91Ó°ÊÓ

X and Y have a continuous joint distribution.

The joint probability density function is

\(f\left( {x,y} \right) = \left\{ \begin{array}{l}k\;for\;a \le x \le b\;and\;c \le y \le d\\0\;otherwise\end{array} \right.\)

Here a<b, c<d and k>0

We have to calculate the value of k, so, we know that the information,

\(\begin{array}{l}\int\limits_{x = - \infty }^\infty {\int\limits_{y = - \infty }^\infty {f\left( {x,y} \right)dxdy} } = 1\\ \Rightarrow \int\limits_a^b {\int\limits_c^d {kdxdy} } = 1\\ \Rightarrow \left( {b - a} \right)\left( {d - c} \right)k = 1\\ \Rightarrow k = \frac{1}{{\left( {b - a} \right)\left( {d - c} \right)}}\end{array}\)

The marginal distribution of X is,

\(\begin{array}{c}f\left( x \right) = \int\limits_c^d {kdy} \\ = k\left( y \right)_c^d\\ = \frac{{\left( {d - c} \right)}}{{\left( {b - a} \right)\left( {d - c} \right)}}\\ = \frac{1}{{\left( {b - a} \right)}}\;;\;\forall a \le x \le b\end{array}\)

Thus, we can conclude that the marginal distribution of X follows uniform distribution with the interval [a,b].

The marginal distribution of Y is,

\(\begin{array}{c}f\left( y \right) = \int\limits_a^b {kdx} \\ = k\left( y \right)_a^b\\ = \frac{{\left( {b - a} \right)}}{{\left( {b - a} \right)\left( {d - c} \right)}}\\ = \frac{1}{{\left( {d - c} \right)}}\;;\;\forall c \le x \le d\end{array}\)

Thus, we can conclude that the marginal distribution of Y follows uniform distribution with the interval [c,d].