty} g(x) dx. $$

  • $\begingroup$ @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
  • 1
    $\begingroup$ LOTUS = Law of the Unconscious Statistician. OK. Now I see it. And I appreciate your help (+1). $\endgroup$ Dec 2, 2016 at 18:43

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.