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

Community
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Antoni Parellada
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1 Answers1

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

user251257
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  • @AntoniParellada: It has nothing to do with homework or not. What remains unclear to you? I will add more details. – user251257 Dec 02 '16 at 18:21
  • 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. – Antoni Parellada Dec 02 '16 at 18:28
  • Just replace $f_X(x)$ by $\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$, then. – Clement C. Dec 02 '16 at 18:31
  • There is no multiplication of pdf's. Do you agree that if $Y=X^2$, then $XY=X^3$? And therefore that $\mathbb{E}[XY]=\mathbb{E}[X^3]$? – Clement C. Dec 02 '16 at 18:36
  • Yes, I follow this part well. But after that, is the formula as you set it up an inner product? What is the set up for the formula? I see that the definition is $E[XY]$, but I thought, from my previous question that this was a bit more complicated than multiplying the random variables, and using the pdf of $X$ to integrate. – Antoni Parellada Dec 02 '16 at 18:39
  • @AntoniParellada: It is the definition of pdf that $E(\phi(X)) = \int_\mathbb R \phi(x) f_X(x) dx$ holds for every borel measurable function $\phi:\mathbb R\to \mathbb R$. – user251257 Dec 02 '16 at 18:41
  • ... LOTUS strike again? – Antoni Parellada Dec 02 '16 at 18:42
  • @AntoniParellada: pardon? – user251257 Dec 02 '16 at 18:42
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    LOTUS = Law of the Unconscious Statistician. OK. Now I see it. And I appreciate your help (+1). – Antoni Parellada Dec 02 '16 at 18:43
  • @AntoniParellada the definition part refers to $f_X , dx = dF_X$ not the transformation with $\phi$ part :p – user251257 Dec 02 '16 at 18:49
  • Can you clarify your last comment? – Antoni Parellada Dec 02 '16 at 19:03
  • @AntoniParellada it is just a joke. You are right, that expected value formula isn't the definition of a PDF. – user251257 Dec 02 '16 at 19:07
  • I am reading this as the expectation of the inner product of the pdf's of $X$ and $Y$. Now the inner product of two continuous functions (pdf's) involves an integral, and the $\mathbb E[\cdot]$ of a continuous random variable (now $XY$) also includes an integral. This latter integral is there, while the inner product is reduced to $X^3$... I know your post is correct, but I wonder if you could insert a sentence linking the idea of inner product and the first $XY=X^3$ expression. Is it the inner product? – Antoni Parellada Dec 02 '16 at 19:51
  • @AntoniParellada: remember that random variables are just functions. The overly formal way is: $E(XY) = \int_\Omega X(\omega) Y(\omega) , dP(\omega) = \int_\Omega X(\omega) X^2(\omega) , dP(\omega) = \int_\Omega X^3 (\omega) , dP(\omega) = \int_\mathbb R x^3 f_X(x) , dx$. – user251257 Dec 02 '16 at 20:13