1. ## Conditional Expectations

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?
Thank you!

2. Originally Posted by mistykz
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?
Thank you!
I assume X and Y are independent (the answers you quote suggest that they must be).

I will set one up for you:

$E(X \, | \, Y = y) = \int_0^y x \, f(x \, | \, y) \, dx$ where $f(x \, | \, y) = \frac{f(x, y)}{f(y)} = \frac{1}{y}$ since $f(x, y) = (1) (1) = 1$ for $0 \leq x \leq y \leq 1$ (assuming independence) and $f(y) = \int_0^y 1 \, dx = y$.

Do the calculation and you will get the answer you mention. You should now attempt finding $E(Y \, | \, X = x)$.

3. 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...

4. Originally Posted by mistykz
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) Mr F says: See first comment below.
problem here is that this puts the (1-x) in the denominator of my original function, and it needs to be in the numerator...
$f(x, y) \neq 2$. $f(x, y) = 1$. I said this in my first post. So $f(x) = 1 - x$.

Then $E(Y \, | \, X = x) = \int_{y=x}^{y=1} \frac{y}{1 - x} \, dy = \frac{1+x}{2} = x + \frac{1-x}{2}$.

5. Based on...
'Suppose x,y are uniformly distributed over the region 0=< X =< Y =< 1'
the random variables are NOT independent.
$P(X>.5|Y=.5)\not= P(X>.5|Y=.3)$
If they are indep, then the region must be a rectangle.
This is a triangle.
$f(x,y)=2$ on $0\le x\le y \le 1$.

6. The joint density is $f(x,y)=2$ on $0.
So the marginal of Y is $\int_0^y 2 dx=2y$ on $0.
Hence $f_{X|Y}=1/y$ on $0.
Giving us $E(X|Y)=\int_0^y x/y dx=y/2$.

7. The second aswer is correct, but it reduces to (x+1)/2.
Since the joint density is 2 on that triangle, we have

$f(x)=2\int_x^1 dy=2(1-x)$.

Thus the conditional density of Y given X is 1/(1-x) on that same triangle.
Therefore
$E(Y|X)=\int_x^1 {y\over 1-x}dy={1-x^2\over 2(1-x)}={1+x\over 2}$.

8. Originally Posted by matheagle
Based on...
'Suppose x,y are uniformly distributed over the region 0=< X =< Y =< 1'
the random variables are NOT independent.
$P(X>.5|Y=.5)\not= P(X>.5|Y=.3)$
If they are indep, then the region must be a rectangle.
This is a triangle.
$f(x,y)=2$ on $0\le x\le y \le 1$.
That's what happens when my mind is on the tour .....