# Quotient Distribution of Positive Independent Random Variables

Suppose $$X$$ and $$Y$$ are independent positive random variables with probability density functions $$f_X$$ and $$f_Y$$ respectively. Show that $$Z=X/Y$$ is absolutely continuous and find its probability density function.

$$\textbf{My Thought:}$$ In order to show that $$Z$$ is absolutely continuous, I need to show that it has a commulative distribution function. Then I can also differentiate the latter to obtain the probability density function of $$Z$$. I have the following calculation \begin{align*} P(X/Y\leq z)&=P(X\ge zY,Y<0)+P(X\le zY,Y>0)\\ &=\int_{-\infty}^{0}\left(\int_{yz}^{\infty}f_{X}(x)dx\right)f_{Y}(y)dy+\int_{0}^{\infty}\left(\int_{-\infty}^{yz}f_{X}(x)dx\right)f_{Y}(y)dy\\ &=\int_{0}^{\infty}\left(\int_{0}^{yz}f_{X}(x)dx\right)f_{Y}(y)dy. \end{align*} Also, if $$z<0$$, then we have that $$F_Z(z)=0,$$ since $$X,Y$$ are positive random variables. Therefore the random variable $$Z$$ is continuous, and we can differentiate it to obtain the probability distribution function of $$Z$$. We have that $$f_Z(z)=0$$ if $$z\leq 0,$$ and otherwise we have $$f_Z(z)=\int_{0}^{\infty}yf_{X}(yz)f_Y(y)dy.$$

Is my reasoning above correct?

Any feedback is much appreciated.

• Please check my edit where the last $dz$ is interchanged by $dy$. – drhab Mar 29 '19 at 9:16

For a fixed positive $$z$$ we find:
$$P\left(Z\leq z\right)=P\left(X\leq zY\right)=\int_{0}^{\infty}\int_{0}^{zy}f_{X}\left(x\right)f_{Y}\left(y\right)dxdy=\int_{0}^{\infty}\int_{0}^{z}f_{X}\left(uy\right)yf_{Y}\left(y\right)dudy=$$$$\int_{0}^{z}\int_{0}^{\infty}f_{X}\left(uy\right)yf_{Y}\left(y\right)dydu$$
where the third equality rests on the substitution $$x=uy$$.
This shows directly that functiont $$f_{Z}$$ prescribed by: $$z\mapsto\int_{0}^{\infty}f_{X}\left(zy\right)yf_{Y}\left(y\right)dy$$ if $$z>0$$ and $$z\mapsto 0$$ otherwise serves as PDF of positive random variable $$Z$$.