Question

Assume that X and Y are jointly continuous random variables with joint probability density function given by f(x,y)=\begin{cases}\frac{1}{36}(3x-xy+4y)\ if\ 0 < x < 2\ and\ 1 < y < 3\\0\ \ \ \ \ othrewise\end{cases} Find the marginal density functions for X and Y .

Random variables
ANSWERED
asked 2021-06-03
Assume that X and Y are jointly continuous random variables with joint probability density function given by
\(f(x,y)=\begin{cases}\frac{1}{36}(3x-xy+4y)\ if\ 0 < x < 2\ and\ 1 < y < 3\\0\ \ \ \ \ othrewise\end{cases}\)
Find the marginal density functions for X and Y .

Expert Answers (1)

2021-06-04
Step 1
The joint density function for X and Y is given below:
\(f(x,y)=\begin{cases}\frac{1}{36}(3x-xy+4y)\ if\ 0 < x < 2\ and\ 1 < y < 3\\0\ \ \ \ \ othrewise\end{cases}\)
The marginal density function of X, g(x) is obtained as given below:
\(g(x)=\int_{1}^{3}\frac{1}{36}(3x-xy+4y)dy\)
\(=\frac{1}{36}(3xy-x\frac{y^{2}}{2}+4\frac{y^{2}}{2})_{1}^{3}\)
\(=\frac{1}{36}(2x+16)\)
\(=\frac{2}{36}(x+8)\)
\(=\frac{1}{18}(x+8)\)
\(=\begin{cases}\frac{1}{18}(x+8),\ for\ 0< x < 2\\0, \ \ \ otherwise\end{cases}\)
The marginal density function of y, h(y) is obtained as given below:
\(h(y)=\int_{0}^{2}\frac{1}{36}(3x-xy+4y)dx\)
\(=\frac{1}{36}(3\frac{x^{2}}{2}-\frac{x^{2}}{2}y+4xy)_{0}^{2}\)
\(=\frac{6}{36}(1+y)\)
\(=\frac{1}{6}(y+1)\)
\(=\begin{cases}\frac{1}{6}(y+1),\ for\ 1< y < 3\\0, \ \ \ otherwise\end{cases}\)
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