Answer to Question #270913 in Statistics and Probability for Neil

Question

Let the joint probability functions of X and Y be

f(x,y) = 2(x + y - 3xy²) , if 0 < x < 1, 0 < y < 1

0 , Otherwise

Find (i) Marginal probability density function of X and Y

(ii) Conditionally density functions

Answer 1

(i)

$$f_X(x)=\displaystyle\int_{-\infin}^{\infin}f(x, y)dy=\displaystyle\int_{0}^{1} 2(x + y - 3xy^2)dy$$

$$=2[xy+\dfrac{y^2}{2}-xy^3]\begin{matrix} 1 \\ 0 \end{matrix}=2x+1-2x=1$$

$$f_X(x)=\begin{cases} 1 & 0<x<1\\ 0 & otherwise \end{cases}$$

$$f_Y(y)=\displaystyle\int_{-\infin}^{\infin}f(x, y)dx=\displaystyle\int_{0}^{1} 2(x + y - 3xy^2)dx$$

$$=2[\dfrac{x^2}{2}+xy-\dfrac{3x^2y^2}{2}]\begin{matrix} 1 \\ 0 \end{matrix}=1+2y-3y^2$$

$$f_Y(y)=\begin{cases} 1+2y-3y^2 & 0<y<1\\ 0 & otherwise \end{cases}$$

(ii)

The conditional probability density function of $Y$ given that $X = x$ is

$$f_{Y|X}=\dfrac{f(x, y)}{f_X(x)}=\dfrac{2(x + y - 3xy^2)}{1}$$

$$=2(x + y - 3xy^2),0 < x < 1, 0 < y < 1$$

The conditional probability density function of $X$ given that $Y=y$ is

$$f_{X|Y}=\dfrac{f(x, y)}{f_Y(y)}=\dfrac{2(x + y - 3xy^2)}{1+2y-3y^2} ,$$

$$0 < x < 1, 0 < y < 1$$