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.
- 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*}
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.