Q.6.5 - A First Course in Probability

Question

If X and Y are independent continuous positive random variables, express the density function of (a) Z = X/Y and (b) Z = XY in terms of the density functions of X and Y. Evaluate the density functions in the special case where X and Y are both exponential random variables 

Step-by-Step Solution

Answer

In order to obtain pdf of required variable, find the CDF first and then use derivation.

Step 1: Content Introduction

The derivative of the CDF is the probability density function f(x), abbreviated PDF if it exists. A distribution function FX describes each random variable X. (x).

Step 2: Explanation (Part a)

Take any z > 0. Lets find the CDF of Z. We have that,

$$\begin{gathered} F_Z\left(z\right)=P( Z\ge z)=P (\frac{X}{Y}\le z) \\ =P( X \le zY)={\int }_0^{\infty }{f}_y\left(y\right){\int }_0^{zy}{f}_x\left(x\right)dxdy \\ ={\int }_0^{\infty }{F}_x\left(zy\right)f_y\left(y\right)dy \end{gathered}$$

Now that we have,

Step 3: Explanation (Part b)

Take any z > 0. Lets find the CDF of Z. We have that

$$\begin{gathered} F_z\left(z\right)=P(Z\le z)=P(XY\le z) \\ =P(X\le \frac{z}{Y})={\int }_0^{\infty }{f}_Y\left(y\right){\int }_0^{\frac{z}{y}}{f}_x\left(x\right)dxdy \\ ={\int }_0^{\infty }{F}_x\left(\frac{z}{y}\right)f_y\left(y\right)dy \end{gathered}$$

Now we have that

$$\begin{gathered} F_z\left(z\right)=\frac{dF}{dz}\left(z\right)=\frac{d}{dz}{\int }_0^{\infty }{F}_x\left(\frac{z}{y}\right)f_y\left(y\right)dy \\ ={\int }_0^{\infty }\frac{d}{dz}{F}_x\left(\frac{z}{y}\right)f_y\left(y\right)dy \end{gathered}$$

If $X~Expo\left(\lambda \right) and Y~Expo\left(\mu \right),$we have in (a)

$$\begin{gathered} {\int }_0^{\infty }fx\left(zy\right)yfy\left(y\right)dy={\int }_0^{\infty }\lambda e^{-\lambda zy}y\mu e^{-\mu y}dy \\ =\lambda \mu {\int }_0^{\infty }y{e}^{-(\lambda z+\mu )y}dy \\ =\frac{\lambda \mu }{{(\lambda z+\mu )}^2} \end{gathered}$$