# independent rv expectation

If $X, Y, Z$ are independent random variables, then $E[XYZ] = E[X] E[Y] E[Z]$

But how can I say that $E[X^2Y^2Z] = E[X^2]E[Y^2]E[Z]$? That being said, how can I prove that $X^2, Y ^2, Z$ are independent random variables?

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If $X$ and $Y$ are independent, $f(X)$ and $g(Y)$ are also independent. \begin{align} P(X=x\;|Y=y)&=P(X=x)\\P(X^2=u\;|\;Y^2=v)&=P(X^2=u) \end{align}

Of course, I am not being as rigorous here as I'd like. When I say $f(\bullet)$ and $g(\bullet)$ , I am ignoring the intricate details of measurability.

You may also look at Are functions of independent variables also independent? which asks a similar question.

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... and this extends also to any collection of random variables. This is best seen using the following definition of independence: $X_1, \ldots, X_n$ are independent iff for all Borel subsets $B_1, \ldots, B_n$ of $\mathbb R$ we have $P\left(\bigcap_{i=1}^n (X_i \in B_i) \right) = \prod_{i=1}^n P(X_i \in B_i)$ –  Robert Israel Feb 10 '13 at 19:39
@RobertIsrael. Perfect. I didn't want to involve $\sigma$-algebra into the picture since I think the OP is asking from a Probability 101 standpoint. –  Inquest Feb 10 '13 at 19:40
Thanks! Also I don't understand the meaning of adding and multiplying two random variables. Could you also explain this? –  user61676 Feb 10 '13 at 21:35
@user61676 What you mean by "don't understand the meaning of adding and multiplying two random variables". Although random variables aren't really "variables", you can think of addition and multiplication to be defined almost the same way as defined on reals. If $X$ is a RV corresponding to the face on the 5th roll of dice and $Y$ corresponds to the number of heads in 5 flips of a coin, You can think of $X+Y$ to be the sum of number of heads and the face on the fifth roll of dice. (Why you'd need such a sum, I leave to your imagination). Similar for multiplication. –  Inquest Feb 10 '13 at 21:44

Lets assume that $X^2$ and $Y^2$ are not independent,then $\exists$ sets $A\in \mathbb{R}, B\in \mathbb{R}$ such that,

$P(X^2\in A,Y^2\in B)\neq P(X^2\in A)P(Y^2\in B)$

Take $A_1=\{x\in \mathbb{R}|x^2\in A\},B_1=\{y\in \mathbb{R}|y^2\in B\}$

$\Rightarrow P(X^2\in A,Y^2\in B)=P(X\in A_1,Y\in B_1)\neq P(X^2\in A)P(Y^2\in B)=P(X\in A_1)P(Y\in B_1)$

This leads us to a contradiction hence proving that our assumption was wrong.

Similar arguement shows the independence of others too.

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