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Let X and Y be independent variables with densities f and g concentrated on $(0, \infty)$. If E(X) < $\infty$ , show that the ratio X/Y has a finite expectation iff

$ \int_0^1 \frac{1}{y} g(y)dy < \infty $

I know that I have show both sides. Can I just use the expectation formula for continuous variables

$ \int x f(x) dx $ for a density f(x) of the variable X?

Thanks!

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2 Answers 2

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Let $Z=Y^{-1}X$.

(1) Since $X$ and $Y$ are almost surely positive and independent, $E(Z)=E(X)E(Y^{-1})$, whether both sides are finite or not.

(2) But $E(X)$ is positive and finite. Hence $Z$ is integrable if and only if $Y^{-1}$ is.

(3) Now, $Y^{-1}=S+T$ where $S=\mathbf{1}_{Y\le1}\cdot Y^{-1}$ and $T=\mathbf{1}_{Y>1}\cdot Y^{-1}$, and $T$ is always integrable since $T<1$ almost surely (either $Y\le1$, and then $T=0$, or $Y>1$, and then $T=Y^{-1}<1$).

(4) Hence $Y^{-1}$ is integrable if and only if $S$ is.

(5) Since $E(S)$ is the integral of $y^{-1}g(y)$ over $(0,1)$ which you wrote, you are done.

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    @qed These are [indicator functions](http://en.wikipedia.org/wiki/Indicator_function).2013-09-05
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So we want to show the following: $E\left(\frac{X}{Y}\right) < \infty \Longleftrightarrow \int_{0}^{1} \frac{1}{y} f_{Y}(y) \ dy < \infty$

Let $Z = X/Y$. Then $f_{Z}(z) = \int_{0}^{\infty} f_{X}(yz) f_{Y}(y) |y| \ dy$ for $z \in (0, \infty)$. Then I think you can apply the usual definition of expectation to get the desired results (in both directions).

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    PEV: I would appreciate if you could make precise the ways in which your post may be helpful to the OP in any way whatsoever.2011-08-26