Q. 6.41 - A First Course in Probability

Question

The joint density function of X and Y is given by 

$f(x,y)=x{e}^{-x(y+1)} x>0,y>0$

(a) Find the conditional density of X, given Y = y, and that of Y, given X = x. 

(b) Find the density function of Z = XY. 

Step-by-Step Solution

Answer

The conditional density of X is $x(y+1{)}^2{e}^{-x(y+1)} and x{e}^{-xy}$ when Y = y, and X = x respectively.

Step 1: Given information (part a)

$f(x,y)=x{e}^{-x(y+1)} x>0,y>0$

Step 2: Explanation

the conditional density of X and Y = y is 

$$\begin{gathered} f_{xy}(x\mid y)=\frac{f(x,y)}{f_y\left(y\right)} \\ =\frac{x{e}^{-x(y+1)}}{\frac{1}{(y+1{)}^2}} \\ =x(y+1{)}^2{e}^{-x(y+1)} \end{gathered}$$

the conditional density of X when X = x is,

$$\begin{gathered} f_{xy}(y\mid x)=\frac{f(x,y)}{f_y\left(x\right)} \\ =\frac{x{e}^{-x(y+1)}}{e^{-x}} \\ =x{e}^{-xy} \end{gathered}$$

Step 1: Given information (part b)

Density function $Z=XY$

Step 2: Explanation (part b)

The cumulative distributive function of the random variable $Z=XY$ is,

$$\begin{gathered} F_z\left(a\right)=P(Z\le a) \\ =P\left(XY\le a\right) \\ =P\left(X>0,Y\le \frac{a}{x}\right) \\ ={\int }_0^{\infty }{\int }_0^{a/x}f(x,y)dydx \\ ={\int }_0^{\infty }{\int }_0^{a/x}x{e}^{-x(y+1)}dydx \\ ={\int }_0^{\infty }x{e}^{-x}{\int }_0^{a/x}{e}^{-xy}dydx \\ =\int e^{-x}{\left[\frac{e^{-xy}}{-x}\right]}_0^{a/x}dx \\ =\int e^{-x}\left[e^{-x(a/x)}-1\right]dx \\ =\int e^{-x}\left[e^{-a}-1\right]dx \\ =\left(1-e^{-a}\right){\int }_0^{\infty }{e}^{-x}dx \\ =\left(1-e^{-a}\right){\left[\frac{e^{-x}}{-1}\right]}_0^{\infty } \\ =1-e^{-a} \end{gathered}$$