# [SOLVED] Expectation of a Conditional Distribution

• November 25th 2008, 08:06 PM
akolman
[SOLVED] Expectation of a Conditional Distribution
I am stuck in this problem

Let X and Y have the joint pdf $f(x,y)=1, -x zero elsewhere. Show that, on the set of positive probability density, the graph $E(Y|x)$ is a straight line whereas that of $E(X|y)$ 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
$f_{1}(x)=\int_{-x}^{x} f(x,y)dy=\int_{-x}^{x} 1 dy = y|^{-x}_{x}=x-(-x)=2x$ for $0

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

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

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

$f_2(y)=\int ^?_? f(x,y)dx$
• November 26th 2008, 01:54 AM
Quote:

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 :)

Quote:

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.
$-x-y$, and together with x>y this gives x>|y|.
So the limits are |y|<x<1.