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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!

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  • $\begingroup$ Let $W=\frac{1}{Y}$. Is $W$ independent of $X$? $\endgroup$
    – Scortchi ♦
    Commented Feb 5, 2014 at 12:08
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    $\begingroup$ Assuming that the expectations exist (cf. Paul Staab's comment on his answer), the law of the unconscious statistician gives $$\begin{align}E\left[\frac{X}{Y}\right]&=\int\int\frac{x}{y}f_{X,Y}(x,y)\,dx\,dy\\&= \int\int\frac{x}{y}f_X(x)f_Y(y)\,dx\,dy\\&=\int xf_X(x)\,dx\int \frac{1}{y}f_Y(y)\,dy\\&=E[X]E\left[\frac{1}{Y}\right]\end{align}$$ $\endgroup$
    – Dilip Sarwate
    Commented Feb 5, 2014 at 14:41

2 Answers 2

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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.

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  • $\begingroup$ i have assumed your answer in my proof for the independence between $X$ and $Z$. Thanks! $\endgroup$
    – omidi
    Commented Feb 5, 2014 at 12:38
  • $\begingroup$ 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 $Y > 0$. $\endgroup$
    – Paul Staab
    Commented Feb 5, 2014 at 12:45
  • $\begingroup$ g(X) in the first paragraph of the answer should read g(Y) $\endgroup$
    – RossXV
    Commented Feb 7, 2014 at 16:51
  • $\begingroup$ @RossXV right, I corrected it. Thanks for spotting it. $\endgroup$
    – Paul Staab
    Commented Feb 7, 2014 at 17:46
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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?

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  • $\begingroup$ 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. $\endgroup$
    – whuber ♦
    Commented Feb 5, 2014 at 22:39

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