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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 $Z=X/Y$.

Answer:

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

Why do we have to integrate over all possible values of $y$ with non-zero support?

Why don't we integrate over the support of the random variable $Z$?

Why do we have

$$Pr[X\leq yz]f_{Y}(y)dy$$

rather than

$$Pr[X\leq yz]f_{Z}(Z)dz$$

or

$$Pr[X\leq yz]f_{X}(X)\,dx?$$

Also,

Why does one have to use conditioning to solve the problem? Why isn't the solution $$\int_{x=0}^{\infty} Pr[X\leq yz] \, dx?$$

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You can either condition on $X$ or condition on $Y$. But you can't condition on $f_Z(Z)$ as that is the quantity that you would like to find.

If you would like to condition on $X$ assuming that $X$ and $Y$ are independent,

\begin{align} Pr(Z \le z) &= Pr\left(\frac{X}{Y} \le z\right) \\ &=\int_0^\infty Pr(Y \ge \frac{X}{z}|X=x)f_X(x) \, dx \\ &= \int_0^\infty Pr(Y \ge \frac{x}{z})f_X(x) \, dx \\ \end{align}

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  • $\begingroup$ Why does one have to use conditioning to solve the problem? Why isn't the solution $$\int_{x=0}^{\infty} Pr[X\leq yz] dx$$? $\endgroup$
    – user271077
    Feb 16, 2020 at 3:51
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    $\begingroup$ Your $x$ doesn't appear in your integral. also, $y$ is not defiend in your term. Also, we have not address the randomness of $Y$. $\endgroup$ Feb 16, 2020 at 4:01
  • $\begingroup$ May I say that $Pr(Z\leq z)=\int_{0}^{\infty} Pr(Y \geq \frac{X}{z} | Y=y) f_Y(y) dy?$ $\endgroup$
    – user271077
    Feb 18, 2020 at 1:12
  • $\begingroup$ yes, that is fine. Of course, we have assume that $z$ is positive. $\endgroup$ Feb 18, 2020 at 1:22

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