Let $X$ and $Y$ be independent random variables. Suppose $P(X=0)<1$ and $\lim_{z \rightarrow 0}f_{|Y|}(z) > 0$. Prove that $E |\frac{X}{Y}| = \infty$.

I defined a random variable $Z = \frac{X}{Y}$. Then tried to show:

$$\int_{-\infty}^{\infty}zf_{|Z|}(z) dz = \int_{-\infty}^{\infty}z\frac{d}{dz}P(|Z| < z) dz = \int_{-\infty}^{\infty}z\frac{d}{dz}P(|X| < z|Y|) dz = \int_{-\infty}^{\infty}z\frac{d}{dz}\int_{0}^{\infty} \int_{0}^{zy} dx dy$$

But I got stuck here. I'm also not convinced that this is the best way to approach the problem. Any help would be really appreciated


$$E(|\frac X Y|)=E(|X|)E(|\frac 1 Y|)$$ by independence. The first term is positive because $P(|X|>0)>0$ and the second term is $\infty$ because $\lim_{t\to 0}f_{|Y|}(t)>0$.

  • $\begingroup$ Wow this is definitely way simpler than what I was trying. The only thing a little unclear to me is why $\lim_{t\to 0}f_{|Y|}(t)>0$ implies that the second expectation is infinite since $E[|\frac{1}{Y}|] = \int_0^{\infty} tf_{\frac{1}{|Y|}}(t) dt $. Or am I missing something simple? $\endgroup$
    – Brenton
    Oct 28 '15 at 1:48
  • $\begingroup$ @Brenton Think of a simpler way to express $E(|1/Y|)$ in terms of integral involving $f_{|Y|}$ bypassing $f_{1/|Y|}$. $\endgroup$
    – A.S.
    Oct 28 '15 at 1:53
  • $\begingroup$ Am I allowed to do $E[|\frac{1}{Y}|] = \int_0^{\infty} \frac{1}{t}f_{|Y|}(t) dt$? And then since the behavior is like $\frac{1}{t}$ because $f_{|Y|}(t) > 0$ as $t$ approaches $0$, I know the integral is $\infty$? $\endgroup$
    – Brenton
    Oct 28 '15 at 1:57
  • $\begingroup$ Yes you are. Look up "Law of the unconscious statistician". $f_{|Y|}>0$ is not enough - for example $f_{|Y|}(t)\sim ct$. You need the positive limit condition given. $\endgroup$
    – A.S.
    Oct 28 '15 at 2:48
  • $\begingroup$ Ah yes ok I think I understand now. Thank you very much :) $\endgroup$
    – Brenton
    Oct 28 '15 at 3:19

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