TITLE: How to (dis-)prove that this symmetric matrix is P.S.D.?
QUESTION [5 upvotes]: for two random $n$-vectors $(X_1,X_2)$ define 
$$f(X_1,X_2)=\text{med}((X_1+X_2)^2)-\text{med}((X_1-X_2)^2)$$
Now define the matrix $U$ with entries 
$$U_{ij}=\frac{n}{4}f(X_i,X_j)$$
Based on a large number of computer simulations using many nearly collinear matrices $X=(X_1,\ldots,X_p)\in\mathbb{R}^{n\times p}$ with $n>p=100$, I've come to suspect that if the members of $X$ are in general linear position, then $U$ is P.S.D..
To put the matter to rest, I'm trying to prove this statement (or find a counter example). My question is how would you go about attacking this problem?
I've been trying to prove that $U$ is diagonally dominant (it is easy to show that $U_{ii}>U_{ij}\forall i\neq j$, but I have not been able to make any statement about the sum of the absolute values of the off-diagonal elements).

REPLY [3 votes]: (This is too long for a comment.) I don't have an answer to your question. All I can say is that your conjecture is not true in general. For a counterexample, consider the case where $n=3,\,p=2$ and
$$
(X_1, X_2) = \begin{bmatrix}
1+\varepsilon&1-\varepsilon\\
1-\varepsilon&1+\varepsilon\\
1-\varepsilon&1-\varepsilon
\end{bmatrix}
$$
where $\varepsilon>0$ is a small number. Then
\begin{align*}
f(X_1,X_1) &= f(X_2,X_2) = 4(1-\varepsilon)^2,\\
f(X_1,X_2) &= 4 - 4\varepsilon^2 = 4(1-\varepsilon^2)
\end{align*}
and the resulting $U$ has a negative determinant and hence is not positive semidefinite.
Yet this counterexample does not generalize well for larger $n$ and $p$, and your statement is perhaps true in an overwhelming number of cases for some good statistical reasons, but I am totally clueless about them.