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 $\alpha \in (0,1)$.

  • $\begingroup$ Given two r.v.'s $X,Y$, we say that $X$ is greater than $Y$ in the convex order (denoted $X \succeq_c Y$) if $$Eg(x) \ge Eg(Y)$$ for all convex functions $g: \mathbf{R} \rightarrow \mathbf{R}$. The convex order is a partial order between r.v.s'. The convex order is closed under transformation by increasing convex functions: if $X \succeq_c Y$, then $g(X) \succeq_c g(Y)$ for any increasing convex function $g$. [...] $\endgroup$
    – mlc
    Commented Mar 30, 2017 at 16:48
  • $\begingroup$ [...] Suppose you can prove that the r.v. $W=X-Y$ (when $X,Y$ are Bernoulli on $\{0,1\}$) is a maximal element in the convex partial order restricted to the class ${\cal S}$ of all r.v.'s generated by the difference of two i.i.d. r.v.'s. Then, by the closure property, $E|W|^a$ remains a maximal element in the class ${\cal S}$ for any increasing convex transformation. So $Eg(|X-Y|)$ is maximised in ${\cal S}$ for any increasing convex function $g(x)$, and in particular $g(x) = x^a$ for $a > 1$. $\endgroup$
    – mlc
    Commented Mar 30, 2017 at 16:49
  • 3
    $\begingroup$ The fact that $X,Y \in [0,1]$ implies that $|X-Y|^p \leq |X-Y|$ for all $p \geq 1$. Hence $E[|X-Y|^p] \leq E[|X-Y|] \leq 1/2$ for all $p \geq 1$, so the maximizer is Bernoulli. If $p \in [0,1)$ it may be possible to find nontrivial maximizers. The best bound I can get via Jensen is: $E[|X-Y|^p] \leq E[|X-Y|]^p \leq 2^{-p}$. $\endgroup$
    – shalop
    Commented Nov 22, 2017 at 8:55
  • 2
    $\begingroup$ Actually, if the law of $X$ and $Y$ is atomless, then $|X-Y|^{\alpha} \to 1$ a.s. as $\alpha \to 0$. therefore by DCT $$\lim_{\alpha \to 0} E[|X-Y|^{\alpha}]=1$$which shows that the bound cannot be $1/2$ for very small $\alpha>0$. So, some nontrivial behavior is found in the regime of small $\alpha$. Maybe $2^{-\alpha}$ is optimal, as in the previous comment. $\endgroup$
    – shalop
    Commented Nov 22, 2017 at 10:06
  • 1
    $\begingroup$ @JRichey Yes, take $X,Y \sim U[0,1]$. You can do an easy integral to compute that $E[|X-Y|^{\alpha}] = \frac{2}{(\alpha+1)(\alpha+2)}$. So if you plug in $\alpha=\frac{1}{2}$, you get $E[|X-Y|^{1/2}]=\frac{8}{15}$. $\endgroup$
    – shalop
    Commented Nov 28, 2017 at 7:39

4 Answers 4


Throughout this answer, we will fix $\alpha \in (0, 1]$.

Let $\mathcal{M}$ denote the set of all finite signed Borel measures on $[0, 1]$ and $\mathcal{P} \subset \mathcal{M}$ denote the set of all Borel probability measure on $[0, 1]$. Also, define the pairing $\langle \cdot, \cdot \rangle$ on $\mathcal{M}$ by

$$ \forall \mu, \nu \in \mathcal{M}: \qquad \langle \mu, \nu\rangle = \int_{[0,1]^2} |x - y|^{\alpha} \, \mu(dx)\nu(dy). $$

We also write $I(\mu) = \langle \mu, \mu\rangle$. Then we prove the following claim.

Proposition. If $\mu \in \mathcal{P}$ satisfies $\langle \mu, \delta_{t} \rangle = \langle \mu, \delta_{s} \rangle$ for all $s, t \in [0, 1]$, then $$I(\mu) = \max\{ I(\nu) : \nu \in \mathcal{P}\}.$$

We defer the proof of the lemma to the end and first rejoice its consequence.

  • When $\alpha = 1$, it is easy to see that the choice $\mu_1 = \frac{1}{2}(\delta_0 + \delta_1)$ works.

  • When $\alpha \in (0, 1)$, we can focus on $\mu_{\alpha}(dx) = f_{\alpha}(x) \, dx$ where $f_{\alpha}$ is given by

$$ f_{\alpha}(x) = \frac{1}{\operatorname{B}(\frac{1-\alpha}{2},\frac{1-\alpha}{2})} \cdot \frac{1}{(x(1-x))^{\frac{1+\alpha}{2}}}, $$

Indeed, for $y \in [0, 1]$, apply the substitution $x = \cos^2(\theta/2)$ and $k = 2y-1$ to write

$$ \langle \mu_{\alpha}, \delta_y \rangle = \int_{0}^{1} |y - x|^{\alpha} f_{\alpha}(x) \, dx = \frac{1}{\operatorname{B}(\frac{1-\alpha}{2},\frac{1-\alpha}{2})}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left| \frac{\sin\theta - k}{\cos\theta} \right|^{\alpha} \, d\theta. $$

Then letting $\omega(t) = \operatorname{Leb}\left( \theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \ : \ \left| \frac{\sin\theta - k}{\cos\theta} \right| > t \right)$, we can check that this satisfies $\omega(t) = \pi - 2\arctan(t)$, which is independent of $k$ (and hence of $y$). Moreover,

$$ \langle \mu_{\alpha}, \delta_y \rangle = \frac{1}{\operatorname{B}(\frac{1-\alpha}{2},\frac{1-\alpha}{2})} \int_{0}^{\infty} \frac{2t^{\alpha}}{1+t^2} \, dt = \frac{\pi}{\operatorname{B}(\frac{1-\alpha}{2},\frac{1-\alpha}{2})\cos(\frac{\pi\alpha}{2})} $$

Integrating both sides over $\mu(dy)$, we know that this is also the value of $I(\mu_{\alpha})$.

So it follows that

\begin{align*} &\max \{ \mathbb{E} [ |X - Y|^{\alpha}] : X, Y \text{ i.i.d. and } \mathbb{P}(X \in [0, 1]) = 1 \} \\ &\hspace{1.5em} = \max_{\mu \in \mathcal{P}} I(\mu) = I(\mu_{\alpha}) = \frac{\pi}{\operatorname{B}(\frac{1-\alpha}{2},\frac{1-\alpha}{2})\cos(\frac{\pi\alpha}{2})}. \end{align*}

Notice that this also matches the numerical value of Kevin Costello as

$$ I(\mu_{1/2}) = \frac{\sqrt{2}\pi^{3/2}}{\Gamma\left(\frac{1}{4}\right)^2} \approx 0.59907011736779610372\cdots. $$

The following is the graph of $\alpha \mapsto I(\mu_{\alpha})$.

$\hspace{8em} $enter image description here

Proof of Proposition. We first prove the following lemma.

Lemma. If $\mu \in \mathcal{M}$ satisfies $\mu([0,1]) = 0$, then we have $I(\mu) \leq 0$.

Indeed, notice that there exists a constant $c > 0$ for which

$$ \forall x \in \mathbb{R}: \qquad |x|^{\alpha} = c\int_{0}^{\infty} \frac{1 - \cos (xt)}{t^{1+\alpha}} \, dt $$

holds. Indeed, this easily follows from the integrability of the integral and the substitution $|x|t \mapsto t$. So by the Tonelli's theorem, for any positive $\mu, \nu \in \mathcal{M}$,

\begin{align*} \langle \mu, \nu \rangle &= c\int_{0}^{\infty} \int_{[0,1]^2} \frac{1 - \cos ((x - y)t)}{t^{1+\alpha}} \, \mu(dx)\nu(dy)dt \\ &= c\int_{0}^{\infty} \frac{\hat{\mu}(0)\hat{\nu}(0) - \operatorname{Re}( \hat{\mu}(t)\overline{\hat{\nu}(t)} )}{t^{1+\alpha}} \, dt, \end{align*}

where $\hat{\mu}(t) = \int_{[0,1]} e^{itx} \, \mu(dx)$ is the Fourier transform of $\mu$. In particular, this shows that the right-hand side is integrable. So by linearity this relation extends to all pairs of $\mu, \nu$ in $\mathcal{M}$. So, if $\mu \in \mathcal{M}$ satisfies $\mu([0,1]) = 0$ then $\hat{\mu}(0) = 0$ and thus

$$ I(\mu) = -c\int_{0}^{\infty} \frac{|\hat{\mu}(t)|^2}{t^{1+\alpha}} \, dt \leq 0, $$

completing the proof of Lemma. ////

Let us return to the original proof. Let $m$ denote the common values of $\langle \mu, \delta_t\rangle$ for $t \in [0, 1]$. Then for any $\nu \in \mathcal{P}$

$$ \langle \mu, \nu \rangle = \int \left( \int_{[0,1]} |x - y|^{\alpha} \, \mu(dx) \right) \, \nu(dy) = \int \langle \mu, \delta_y \rangle \, \nu(dy) = m. $$

So it follows that

$$ \forall \nu \in \mathcal{P} : \qquad I(\nu) = I(\mu) + 2\underbrace{\langle \mu, \nu - \mu \rangle}_{=m-m = 0} + \underbrace{I(\nu - \mu)}_{\leq 0} \leq I(\mu) $$

as desired.

  • $\begingroup$ How do you know this is also the maximum over all distributions, not just distributions that have a density? And how did you get that solution? $\endgroup$
    – J Richey
    Commented Nov 29, 2017 at 3:07
  • $\begingroup$ Right, I had guessed in one of the comments above that the correct answer should be Beta distributed, but I had no justification other than the idea that it should be symmetric about $1/2$, have smooth density, and converge weakly to Bernoulli(1/2) as $\alpha \to 1$. This seems like a good justification, even if it's still a bit informal. +1. $\endgroup$
    – shalop
    Commented Nov 29, 2017 at 3:31
  • $\begingroup$ Perhaps there are some information theory techniques that might apply here... $\endgroup$
    – J Richey
    Commented Nov 29, 2017 at 3:46
  • 1
    $\begingroup$ I added the proof as promised. :) $\endgroup$ Commented Nov 29, 2017 at 7:32
  • 1
    $\begingroup$ @JRichey, One idea is the variational method as in Kevin Costello's answer. This tells us that if $\mu$ is an 'interior maximum point', then it must satisfy $$\langle\mu,\delta_s\rangle=\text{constant}$$ as in Kevin Costello's answer. But instead of checking whether such interior maximum is possible, I took the logic backward by checking that this condition is enough to guarantee that $\mu$ is a maximum point. The other idea is the lemma, which I was motivated from my previous answer. $\endgroup$ Commented Dec 3, 2017 at 14:12

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.

  • $\begingroup$ So, in the limit $n \to \infty$, supposedly one would obtain the true optimum. Is it possible to explicitly compute the limit of $A^{-1} 1$? $\endgroup$
    – J Richey
    Commented Nov 29, 2017 at 1:47
  • $\begingroup$ Corrected. Thank you! $\endgroup$ Commented Nov 29, 2017 at 5:51

The following result has some bearing on the problem. It shows that the maximising distribution will be symmetrical and can be used to give an elementary proof in the case $\alpha=1$.

For any two real numbers $x$ and $y$ such that $0\leq y<x\leq\frac{1}{2}$ consider a random variable with distribution $$p(X=x)=p+a,p(X=y)=q+b,p(X=-x)=p-a,p(X=-y)=q-b,$$ for $p+q=\frac{1}{2}$.

Then $E[|X-Y|^{\alpha}]=(2x)^{\alpha}p^2+(2y)^{\alpha}q^2+((x+y)^{\alpha}+(x-y)^{\alpha})2pq-F(x,y)$ where $$F(x,y)=(2x)^{\alpha}a^2+((x+y)^{\alpha}-(x-y)^{\alpha})2ab+(2y)^{\alpha}b^2$$ For $\alpha\leq 1$, $(x+y)^{\alpha}-(x-y)^{\alpha}\leq2(xy)^{\frac{\alpha}{2}}$ and so the minimum value of $F(x,y)$ is $0$. This is attained when $a=b=0$ and then $$E[|X-Y|^{\alpha}]=(2x)^{\alpha}p^2+(2y)^{\alpha}q^2+((x+y)^{\alpha}+(x-y)^{\alpha})2pq.$$

The original distribution can, if necessary, be approximated as closely as one likes (in terms of the value of $E[|X-Y|^{\alpha}])$ by a discrete distribution and then one can look at the conditional value of the expectation when $X$ is restricted to pairs of values symmetrically placed about $x=\frac{1}{2}$.The above result can then be used to formally prove that the original distribution has to be symmetrical about $x=\frac{1}{2}$.

It is also straightforward to maximise $(2x)^{\alpha}p^2+(2y)^{\alpha}q^2+((x+y)^{\alpha}+(x-y)^{\alpha})2pq$ for $p+q=\frac{1}{2}$. When $\alpha=1$ this immediately gives $q=0$, giving another proof of the already known result in this case.

  • $\begingroup$ What are $a$ and $b$? Also, you should be clearer about shifting to $X \in [-1/2, 1/2]$, which is equivalent for $\alpha = 1$, but not for arbitrary $\alpha$! $\endgroup$
    – J Richey
    Commented Nov 28, 2017 at 23:46
  • $\begingroup$ @JRichey It is equivalent for arbitrary $\alpha$ since $X-Y = (X-\frac{1}{2})-(Y-\frac{1}{2})$. $\endgroup$
    – shalop
    Commented Nov 29, 2017 at 1:59
  • $\begingroup$ @Shalop Does the absolute value mess that up? $\endgroup$
    – J Richey
    Commented Nov 29, 2017 at 2:01
  • $\begingroup$ @JRichey No, of course not. Shifting never changes distances, even if distance is raised to $\alpha$ power. $\endgroup$
    – shalop
    Commented Nov 29, 2017 at 2:05
  • $\begingroup$ @Shalop Right, duh. I am on board. $\endgroup$
    – J Richey
    Commented Nov 29, 2017 at 2:08

Experimentally, the distribution with pdf $12(x-\frac{1}{2})^2$ looks good for $\alpha=1/2$. This gives an expectation of $$\frac{18(\alpha^2+ \alpha+2)}{(\alpha+1)(\alpha+2)(\alpha+3)(\alpha+6)}.$$ For $\alpha$ close to 0 this is close to 1, and for $\alpha=1/2$, this is 264/455 or roughly 0.58.

Update: the pdf $$\left(\frac{1+\sqrt{2}}{4}\right) \left| \frac{1}{\sqrt{x}}- \frac{1}{\sqrt{1-x}} \right|$$ gives a better result for $\alpha=1/2$, namely 0.594. This suggests difficulties for numerical solutions, since the optimal distribution may have unbounded probabilities.

  • $\begingroup$ Probabilities are always bounded between $0$ and $1$, so I'm not sure what you mean. Simulations will certainly be difficult, since the optimization is over a function space, which is infinite dimensional! @KevinCostello gives a good idea about how to deal with this in his answer. $\endgroup$
    – J Richey
    Commented Nov 29, 2017 at 0:26
  • $\begingroup$ Probabilities in the sense of the values of pdfs can be greater than 1. $\endgroup$
    – user210229
    Commented Nov 29, 2017 at 0:51

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