Positive definitness of the expectation of a random matrix

Question

Let $X$ be a non-negative random variable. We know that if $E[X]= 0$ then $X=0$ a.s. A consequence of this is that $P(X\neq 0)=1$ implies $E[X]>0$.

Now, let $A$ be a $n\times n$ symmetric and non-negative definite random matrix, i.e., for any $x\in\mathbb{R}^n$ we have that $x^TAx\geq 0$ a.s. Analogous to the real case, can we conclude that $P(A\neq 0)=1$ implies $E[A]$ positive definite, i.e., for all $x\in\mathbb{R}^n$, $x^TE[A]x>0$?

I think that this is not true, but I can not think of a counter-example.

Answer 1

Let $A = \begin{bmatrix} 1 & 0\\ 0 & 0\\ \end{bmatrix}$ (i.e. the deterministic matrix chosen with probability $1$). Then $A$ is non-negative definite but $E[A] = A$ is not positive definite.