# expectation of a ratio of variables proof

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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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