91Ó°ÊÓ

The function of joint probability density X, as well as Y, is,

\(f\left( {x,y} \right) = \left\{ \begin{array}{l}\frac{{15}}{4}{x^2}\;for\;0 \le y \le 1 - {x^2}\\0\;otherwise\end{array} \right.\)

The marginal probability density function of X is,

\(\begin{array}{c}{f_X}\left( x \right) = \int\limits_0^{1 - {x^2}} {{f_{X,Y}}\left( {x,y} \right)dy} \\ = \int\limits_0^{1 - {x^2}} {\left( {\frac{{15}}{4}} \right){x^2}dy\;\forall - 1 \le x \le 1} \\ = \left. {\left( {\frac{{15}}{4}} \right){x^2}y} \right|_0^{1 - {x^2}}\\ = \left( {\frac{{15}}{4}} \right){x^2}\left( {1 - {x^2}} \right)\end{array}\)

Thus, the marginal p.d.f of X is\({f_X}\left( x \right) = \left( {\frac{{15}}{4}} \right){x^2}\left( {1 - {x^2}} \right)\).

Now for Y, the range is,\(0 \le y \le 1 - {x^2}\).

So,

\(\begin{array}{c}{x^2} \le 1 - y\\ \Rightarrow - \sqrt {1 - y} \le x \le \sqrt {1 - y} \end{array}\)

So, the marginal probability density function of Y is,

\(\begin{array}{c}{f_Y}\left( y \right) = \int\limits_{ - \sqrt {1 - y} }^{\sqrt {1 - y} } {{f_{X,Y}}\left( {x,y} \right)dx} \\ = \int\limits_{ - \sqrt {1 - y} }^{\sqrt {1 - y} } {\left( {\frac{{15}}{4}} \right){x^2}dx\;\forall - \sqrt {1 - y} \le x \le \sqrt {1 - y} } \\ = \left. {\left( {\frac{5}{4}} \right){x^3}} \right|_{ - \sqrt {1 - y} }^{\sqrt {1 - y} }\\ = \left( {\frac{5}{4}} \right)\left\{ {{{\left( {1 - y} \right)}^{\frac{3}{2}}} + {{\left( {1 - y} \right)}^{\frac{3}{2}}}} \right\}\\ = \frac{5}{2}{\left( {1 - y} \right)^{\frac{3}{2}}}\end{array}\)

Thus, the marginal p.d.f for Y is, \(\frac{5}{2}{\left( {1 - y} \right)^{\frac{3}{2}}}\).

We can see the limits of the integration for the joint p.d.f of (X, Y), which is\(0 \le y \le 1 - {x^2}\)The limits cannot be used separately for Xand Y. So, we can conclude that X and Y are not independent.

Another way to check independence is thatthe support of X and Y is not equal to the product of support of individual X and Y. So, x and Yare not independent.