1
$\begingroup$

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.

Thank you for your time.

$\endgroup$
  • $\begingroup$ Please check my edit where the last $dz$ is interchanged by $dy$. $\endgroup$ – drhab Mar 29 at 9:16
1
$\begingroup$

It might be that the PDF cannot be obtained as derivative of the CDF, simply because the CDF is not necessarily differentiable (everywhere).

Fortunately that is not fatal here because there is a more direct route.

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 $Z$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.