Take the 2-minute tour ×
Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. It's 100% free, no registration required.

Suppose I have $X,Y$, which are independent random variables.

Why is it that $E(\frac{X}{Y}) = E(X)E(\frac{1}{Y})$?

Also, why is it that $E(X^2Y^2)=E(X^2)E(Y^2)$? How is it that the square of an independent random variable is also independent in relation to $Y$ or $Y^2$? Thanks!

share|improve this question
    
Let $W=\frac{1}{Y}$. Is $W$ independent of $X$? –  Scortchi Feb 5 at 12:08
1  
Assuming that the expectations exist (cf. Paul Staab's comment on his answer), the law of the unconscious statistician gives $$\begin{align}E\left[\frac{X}{Y}\right]&=\int\int\frac{x}{y}f_{X,Y}(x,y)\,dx\,d‌​y\\&= \int\int\frac{x}{y}f_X(x)f_Y(y)\,dx\,dy\\&=\int xf_X(x)\,dx\int \frac{1}{y}f_Y(y)\,dy\\&=E[X]E\left[\frac{1}{Y}\right]\end{align}$$ –  Dilip Sarwate Feb 5 at 14:41
add comment

2 Answers

up vote 4 down vote accepted

Basically, if $X$ and $Y$ are independent, then also $f(X)$ and $g(Y)$ are independent if $f$ and $g$ are measurable functions:

$$\eqalign{ P(f(X) \in A,\ g(Y) \in B) &= P\left(X \in f^{-1}(A),\ Y \in g^{-1}(B)\right) \\ & = P\left(X \in f^{-1}(A)\right) \ P\left(Y \in g^{-1}(B)\right) \\ & = P\left(f(X) \in A\right) \ P\left(g(Y) \in B\right). }$$

In particular all continuous functions (like the $f(x)=1/x$ and $f(x)=x^2$ in your examples) are Borel-measurable, and hence also $X$ and $1/Y$ as well as $X^2$ and $Y^2$ are independent.

share|improve this answer
    
i have assumed your answer in my proof for the independence between $X$ and $Z$. Thanks! –  omidi Feb 5 at 12:38
    
Of course, this only holds when 'independence' and $1/Y$ are well defined, e.g. $X$ and $Y$ as well as $f(X)$ and $g(Y)$ live in the same space and $Y > 0$. –  Paul Staab Feb 5 at 12:45
    
g(X) in the first paragraph of the answer should read g(Y) –  RossXV Feb 7 at 16:51
    
@RossXV right, I corrected it. Thanks for spotting it. –  Paul Staab Feb 7 at 17:46
add comment

let $Z=\frac{1}{Y}$, then we have:

\begin{equation} E(XZ) = \int \int XZ p(X,Z) \mathrm{d}x \mathrm{d}z \end{equation}

but $X$ and $Z$ are independent, so $p(X,Z) = p(X)p(Z)$ we have

\begin{equation} E(XZ) = \int \int x .z . p(X=x).p(Z=z) \mathrm{d}x \mathrm{d}z \end{equation}

which can be arranged as: \begin{equation} E(XZ) = \int x . p(X=x) \mathrm{d}x \int z.p(Z=z) \mathrm{d}z = E(X)E(Z) \end{equation}

using the same argument you can show that $E(X^2Y^2) = E(X^2).E(Y^2)$

does that help?

share|improve this answer
    
I believe that "but $X$ and $Z$ are independent" is the statement the O.P. is asking about. The answer by Paul Staab directly addresses that issue in full generality. –  whuber Feb 5 at 22:39
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.