$x$ is a random vector variable and matrix $P$ is symmetric and deterministic. If $\mathbb{E}\{x^T P x\}>0$ for any non-zero $x$, can we say $P$ is positive definite? If it is not, can you give me a counterexample?
$\mathbb{E}$ is the expectation operator.
I think you can conclude that $P$ is positive definite. Any non-zero $x$ in your question allows deterministic distributions for every $x_0 \in \mathbb{R}^n$ which always result in $x_0$. Thus you have $x_0^TPx_0>0$ for all $x_0\in \mathbb{R}^n\backslash \{0\}$, which is the definition of positive definite.
