Question

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

Answer 1

The answer is "1.6 and
`(11)/(12)`".

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 - (x)/(4)) dx, dy =1\\\\ \to k \int^(1)_(y=0) 2y - (x^2)/(8) \int^(2)_(1) dy=1\\\\\to k \ \ y^2 \int^1_(0) - (1)/(8) (3) =1 \\\\ \to k \ (1-(3)/(8)) =1\\\\\to k= (8)/(5)\\\\`

`= 1.6`

In point b:

In the original pdf of x

`\to fx(x)= k \int^(1)_(y=0) (2y-(x)/(4)) dy =k`

`= k \int^(2)_(1) (x-(x^2)/(4)) dx =k\\\\= k (x^2)/(2) - (x^3)/(12))^2_(1) =k\\\\= k (1)/(2) * 3 - (1)/(12) * 7 =k \\\\ =(3)/(2) - (7)/(12)\\\\= (11)/(12)`