Two points :A and B are picked randomly (uniformly) on [0,1] independent of one another. What is the expected value of length AB?

Question

Two points :A and B are picked randomly (uniformly) on [0,1] independent of one another. What is the expected value of length AB?

E[A]=E[B]= $\frac12$ since A,B ~ U(0,1)

Now using conditional expectation:

$E[AB]=E[E[AB|E[A]=E[B]]$

This is where I'm confused. Why aren't we integrating $\frac{1}{4}$ from 0 to 1?

I'd appreciate some insight as to why the solution is $\frac{1}{3}$ and not $\frac15$

Answer 1

$\int_{x=0}^1 \int_{y=0}^1 |x-y| dx dy = 2\int_{x=0}^1 \int_{y=0}^x (x-y) dx dy = 2\int_{x=0}^1 {x^2 \over 2} dx = {1 \over 3}$.