Question

Consider two continuous random variables X and Y with joint pdf f(x,y)= k(2y- x 4 ) for 1

Answer 1

Answer:

The answer is "1.6 and $\frac{11}{12}$".

Step-by-step explanation:

The joint pdf by x and y

$1< x<2\\ 0< x< 1$

In point a:

$\to K \int^1_{y=0} \int^2_{x=1}  (2y - \frac{x}{4}) dx, dy =1\\\\ \to k \int^{1}_{y=0} 2y - \frac{x^2}{8} \int^{2}_{1} dy=1\\\\\to k \ \ y^2 \int^1_{0} - \frac{1}{8} (3) =1 \\\\ \to k \ (1-\frac{3}{8}) =1\\\\\to k= \frac{8}{5}$

       $= 1.6$

In point b:

In the original pdf of x

$\to fx(x)= k \int^{1}_{y=0} (2y-\frac{x}{4}) dy =k$

$= k \int^{2}_{1} (x-\frac{x^2}{4}) dx =k\\\\= k \frac{x^2}{2} - \frac{x^3}{12})^2_{1} =k\\\\= k \frac{1}{2} \times 3 - \frac{1}{12} \times 7 =k \\\\ =\frac{3}{2} - \frac{7}{12}\\\\= \frac{11}{12}$