Solve $\inf_{ X: |X| \le a \text{ a.s.}} E \left[ \frac{1}{1+(X-X^\prime)^2} \right] $

Question

I want to solve the following optimization problem

\begin{align} \inf_{ X: |X| \le a \text{ a.s.}} E \left[ \frac{1}{1+(X-X^\prime)^2} \right] \end{align} where $X^\prime$ is an independent copy of $X$ and $a>0$ is some constant.

How would one approach such a problem? Is the solution easy to find?

At some point I thought that the optimal distribution is given by $X=\{-a,a\}$ equally likely. In which case, the solution is given by \begin{align} \inf_{ X: |X| \le a \text{ a.s.}} E \left[ \frac{1}{1+(X-X^\prime)^2} \right] \le \frac{1}{2}\frac{1}{1+4a^2}+\frac{1}{2}. \end{align} However, I don't have any supporting arguments for this.

The following might be useful. Note that by Jensens' inequality

\begin{align} E \left[ \frac{1}{1+(X-X^\prime)^2} \right] \ge \frac{1}{1+E[(X-X^\prime)^2]} =\frac{1}{1+2Var(X)}. \end{align}

Therefore,

\begin{align} \inf_{ X: |X| \le a \text{ a.s.}} E \left[ \frac{1}{1+(X-X^\prime)^2} \right] \ge \frac{1}{1+2 \sup_{ X: |X| \le a \text{ a.s.}} Var(X)}=\frac{1}{1+2a^2}, \end{align}

where in the last optimization step we used \begin{align} Var(X) \le E[X^2] \le a^2, \end{align} which is achievale with $X=\{-a,a\}$ equally likely.

Answer 1

My initial intuition was incorrect. It seems we can sometimes solve this explicitly, and the minimizer does not have full support; my argument is incomplete (as discussed below), but I think it's reasonably convincing anyway. I'm also only going to discuss $a=1$ (initially, I thought that was the general case, but of course that's not true because of the $1$ in the denominator of the kernel; see also Mateusz's answer for more on this). Then it is given by $$ \mu= \frac{5}{12}(\delta_{-1}+\delta_1) + \frac{1}{6}\delta_0 . $$ To see why this is optimal, look at $$ F(x)=\int \frac{d\mu(t)}{1+(t-x)^2} = \frac{5}{12}\left( \frac{1}{1+(x+1)^2} + \frac{1}{1+(x-1)^2} \right) + \frac{1}{6}\frac{1}{1+x^2} . $$ Notice that $F(x)=7/12$ on $x=0,\pm 1$ (the support of $\mu$), and $F(x)\ge 7/12$ for $-1\le x\le 1$ (this is a tedious but elementary calculus exercise).

This means that if we vary the measure slightly, say $\nu=\mu+\epsilon\sigma$ with a signed measure $\sigma$ with $\int d\sigma=0$ and $\epsilon\ll 1$, then the change in first order will be proportional to $\int F(x)\, d\sigma(x)$. However, since $\nu$ must remain a positive measure, the negative part of $\sigma$ can only be supported by $0,\pm 1$. This means that $\int F\, d\sigma\ge 0$, by the properties of $F$ observed above.

This verifies that my $\mu$ is a local minimum. I don't have an argument that shows that it gives the global minimum also, though I'm fairly optimistic that this will be true.

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A few things can perhaps be said in general: First of all, this criterion ($F_{\mu}$ constant on the support of $\mu$ and $F(x)$ at least as large otherwise) is also necessary for local minima. The general scenario that seems likely is that as you increase $a$, additional points enter the support. This would also be consistent with what Mateusz does in his answer.

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Finally, here is a general argument why a measure $d\mu =f\, dx$ with $f>0$ on $-a<x<a$ can not give a minimum, not even a local one. To see this, consider again a small variation $d\nu =(f+\epsilon g)\, dx$. Now $g$ can be an arbitrary (let's say, almost arbitrary, to be safe) function with $\int g=0$, so it now follows that at an extremum, $\int Fg =0$ for all such $g$. This forces $F$ to be constant on $-a<x<a$, but since $F$ is the restriction of a harmonic (on $\mathbb C^+$) function to the line $\textrm{Im}\: z =1$ (your kernel is the Poisson kernel for the upper half plane), this harmonic function would have to be constant on this whole line, which makes $f$ constant on $\mathbb R$, but we need an $f$ that is zero outside $(-a,a)$.

Answer 2

You ask what is the minimiser of the functional $$ F(\mu) = \int_{[-a,a]} \int_{[-a,a]} \frac{1}{1+(x-y)^2} \mu(dx) \mu(dy) . $$ As Christian Remling points out, it is unlikely a closed-form expression exists. However, you can say what happens as $a \to \infty$ and $a \to 0^+$.

Some notation first. Denote by $\nu(E) = \mu(a E)$ a measure rescaled by a factor of $a$. Then $$ G_a(\nu) := F(\mu) = \int_{[-1,1]} \int_{[-1,1]} \frac{1}{1+a^2 (x-y)^2} \nu(dx) \nu(dy) . $$

As $a \to \infty$, $a/(1+a^2(x-y)^2) dx$ converges to the Dirac measure at $y$. Therefore, if $\nu$ has a density $g(x)$, then $a G_a(\nu)$ converges as $a \to \infty$ to $$ \int_{[-1,1]} g(y) \nu(dy) = \int_{[-1,1]} (g(y))^2 dy ,, $$ which is minimised by a constant $g$. Therefore, the minimiser of $F$ can be expected to be close to the uniform distribution if $a$ is large.

When $a \to 0^+$, $G_a(\nu)$ converges to $1$, but one can inspect $(G_a(\nu)-1)/a^2$ instead. We have $$ \frac{G_a(\nu)-1}{a^2} = \int_{[-1,1]} \int_{[-1,1]} \frac{-(x - y)^2}{1+a^2 (x-y)^2} \nu(dx) \nu(dy) , $$ which converges to $$ -\int_{[-1,1]} \int_{[-1,1]} (x - y)^2 \nu(dx) \nu(dy) . $$ This expression is easily evaluated to be $-2 \operatorname{Var}(\nu)$, a negative multiple of the variance of $\nu$, which is maximised by a symmetric two-point distribution $\tfrac{1}{2}\delta_{-a}+\tfrac{1}{2}\delta_a$. Therefore, the minimisers of $F$ should behave roughly as a two-point distribution for small $a$.

Answer 3

I am proposing the following lemma.

For $g$ concave $\inf_{|X|\leq a}\mathbb{E}(g(X-X'))=\inf_{|X|= p\delta_a+(1-p)\delta_{-a},p\in[0,1]}\mathbb{E}(g(X-X'))$

This show that for $a<2/7$, the solution of your problem is exactly $1/2\delta_a+1/2\delta_{-a}$.

Proof of the lemma : replace $X$ by $X+B_T$ where $B$ is a brownian motion and $T$ the stoping time $\inf(t:|X+B_t|=a)$. Define $$ u(t)=\mathbb{E}(g(X+B_{\inf (t,T)}-X'-B_{\inf (t,T')}'))$$ Then because of ito formula : $$ \frac{d}{dt}u(t)=\mathbb{E}(\frac{1}{2}g''(X+B_{\inf (t,T)}-X'-B_{\inf (t,T)}')(1_{t\leq T}+1_{t\leq T'}))\leq 0$$ $u$ is then decreasing and for $t\rightarrow \infty$, $|X+B_{\inf (t,T)}|=a$ with probability one.

For large $a$, One can adapt your perturabative proof : $\int Fg = 0$ for all $g$ such that $\int g =0$ and $g\geq 0$ outside the support of $\mu$. If there exist $x\in supp(\mu)$ and $F(x)>\inf_{[-a,a]} F(x)$ then let $\exists F(x_0)<F(x)$ and set $g=-1/2 \delta_x+1/2 \delta_{x_0}$ which do not satisfy the condition.

Conclusion : $supp(\mu)\subset [x: F(x)=\inf F]$. Because $F$ is harmonic, then $supp(\mu)$ is discreet and $\mu $ is therefore a sum of dirac mass. We can see the dirac like particles which repel each other and at the limit $a \rightarrow \infty$ because of Mateusz remark will form a regular cristal.