Let $\ X, Y \sim U(0,1) $ be random independent variables.
How do I prove that: $$\ \text{E}[|X-Y|^a] = \frac{2}{(a+1)(a+2)} $$
my attempt is to split it to two parts in order to get rid of absolute value function. My problem is setting the integral limits.
$$\ E[|X-Y|^a] = \int \int (x-y)^a dy dx + \int \int (y-x)^a dx dy $$
Where first part is where $\ x > y $ and the second part is $\ y > x $.
$$\ \int \int (x-y)^a f_y f_x \ dy \ dx \ \\ $$
I tried to set the $\ y $ integral from $\ 0 $ to some value $\ k $ and then $\ x $ integrate from $\ k $ to $\ 1 $ and then do vice versa on the second part.
Didn't work though..
You are correct in splitting the integral. The only error that I see is in the limits for $x$. This is what I think should lead you to the correct answer.
$$I_1 = \int\limits_0^1 \int\limits_0^x(x-y)^a dy dx$$
Since $y<x$, $y$ integrates from $0$ to $x$. However, when you integrate $x$, it should be allowed to take all the values between $0$ and $1$. Hence, $I_1 = \frac{1}{(a+1)(a+2)}$ and so is the other integral. Hope this helps.