Expected absolute difference between two iid variables Suppose $X$ and $Y$ are iid random variables taking values in $[0,1]$, and let $\alpha > 0$. What is the maximum possible value of $\mathbb{E}|X-Y|^\alpha$? 
I have already asked this question for $\alpha = 1$ here: one can show that $\mathbb{E}|X-Y| \leq 1/2$ by integrating directly, and using some clever calculations. Basically, one has the useful identity $|X-Y| = \max{X,Y} - \min{X,Y}$, which allows a direct calculation. There is an easier argument to show $\mathbb{E}|X - Y|^2 \leq 1/2$. In both cases, the maximum is attained when the distribution is Bernoulli 1/2, i.e. $\mathbb{P}(X = 0) = \mathbb{P}(X = 1) = 1/2$. I suspect that this solution achieves the maximum for all $\alpha$ (it is always 1/2), but I have no ideas about how to try and prove this. 
Edit 1: @Shalop points out an easy proof for $\alpha > 1$, using the case $\alpha = 1$. Since $|x-y|^\alpha \leq |x-y|$ when $\alpha > 1$ and $x,y \in [0,1]$, 
$E|X-Y|^\alpha \leq E|X-Y| \leq 1/2$.  
So it only remains to deal with the case when $\alpha \in (0,1)$.

This isn't a full solution, but it's too long for a comment. 
For fixed $0<\alpha<1$ we can get an approximate solution by considering the problem discretized to distributions that only take on values of the form $\frac{k}{n}$ for some reasonably large $n$.  Then the problem becomes equivalent to 
$$\max_x x^T A x$$
where $A$ is the $(n+1) \times (n+1)$ matrix whose $(i,j)$ entry is $\left(\frac{|i-j|}{n}\right)^{\alpha}$, and the maximum is taken over all non-negative vectors summing to $1$.  
If we further assume that there is a maximum where all entries of $x$ are non-zero, Lagrange multipliers implies that the optimal $x$ in this case is a solution to 
$$Ax=\lambda {\mathbb 1_{n+1}}$$
(where $1_{n+1}$ is the all ones vector), so we can just take $A^{-1} \mathbb{1_{n+1}}$ and rescale.  
For $n=1000$ and $n=\frac{1}{2}$, this gives a maximum of approximately $0.5990$, with a vector whose first few entries are $(0.07382, 0.02756, 0.01603, 0.01143)$.

If the optimal $x$ has a density $f(x)$ that's positive everywhere, 
and we want to maximize 
$\int_0^1 \int_0^1 f(x) f(y) |x-y|^{\alpha}$
the density "should" (by analogue to the above, which can probably be made rigorous) satisfy
$$\int_{0}^1 f(y) |x-y|^{\alpha} \, dy= \textrm{ constant independent of } x,$$
but I'm not familiar enough with integral transforms to know if there's a standard way of inverting this.  
