Thread: [SOLVED] Expectation of a Conditional Distribution

1. [SOLVED] Expectation of a Conditional Distribution

I am stuck in this problem

Let X and Y have the joint pdf $\displaystyle f(x,y)=1, -x<y<x, 0<x<1,$ zero elsewhere. Show that, on the set of positive probability density, the graph $\displaystyle E(Y|x)$ is a straight line whereas that of $\displaystyle E(X|y)$ is not a straight line.

Here's what I have so far, I am not sure if it's right though

The marginal pdf of X is
$\displaystyle f_{1}(x)=\int_{-x}^{x} f(x,y)dy=\int_{-x}^{x} 1 dy = y|^{-x}_{x}=x-(-x)=2x$ for $\displaystyle 0<x<1$

So the conditional $\displaystyle f(Y|x)$ is
$\displaystyle f(Y|x)=\frac{1}{2x}$ for $\displaystyle 0<x<1$

$\displaystyle E(Y|x)=\int^{x} _{-x} \frac{y}{2x}dy=\frac {y^2}{4x} |^{x} _{-x}=\frac{x^2-x^2}{4x}=0$???

What are the limits of integration for the marginal pdf of Y??

$\displaystyle f_2(y)=\int ^?_? f(x,y)dx$

2. 0<x<1

E(Y|x)=\int^{x} _{-x} \frac{y}{2x}dy=\frac {y^2}{4x} |^{x} _{-x}=\frac{x^2-x^2}{4x}=0???
Correct

What are the limits of integration for the marginal pdf of Y??

f_2(y)=\int ^?_? f(x,y)dx
if you know that Y = y, then we have the restriction -x<y<x as well as 0<x<1.
$\displaystyle -x<y \implies x>-y$, and together with x>y this gives x>|y|.
So the limits are |y|<x<1.