Suppose $M$ is a random symmetric $n\times n$ matrix satisfying $\lambda_{\min}(\mathbb{E}M)>0$, where $\lambda_{\min}$ denotes the smallest eigenvalue.
My question: What can we say about the random variable $\lambda_{\min}(M)$? For example, do we have $\mathbb{P}(\lambda_{\min}(M)\ge c)>0$ for some $c>0$ and $\mathbb{E}\lambda_{\min}(M)>0$? Of course, the latter would imply the former because $$ \max\{0,\mathbb{E}\lambda_{\min}(M)\} = \int_0^{\infty}\mathbb{P}(\lambda_{\min}(M)>t)\,dt. $$
My thoughts: Clearly, $$ \mathbb{E}\lambda_{\min}(M)=\mathbb{E}\min_{\|x||_2=1}x^\top Mx\le\min_{\|x\|_2=1}x^\top\mathbb{E}Mx=\lambda_{\min}(\mathbb{E}M). $$ But I am not sure how to use $\lambda_{\min}(\mathbb{E}M)>0$. In addition, I don't know how to analyze the probability $$ \mathbb{P}(\lambda_{\min}(M)\ge c)=\mathbb{P}\left(\bigcap_{\|x\|_2=1}\{x^\top Mx\ge c\}\right) $$ which involves an uncountable intersection of events.
I have just found a counterexample. Suppose $z\sim\mathcal{N}(0,I_2)$, where $I_2$ is the $2\times 2$ identity matrix. Let $M=zz^\top$. Then $\mathbb{E}M=I_2$ while $\lambda_{\min}(M)=0$ always.