Probability – Expected Absolute Difference Between Two IID Variables

Question

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

Answer 1

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} $

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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.