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Stuck on this question, and I have no idea how to proceed:

Let $X$ have the probability density $f_{X}(x)=\lambda e^{-\lambda x}, \;\; x>0$ and let $Y$ have the probability density $f_{Y}(y)=\lambda e^{-\lambda x},\;\; y>0.$ Find the probability density of $Z=X/Y$. Carry out reasonability checks of your answer by (i) verifying that it is a legitimate probability density, and (ii) using it to find the probability that $X>Y$.

Now I'm stuck on the first part; So far I have $$Z=\frac{f_X(x)}{f_Y(y)}=e^{-\lambda(x-y)}$$ And then I can't seem to find the bounds $(\alpha_{1,2}$ and $\beta_{1,2})$ of the integral such that $$\int_{\alpha_1}^{\beta_1}\int_{\alpha_2}^{\beta_2} e^{-\lambda(x-y)}dydx=1$$ I've tried from $0$ to $\infty$ for both bounds but that doesn't give an integral which converges. Am I doing this right? I'm starting to think my actual function that I'm integrating is wrong.

George1811
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2 Answers2

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Hint: If you have $Z = X/Y$, then the probability density of $p_Z(z)$ is not necessarily related to $f_X(x) / f_Y(y)$. It is instead, for a particular $Z = z$, the integral over all the ways that you can get $X/Y = z$, with the appropriate density functions for $X$ and $Y$ that give you that $z$.

user2566092
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$$\Pr[Z \le z] = \Pr[X/Y \le z] = \int_{y=0}^\infty \Pr[X \le yz] f_Y(y) \, dy = \int_{y=0}^\infty F_X(yz) f_Y(y) \, dy.$$

heropup
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  • @heropop In the second and third element of the equality, why do you need to integrate over all the values of $y$? –  Feb 16 '20 at 02:26