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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?


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To quote from the FAQ When you have decided which answer is the most helpful to you, mark it as the accepted answer by clicking on the check box outline to the left of the answer. This lets other people know that you have received a good answer to your question. Doing this is helpful because it shows other people that you’re getting value from the community. (If you don’t do this, people will often politely ask you to go back and accept answers for more of your questions!) – Did May 2 '11 at 8:40

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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Could you please further explain step 3? I don't quite follow your notation. Thanks! – qed Sep 5 '13 at 15:33
@qed Is it better like this? – Did Sep 5 '13 at 15:40
Sorry for not having been specific. I mean the $1_{Y\le 1}$ and $1_{Y\ge 1}$. – qed Sep 5 '13 at 17:13
@qed These are indicator functions. – Did Sep 5 '13 at 17:24

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. – Did Aug 26 '11 at 14:25

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