Question

Random variables z and y described by the PDF if x-+ yo 1 and x.> 0 and y, > 0 0 otherwise a Are x and y independent random variables? b Are they conditionally independent given max(x,y) S 0.5? c Determine the expected value of random variabler, defined byr xy.

Answer 1

The pdf of random variables is

$f_{x,y}\left ( x_0,y_0 \right )=\left\{\begin{matrix} K, & x_0+y_0\leqslant 1,x_0,y_0>0\\ 0, & \textup{ otherwise} \end{matrix}\right.$

The region is the triangle in the first quadrant between $x_0+y_0\leqslant 1,x_0,y_0>0$ whose area is $\frac{1}{2}\left ( 1\times 1 \right )=0.5$ . Hence $K=1/0.5=2$

The marginal pdfs are

f (zo) 1-10 fr (xo)-Kdyo

Similarly,

$f_y\left ( y_0 \right )=K\left ( 1-y_0 \right ),1\leqslant y_0\leqslant 1$

a) We can see that $f_x\left ( x_0 \right )f_y\left ( y_0 \right)\neq f_{x,y}\left ( x_0,y_0 \right )$ , are not independent.

b) The probability $P\left ( \max\left ( x_0,y_0 \right ) \leqslant 0.5\right )=\frac{0.5^2}{0.5}=0.5$ . Since the region $\max\left ( x_0,y_0 \right ) \leqslant 0.5$ is square of area 0.25.

$P\left (x_0\leqslant 0.5\right )=K\int_{0}^{0.5}\left ( 1-x_0 \right )dx_0\\ P\left (x_0\leqslant 0.5\right )=\frac{3K}{8}=\frac{3}{4}\\$

Similarly,

$P\left (y_0\leqslant 0.5\right )=\frac{3K}{8}=\frac{3}{4}\\$

We can see that  $P\left (x_0\leqslant 0.5\right )P\left (y_0\leqslant 0.5\right )\neq P\left ( \max\left ( x_0,y_0 \right ) \leqslant 0.5\right )$, $x_0,y_0$ are not conditionally independent given $\max\left ( x_0,y_0 \right ) \leqslant 0.5$ .

c) The expected value

$E\left ( xy \right )=K\int_{0}^{1}\int_{0}^{1-x_0}x_0y_0dy_0dx_0\\ E\left ( xy \right )=K\int_{0}^{1}x_0\left [ \frac{y_0^2}{2} \right ]_{y_0=0}^{1-x_0}dx_0\\ E\left ( xy \right )=K\int_{0}^{1}x_0\left [ \frac{\left ( 1-x_0 \right )^2}{2} \right ]dx_0\\$

$E\left ( xy \right )=0.5K\left [\frac{x_0^2}{2} -\frac{2x_0^3}{3}+\frac{x_0^4}{4} \right ]_0^1\\ {\color{Blue}E\left ( xy \right )=\frac{1}{12}}$