I am trying to walk through the following problem, which has been discussed in this thread: Expectation of absolute difference of two uniform random variables
The problem is as follows:
Let ∼(0,1) and ∼(0,2) be independent random variables. Find [|−|]
The answer in the responses states that:
$$(|−|)=∫_0^2∫_0^1\frac{|x-y|}{2}=\frac{2}{3}.$$
In the comments below this answer, the following formula is provided to solve the above integral: $$∫_0^1|−|=∫_0^y(−)+∫_^1(−)$$
However, when I try to solve the above integral using this formula, I get a different answer. Here is my work:
$$∫_0^2∫_0^1\frac{|x-y|}{2}$$ $$=\frac{1}{2}∫_0^2∫_0^1|x-y|$$ $$=\frac{1}{2}∫_0^2(∫_0^y(−)+∫_^1(−))$$ $$=\frac{1}{2}∫_0^2(yx-\frac{x^2}{2})\Big|_0^y + (\frac{x^2}{2}-xy)\Big|_y^1$$ $$=\frac{1}{2}∫_0^2((y^2-\frac{y^2}{2}) - (y\cdot0 - \frac{0^2}{2}) + ((\frac{1^2}{2}-1\cdot y) - (\frac{y^2}{2} - y^2)$$ $$=\frac{1}{2}∫_0^2((\frac{y^2}{2}) + (\frac{1}{2} - y) + \frac{y^2}{2}$$ $$=\frac{1}{2}∫_0^2(y^2 - y + \frac{1}{2})$$ $$=\frac{1}{2}(\frac{y^3}{3} - \frac{y^2}{2} + \frac{y}{2})\Big|_0^2$$ $$=\frac{1}{2}(\frac{8}{3} - \frac{4}{2} + \frac{1}{2} - (0 - 0 + 0))$$ $$=\frac{1}{2} \cdot \frac{5}{3}$$ $$=\frac{5}{6}$$
Is there something wrong with my math here? Or is the formula from the comment I'm using to solve this incorrect? If so, what is the correct formula?
The comment you linked suggests that the formula for the integral of the absolute difference is valid when $y$ is in the range $(0,1)$, which is indeed the case. For $y > 1$ it no longer holds. So in order to apply this formula to solve the double integral you should first split the $y$-integration into two integrals, one from $0$ to $1$ and one from $1$ to $2$.
For the second integral notice that when $y>1$ and $x\in (0,1)$ you always have $|x-y| = y-x$. I'll leave the details for you to fill out.