Thread: [SOLVED] Conditional Expectation- Point me in right direction

1. [SOLVED] Conditional Expectation- Point me in right direction

I am having trouble getting started in the right direction with the following problem:

Random variable X~U[0,1] and the conditional distribution of Y is Y|X=x~U[0,x], what is E[y].

This is what I have:

1.) Definition of conditional expectation
$
f(Y|X=x) = \frac{f_{XY} (x,y)}{f_x (x)}
$

2.) Get fx,y(x,y)

3.) get fx(x)

I know the general steps, but I don't know how to set them up. Can somebody please put these steps in working form and get me started. Thanks

2. $f_{Y|X}(y|x)= 1/x$ on 0<y<x

$f_X(x)= 1$ on 0<x<1

Thus the joint density is the product... $f_{X,Y}(x,y)= 1/x$ on 0<y<x<1.

you have the density of X, do you want the density of Y?

$f_Y(y)= \int_y^1 {dx\over x}=-\ln y$ on 0<y<1, which is a valid density via l'hopital's rule.

So, $E(Y)= -\int_0^1 y\ln y dy$.

3. Thanks a million.

4. Originally Posted by spearfish
Thanks a million.

as in dollars? or ruppees?