Suppose x,y are uniformly distributed over the region 0=< X =< Y =< 1
Prove that E(x|Y=y) = y/2 and E(Y|X=x) = x + (1-x)/2
So, I found the density function f(x,y) to be equal to 2, as long as x and y are within the bounds.
I set up E(x|Y=y) = integral from 0 to y of x*f(x|y) dx
f(x|y) = f(x,y)/f(y)
f(y) = integral from 0 to 1 of 2 dx, therefore 2.
Then f(x|y) = 2/2 = 1
And E(x|Y=y) = integral from 0 to y of x * dx, but then that would give (y^2)/2, which doesn't appear to be correct...can anyone help me please? I'm slightly befuddled. Also, would the second one go the same way?
I got kind of stuck on the second one. I used E(Y|X=x) = integral from x to 1 of y*f(y|x) dy, with f(y|x) = f(y,x)/f(x)
i got f(x) = integral x to 1 of 2dy = 2(1-x)
problem here is that this puts the (1-x) in the denominator of my original function, and it needs to be in the numerator...