Question

The joint pdf fr (x)) of two random variables X and Y is given by fo (x,y)=cx2y for x +y s1. Determi use them to determine whether or not the two random variables are statistically independent. ne the constant c. Determine the marginal pdfs "Ax) and f, (y) and

Answer 1

The support is $\fn_jvn D=x^2+y^2\leqslant 1$ . This can be expressed as $\fn_jvn D=-\sqrt{1-x^2}\leqslant y\leqslant \sqrt{1-x^2},-1\leqslant x\leqslant 1$

The condition for PDF is $\fn_jvn \int \int _Df_{XY}\left ( x,y \right )dydx=1$ . Or

$\fn_jvn \int_{-1} ^1\int _{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}cx^2y^2dydx=1\\ 4\int_{0} ^1\int _{0}^{\sqrt{1-x^2}}cx^2y^2dydx=1\\ 4\int_{0} ^1cx^2\int _{0}^{\sqrt{1-x^2}}y^2dydx=1\\ 4\int_{0} ^1cx^2\left [y^3/3 \right ] _{0}^{\sqrt{1-x^2}}dx=1\\$

$\fn_jvn \frac{4}{3}\int_{0} ^1cx^2\left ( 1-x^2 \right )^{3/2}dx=1\\$

To evaluate the integral substitute $\fn_jvn x=\sin \theta$ . Thus  $\fn_jvn {\color{Blue} c=\frac{24}{\pi }}$.

The marginal PDFs are

$\fn_jvn f_X\left ( x \right )=\int _{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}f_{XY\left ( x,y \right )}dy\\ f_X\left ( x \right )=\int _{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}cx^2y^2dy\\ f_X\left ( x \right )=2\int _{0}^{\sqrt{1-x^2}}cx^2y^2dy\\ f_X\left ( x \right )=\frac{2}{3}cx^2\left ( 1-x^2 \right )^{3/2}\\ f_X\left ( x \right )=\frac{2}{3}\left ( 24/\pi \right )x^2\left ( 1-x^2 \right )^{3/2}\\ {\color{Blue} f_X\left ( x \right )=\frac{16}{\pi }x^2\left ( 1-x^2 \right )^{3/2};-1<x<1}$

Similarly, the marginal PDF of $\fn_jvn Y$ is

$\fn_jvn {\color{Blue} f_Y\left ( y \right )=\frac{16}{\pi }y^2\left ( 1-y^2 \right )^{3/2};-1<y<1}$

Since $\fn_jvn f_X\left ( x \right ) f_Y\left ( y \right )\neq f_{XY}\left ( x,y \right )$ , the two random variables $\fn_jvn X,Y$ are not statistically independent.