Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

2 Answers

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.

share|improve this answer
1  
... 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
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.