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Let $X \sim \mathcal{N}(0,1)$ and $Y \sim \mathcal{N}(0,1)$ independent from $X$. I know that $E(\frac{X}{Y})=E(X)E(\frac{1}{Y})$; the first term is zero and the second infinity? But do not know how to proceed from there, or even if it the expected value exists.

econ86
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    What have you tried? Just posting a question statement will likely result in mass down-votes followed by the question being closed. It also won't help you in the long run, since how can anyone give you a good answer to cater for your current understanding if they don't know what you currently know? It also does not help future people that may want help with a similar / the same question - the more context the better – Riemann'sPointyNose Aug 26 '20 at 21:53
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    You're right, thank you – econ86 Aug 26 '20 at 22:11
  • It is not correct that $E[1/Y]>1/E[Y]$. The function $1/y$ is discontinuous and not convex. Also, how do you know if you should treat $1/0$ as $\infty$ instead of $-\infty$? – Michael Aug 26 '20 at 22:15
  • I see, thank you. But is it then $E(1/Y)$ unbounded still correct? – econ86 Aug 26 '20 at 22:23
  • If it exists it must equal 0 by symmetry. – Nate Eldredge Aug 26 '20 at 22:30
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    You cannot simply conclude that $E(X/Y)$ does not exist if $E(1/Y)$ does not exist (consider $X=0$ a.s.). You have to use a property of $X$. – LinAlg Aug 26 '20 at 22:37
  • Hmm, for whatever reason, my first thought was to convert the integral to polar coordinates where the joint distribution measure is $\frac{1}{2\pi} r e^{-r^2/2} , dr , d\theta$ or something along those lines, and then observe that $\cot\theta$ not being in $L^1[0, 2\pi]$ implies $X/Y$ is not an integrable random variable. – Daniel Schepler Aug 26 '20 at 23:10

1 Answers1

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Hint: by independence, $$\mathbb{E}\left|\frac{X}{Y}\right| = E|X| \cdot E\left[\frac{1}{|Y|}\right].$$ Prove that $E\left[\frac{1}{|Y|}\right] = \infty$ (you can write down an appropriate integral and compare it to $\int_{-1}^{1} \frac{1}{|y|}\,dy$) and that $E|X| > 0$.

Nate Eldredge
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  • Thanks, very helpful. What do you mean by comparing it to $\int_{-1}^{1} \frac{1}{|y|} dy$? – econ86 Aug 26 '20 at 23:17
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    @econ86: Write down the integral expression for $E\left[\frac{1}{|Y|}\right]$, and try to show that the integrand is greater than a certain constant multiple of $\frac{1}{|y|}$ on the interval $[-1,1]$. Since $\int_{-1}^1 \frac{1}{|y|},dy = \infty$, this will prove that the expectation is infinite. – Nate Eldredge Aug 26 '20 at 23:26