Why condition on either the r.v. $X$ or $Y$ and integrate over a product of pdfs rather a single pdf to find this probability density?

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 $$Z=X/Y$$.

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

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 $$\int_{x=0}^{\infty} Pr[X\leq yz] \, dx?$$

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}

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