Answer to Question #235587 in Statistics and Probability for Simmj sharma

Question

The joint probability distribution function of two random variable x&y given by f(x, y )=9(1+x+y)/2(1+x)^4(1+y)^4,0<x<∞,0<y ∞ find the marginal distribution

Answer 1

$$f(x, y)=\dfrac{9(1+x+y)}{2(1+x)^4(1+y)^4}$$

$$f_X(x)=\displaystyle\int_{-\infin}^{\infin}f_{XY}(x,y)dy$$

$$=\displaystyle\int_{0}^{\infin}\dfrac{9(1+x+y)}{2(1+x)^4(1+y)^4}dy$$

$$=\displaystyle\int_{0}^{\infin}\dfrac{9}{2(1+x)^4(1+y)^3}dy$$

$$+\displaystyle\int_{0}^{\infin}\dfrac{9x}{2(1+x)^4(1+y)^4}dy$$

$$=\lim\limits_{t\to \infin}\displaystyle\int_{0}^{t}\dfrac{9}{2(1+x)^4(1+y)^3}dy$$

$$+\lim\limits_{t\to \infin}\displaystyle\int_{0}^{t}\dfrac{9x}{2(1+x)^4(1+y)^4}dy$$

$$=\dfrac{9}{2(1+x)^4}\lim\limits_{t\to \infin}\big[-\dfrac{1}{2(1+y)^2}-\dfrac{x}{3(1+y)^3}\big]\begin{matrix} t\\ 0 \end{matrix}$$

$$=\dfrac{9}{2(1+x)^4}(\dfrac{1}{2}+\dfrac{x}{3})=\dfrac{3(3+2x)}{4(1+x)^4}$$

$$f_Y(y)=\displaystyle\int_{-\infin}^{\infin}f_{XY}(x,y)dx$$

$$=\displaystyle\int_{0}^{\infin}\dfrac{9(1+x+y)}{2(1+x)^4(1+y)^4}dx$$

$$=\displaystyle\int_{0}^{\infin}\dfrac{9}{2(1+x)^3(1+y)^4}dx$$

$$+\displaystyle\int_{0}^{\infin}\dfrac{9x}{2(1+x)^4(1+y)^4}dx$$

$$=\lim\limits_{t\to \infin}\displaystyle\int_{0}^{t}\dfrac{9}{2(1+x)^3(1+y)^4}dx$$

$$+\lim\limits_{t\to \infin}\displaystyle\int_{0}^{t}\dfrac{9y}{2(1+x)^4(1+y)^4}dx$$

$$=\dfrac{9}{2(1+y)^4}\lim\limits_{t\to \infin}\big[-\dfrac{1}{2(1+x)^2}-\dfrac{y}{3(1+x)^3}\big]\begin{matrix} t\\ 0 \end{matrix}$$

$$=\dfrac{9}{2(1+y)^4}(\dfrac{1}{2}+\dfrac{y}{3})=\dfrac{3(3+2y)}{4(1+y)^4}$$

$$f_X(x)=\dfrac{3(3+2x)}{4(1+x)^4}$$

$$f_Y(y)=\dfrac{3(3+2y)}{4(1+y)^4}$$