Question

The joint density of X and Y is

$f(x,y)=c\left(x^2-y^2\right)e^{-x}0\le x<\infty ,-x\le y\le x$

Find the conditional distribution of Y, given X = x.

Short answer

For $t\le x:$

$P(Y\le t\mid X=x)=\frac{3}{2x^3}\left(x^2(t+x)-\frac{t^3}{3}+\frac{x^3}{3}\right)$

For $t<-x:$

$P(Y\le t\mid X=x)=0$

For $t>x:$

$P(Y\le t\mid X=x)=1$

Step-by-step solution

Given information

Joint density of X and Y

$f(x,y)=c\left(x^2-y^2\right)e^{-x}0\le x<\infty ,-x\le y\le x$

also $X=x$

Formula to be used :

$f(y\mid x)=\frac{f(x,y)}{f\left(x\right)}$

Explanation

$$\begin{gathered} \int f(x,y)dy={\int }_{-x}^{x}c\left(x^2-y^2\right)e^{-x}dx=\frac{2c}{3}{x}^3{e}^{-x},0<x<\infty \\ Thus, \\ f(y\mid x)=\frac{3}{2x^3}\left(x^2-y^2\right),-x\le y\le x \end{gathered}$$

By integration,

For $t\le x,$

localid="1647240753725" role="math" $P(Y\le t\mid X=x)={\int }_{-x}^{t}f(x,y)dy=\frac{3}{2x^3}\left(x^2(t+x)-\frac{t^3}{3}+\frac{x^3}{3}\right)$

Then for t < -x,

$P(Y\le t\mid X=x)=0$

and for t > x,

$P(Y\le t\mid X=x)=1$