Suppose we have a graph $G$ with treewidth $t$. Let $p \in (0,1)$ be a constant. Then let's independently remove each edge from $G$ with probability $p$. What is the expected treewidth of the resulting graph? I think it should be $\Omega(t)$, but have little clue how to prove it.
For $n \times n$ grid there is another distribution such that each edge has constant probability to be removed and with probability 1 the treewidth of the resulting graph is at least $n/2$. Let's pick the cells of the grid having odd coordinates $(2k+1,2\ell+1)$. If we contract all these cycles we get a $(n/2)\times(n/2)$ grid with doubled edges. Then let our distribution remove exactly one of the edges from each of the squares and exactly one edge of each pair that becomes parallel if the cycles are contracted. The result always has $(n/2)\times(n/2)$ grid minor.
I am interested in such "almost independent" distribution on the sets of the removed edges that does not reduce treewidth too much as well.