Let $X$ and $Y$ are independent random variables. Can I say that $E[XY^2] = E[X]*E[Y^2]$? I know that $E[XY] = E[X]*E[Y]$ but does this still hold true if a random variable is squared?

  • $\begingroup$ What's $*$? And if you know the first equality holds (whatever it means) for any random variables $X$ and $Y$, why can't you just replace $Y$ by $Y^2$ to get the second equality? $\endgroup$
    – Sayantan
    Apr 19, 2018 at 11:17
  • $\begingroup$ Are $X$ and $Y$ independent? There are counter-examples if they are merely uncorrelated $\endgroup$
    – Henry
    Apr 19, 2018 at 11:20
  • $\begingroup$ Yes they are independent. $\endgroup$ Apr 19, 2018 at 11:21
  • 1
    $\begingroup$ I believe the answer is Yes; because of linearity of expectation. $\endgroup$
    Apr 19, 2018 at 11:23

2 Answers 2


\begin{align}E(XY^2) &= \int \int xy^2 P(x,y) \, dx \, dy \\ &= \int \int xy^2 P(x) P(y) \, dx \, dy & \text{ (as the they are independent)}\\ & = \int x P(x) dx\int y^2 P(y)dy \\ & = E(X) E(Y^2) \end{align}


Theorem: If $X_1, X_2,...,X_n$ are independent random variables and for $i=1,2,...,n,$ the expectation $\mathbb{E}[f_i(X_i)]$ exists then:$$\mathbb{E}\left[\prod_{i=1}^{n}f_i(X_i)\right]=\prod_{i=1}^{n}\mathbb{E}[f_i(X_i)].$$ Proof is similar to @kasa's answer. It can be proved more general:

Theorem:$X_1, X_2,...,X_n$ are mutually independent $\Longleftrightarrow$ $$\mathbb{E}\left[\prod_{i=1}^{n}f_i(X_i)\right]=\prod_{i=1}^{n}\mathbb{E}[f_i(X_i)]$$ for eny $n$ function $f_i$ such that the expected values exist and are well-defined


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