TITLE: Prove that matrix is positive definite
QUESTION [11 upvotes]: I faced a hard question in kernel methods theory, which I can't answer for about one week. Initially it was formulated in terms of positive valued functions, but it could be reformulated easier:
Let $\{x_1, \dotsc, x_n\}$ and $\{y_1, \dotsc, y_n\}$ be two sets of real positive numbers. Prove that matrix $A$ is positive definite where $A_{i, j} = \min(x_i\cdot y_j, x_j\cdot y_i)$. Equivalently, I want to prove that for arbitrary $a_1, \dotsc, a_n$ the following inequality holds:
$$
\sum_{i,j=1}^n a_i a_j \min(x_i\cdot y_j, x_j\cdot y_i)\ge 0.
$$
I've run out of ideas on how it could be proved, so any advice is welcome.

REPLY [5 votes]: This is a very similar proof but  based on $\mathbb{E}B_{z_{i}}B_{z_{j}}=\min(z_{i},z_{j})$ where $B_{z}$ is the standard  brownian motion starting at zero: 
$
0\leq \mathbb{E}\left(\sum_{j} a_{j} x_{j} B_{\frac{y_{j}}{x_{j}}} \right)^{2}=\sum_{i,j}a_{i}a_{j}\min(x_{i}y_{j},x_{j}y_{i})
$