X and Y have uniform joint density function:
f(x,y)= d (constant) for 0<x<1 and 0<y<1-x
1. find d
2. find p(y<x)
3. find cov (x,y)
The Attempt at a Solution
1. For this I first graphed x=1 and y=1 and created a square since x can go from 0 to a maximum of 1 and y also. I then graphed y=1-x. The area of the bottom triangle is what I want. That is 1/2. So i then do a double integral (x goes from 0 to 1 and y goes from 0 to 1-x) of (c dy dx). This should equal 1/2. I get 1/2 c equals 1/2, so c=1. Is that correct?
2. For the p(y<x), I graphed y=x and you get two triangles within the larger one. We want the area of the lower one. We could get this through simple geometry or by using integrals. If you want to use integrals, you have to split the triangle into half and you get a right triangle and a left one.
Area of left one is
Double integral (x goes form 0 to 1/2 and y goes from 0 to x) of 1, which is 1/8
Area of right one is:
Double integral (x goes from 1/2 to 1 and y goes from 0 to 1-x) of 1, which is also 1/8
So the area of the triangle is 1/8+1/8 which is 1/4. This is p(y<x)
3. Cov(x,y)= E(xy)-E(x)E(y)
I got that E(xy) is 1/24
So Cov(x,y) is 1/72.
Can someone please tell this poor, hopeless soul if these answers are even close to being correct.