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Find $E\left[|\dfrac{X}{Y}|\right]$, $E\left[\dfrac{|X|}{Y}\right]$ and $E\left[\dfrac{X}{|Y|}\right]$ for $X,Y iid \sim Exp(\lambda)$

Does this differ from solving $E\left[\dfrac{X}{Y}\right]$?

I would solve $E\left[\dfrac{X}{Y}\right]$? by finding the density of Z and then finding the expectation of the random variable Z with this distribution.

  1. Distribution of $Z = Y/X$.

If $X,Y$ are independent exponentials with rates $\lambda,\mu$, then $Y = ZX$ and one way to do it is \begin{align*} f_Z(z) &=\int_0^\infty f_X(x)f_Y(zx)\left|\frac{dy}{dz}\right|dx\\ &= \int_0^\infty \lambda e^{-\lambda x}\cdot \mu e^{-\mu zx}|x|\,dx\\ &= \int_0^\infty \lambda\mu e^{-(\lambda +\mu z)x}|x|\,dx\\ &= \frac{\lambda\mu}{(\lambda+\mu z)^2}. \end{align*}

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I will instead consider the case where $X\sim\mathrm{Expo}(\lambda)$ and $Y\sim\mathrm{Expo}(\mu)$ since it generalizes the result and the computations are much the same.

For $t>0$ we have \begin{align} \mathbb P\left(\frac XY>t\right) &= \iint_{\{(x,y)\in\mathbb R^2\ :\ 0\leqslant ty\leqslant x\}} \lambda\mu e^{-\lambda x}e^{-\mu y}\ \mathsf d(x\times y)\\ &= \int_0^\infty \left(\int_{ty}^\infty \lambda e^{-\lambda x}\ \mathsf dx \right) \mu e^{-\mu y}\ \mathsf dy\\ &= \int_0^\infty e^{-\lambda ty}\mu e^{-\mu y}\ \mathsf dy\\ &= \frac{\mu }{\mu +\lambda t}. \end{align} Since $\mathbb P(X>0) = \mathbb P(Y>0)=1$, we may compute the expectation of $\frac XY$ by integrating the survivor function above over $(0,\infty)$: \begin{align} \int_0^\infty \frac{\mu }{\mu +\lambda t}\ \mathsf dt = +\infty. \end{align} We conclude that this random variable does not have finite expectation.

Math1000
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  • I think you forgot a mu in your integral at the first step but corrected it on subsequent steps. –  Feb 17 '20 at 13:51
  • Are you saying that because $E[\frac{X}{Y}] \geq \int_{0}^{\infty} \frac{\mu}{\mu+\lambda t} dt=\infty$, then $E[\frac{X}{Y}] \geq \infty$ and therefore doesn't exist? –  Feb 17 '20 at 14:06
  • I see no missing $\mu$. In general, if $Y$ is a continuous random variable with density $f_Y$ satisfying $f_Y(0)>0$ then $\frac 1Y$ is not integrable, and so if $X,Y$ are i.i.d. then $\frac XY$ is not integrable. – Math1000 Feb 17 '20 at 16:39
  • How do you know that the range of t is from zero to infinity rather than from x/y to infinity ? Or 0 to x/y? –  Feb 17 '20 at 19:25
  • @Numbers $X$ and $Y$ take values in $(0,\infty)$, and so too does $\frac XY$. For a positive number divided by a positive number is a positive number... – Math1000 Feb 19 '20 at 21:14