Invertibility of $E(X^\top X)$ and $\tilde{X}^\top \tilde{X}$ Consider the linear regression model with 3 regressors
$$
Y=\beta_1 Q+\beta_2 W+\beta_3 Z+\epsilon 
$$
Let $X\equiv (Q, W, Z)$ and $\beta\equiv (\beta_1,\beta_2,\beta_3)^\top$. Also suppose that $Q,W,Z$ are binary variables.
Assume that the distribution of $(Y, X)$ is known and that $E(\epsilon X)=0$. One can show that under these assumptions $\beta$ is identified if $E(X^\top X )$ is full rank.
The proof is as follows:
$$
Y=X\beta+\epsilon\\
X^\top Y=X^\top X \beta +X^\top\epsilon\\
E(X^\top Y)=E(X^\top X)\beta +E(X^\top\epsilon)\\
\beta=E(X^\top X)^{-1} E(X^\top Y)
$$
Now suppose we have an i.i.d. sample $(Y_1,X_1),\dots,(Y_n, X_n)$ where each $(Y_i, X_i)\sim (Y,X)$.
Let $\tilde{X}\equiv (\tilde{Q}, \tilde{W}, \tilde{Z})$ be the $n\times 3$ matrix stacking $X_i\equiv (Q_i, W_i, Z_i)$ in each row.
Suppose that the matrix $\tilde{X}$ has not linearly independent columns because the first column is the sum of the other 2. Therefore the matrix $\tilde{X}^\top \tilde{X}$ is not full rank. Can still the matrix $E(X^\top X )$ be full rank? How do we reconcile the fact that $\tilde{X}^\top \tilde{X}$  is not full rank with the identification proof above?
My intuition is that, under appropriate moment conditions (finiteness of moments), $\frac{1}{n}\tilde{X}^\top \tilde{X} \rightarrow_p E(X^\top X )$ by law of large numbers. Therefore, if $\tilde{X}^\top \tilde{X}$ is not full rank, then its limit in probability cannot be full rank. Is this correct?

No, this is not correct. $\tilde X^T \tilde X$ being rank-deficient for one specific $n$ tells you nothing about the limit. For a trivial example, suppose $n=2$, so $\tilde X^T \tilde X$ is of rank at most 2.
You can make some deductions about limits. The set of matrices of full rank is an open set: if $A$ is in that set, so is a neighbourhood of $A$. For a deterministic sequence $A_n\to A$ then $A_n$ is of full rank for all large enough $n$.  Thus if $A_n\stackrel{a.s.}{\to} A$ (which will be true here), $A_n$ is of full rank for all large enough $n$, and if $A_n\stackrel{p}{\to} A$,
$$P(A_n{\textrm{ is of full rank}})\to 1.$$
So, for $n$ much bigger than the number of variables it's unlikely that $\tilde X^T\tilde X$ is rank-deficient when $E[X^TX]$ isn't, but it's not impossible.
