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.

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.

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+2Cov(X)} \end{align}

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.

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.

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+2Cov(X)} \end{align}