What is the expected value of $E[X^2 Y]$ if X and Y are independent?

Question

Given X, Y independent, how can we compute $E[X^2 Y]$?

Answer 1

$X$ and $Y$ are independent $\Rightarrow X^{2}$ and $Y$ are independent. Let us recall some definitions. Let $(\Omega,\mathcal{F},P)$ be a probability space. Let $\mathcal{G}_{1},\mathcal{G}_{2}\subseteq\mathcal{F}$. We say that $\mathcal{G}_{1},\mathcal{G}_{2}$ are independent if $P(A\cap B)=P(A)P(B)$ for any $A\in\mathcal{G}_{1}$ and $B\in\mathcal{G}_{2}$. Now, if $X$ and $Y$ are random variables, we say that $X$ and $Y$ are independent if $\sigma(X)$ and $\sigma(Y)$ are independent. Note that $\sigma(X^{2})\subseteq\sigma(X)$ (because $X^{2}$ is $\sigma(X)/\mathcal{B}(\mathbb{R})$-measurable and $\sigma(X^{2})$ is the smallest $\sigma$-algebra on $\Omega$ such that $X^{2}$ is measurable). Now, it is clear that $\sigma(X)$ and $\sigma(Y)$ are independent $\Rightarrow$ $\sigma(X^{2})$ and $\sigma(Y)$ are independent.

Answer 2

The short answer is that $$E(X^2 Y) = E(X^2) E(Y)$$ as independence is preserved under transformations. In general, if $X$ and $Y$ are independent, then $f(X)$ and $g(Y)$ will be independent.

Note however that this does not simply any further. We cannot say that $E(X^2) E(Y) = E(X)^2 E(Y)$ as this is untrue in general.

Answer 3

$$\begin{align}\mathsf E(X^2Y)&=\mathsf E(\mathsf E(X^2Y\mid Y))&&\text{Law of Total Expectation}\\&=\mathsf E(\mathsf E(X^2\mid Y)\cdot Y)&&Y\text{ relatively constant to a conditional expectation wrt }Y\\&=\mathsf E(\mathsf E(X^2)\cdot Y)&&\bigstar\\&=\mathsf E(X^2)\cdot \mathsf E(Y)&&\mathsf E(X^2)\text{ is a constant}\\[3ex]&=\big(\mathsf{Var}(X)+\mathsf E(X)^2\big)\cdot \mathsf E(Y)&&\text{by definition for Variance.}\end{align}$$

$^\star:$ $X$ and $Y$ are independent, so the realised value of $Y$ has no influence on the expected value of $X$ nor any monovariate function of $X$ (such as $X^2$).   Therefore the conditional expectation with respect to $Y$ is just the expectation.