Finding $\mathbb{E}(\xi|\eta)$

Question

Vector $(\xi, \eta)$ is evenly distributed in set $D=\{ (x,y): 0\leq x\leq 1, 0\leq y \leq 1, x+y \leq 1 \}$. I need to find $\mathbb{E}(\xi|\eta)$.

So first of all I found the size of $D$. It is $\int_{0}^{1}(1-x)dx= \frac{1}{2}.$

Now I think I need to find the density function $p(x,y)$. How could I find it? Is it 2?

Than am I right $p_1(x) = 2(1-x)$ and $p_2(y)=2y?$

Next step maybe should be finding $p_2(y|x)$ and $p_1(x|y)$ and after than I can find $\mathbb{E}(\xi|\eta)$. Am I right? How to find all those density functions?

Answer 1

The area enclosed by the region $D$ is $\frac{1}{2}$.

Hence the density of $(X,Y)$ uniform on $D$ is

$$f_{X,Y}(x,y)=2\,\mathbf1_{0<x,y<1\,,\,x+y<1}$$

Rewriting the above we see that the density factors as

$$f_{X,Y}(x,y)=\underbrace{\frac{1}{1-y}\mathbf1_{0<x<1-y}}_{f_{X\mid Y}(x)}\,\,\underbrace{2(1-y)\mathbf1_{0<y<1}}_{f_Y(y)}$$

So $X\mid Y$ is uniformly distributed over $(0,1-Y)$, giving $$E(X\mid Y)=\frac{1-Y}{2}$$