Conditional expectations of $E(X+Y|z)$

Question

Given: $$f(x,y,z) = \frac23 (x+y+z), \,\,\, 0<x<1,\,\,\, 0<y<1,\,\,0<z<1$$ zero elsewhere.I was instructed to determine the cumulative df of $x,y,z$. Here is my answer $$F (x,y,z) = \frac {xyz (x+xy+z)} {3} $$ Another problem that I can not answer is this one: Find the conditional distribution of $X$ and $Y$, given $Z=z$, and evaluate $E(X+Y|z)$.

Answer 1

The first thing to notice is that you have mis-interpreted the first question. There is no common notion of the c.d.f. of multiple fvariables; what your professor or book wanted is the c.d.f. of $x$ which is just like the c.d.f. of $y$ or of $z$.

Thus for a value $x = X$, $$ F(X) = \frac{2}{3}\int_{x=0}^{X} dx\int_{y=0}^{1} dy \int_{z=0}^{1} dz (x+y+z) = \frac{X^2+2X}{3} $$

The conditional distribution of $(X,Y)$ given $z=Z$ is obtained by plugging $z=Z$ into $f(x,y,z)$ and then normalizing the resulting expression; this gives $$ f(X,Y |z=Z) = \frac{x+y+Z}{1+Z} $$

Given that, then $$ E(X+Y | Z = z) = \frac{1}{1+z}\int_{x=0}^{1} dx\int_{y=0}^{1} dy (x+y) =\frac{1}{1+z} $$

Finally, to find $E(X|y,z)$ we first find $f(X|y,z)$ by the same business of plugging in and normalizing: $$ f(X|y,z) = \frac{(X+y+z)}{\int_0^1 (x+y+z) dx} = \frac{2}{1+2y+2z}(X+y+z) $$ And then $$ E(X|y,z) = \frac{2}{1+2y+2z} \int_0^1 x(x+y+z)dx =\frac{2+3(y+z)}{3+6(y+z)} $$

Answer 2

Marginal distribution of Z, $$f_{Z}(z) = \int_{0}^{1}\int_{0}^{1} \frac{2}{3} (x+y+z)dxdy$$ $$=\frac{2}{3}(1+z)$$

Conditional distribution of X,Y given (Z=z) $= \frac{f(x,y,z)}{f_{Z}(z)}$

Thus, it is = $\frac{(x+y+z)}{(1+z)}$

Marginal distribution of YZ, $$f_{YZ}(y,z) = \int_{0}^{1}\frac{2}{3}(x+y+z)dx$$ $$= \frac{1}{3}.(1+2y+2z)$$

Conditional distribution of X given (Y=y) and (Z=z) = $= \frac{f(x,y,z)}{f_{YZ}(y,z)}$

Thus it is = $\dfrac{{\frac{2}{3}(x+y+z)}}{\frac{1}{3}.(1+2y+2z)}$

$$=\dfrac{2(x+y+z)}{(1+2y+2z)}$$

E(X+Y/Z=z) $=\int_{0}^{1}\int_{0}^{1}(x+y).\frac{(x+y+z)}{(1+z)}dxdy$

When you evaluate this, you get

$$=\frac{(7+6z)}{6(1+z)}$$