My question
I have a linear regression that contains some regressors that vary only at a group level and some that vary at the individual level.
Slide 8 of this suggests that the coefficients on the group-level regressors are only affected by group-level variation in the individual-level regressors (i.e. variation within groups for the individual-level regressors doesn't matter for the coefficients on the group-level regressors). How do I prove this?
How far I've got
Call the group-level regressors $\mathbf{X_1}$ (a $N \times K_1$ matrix) and the individual-level regressors $\mathbf{X_2}$ (a $N \times K_2$ matrix).
Using the partitioned regression formula, I've been able to show that if there's no group-level variation in $\mathbf{X_2}$, then the coefficients on $\mathbf{X_1}$ do not depend on $\mathbf{X_2}$. To see this, note that the partitioned regression formula states: \begin{equation} \mathbf{b_1} = (\mathbf{X_1}'\mathbf{X_1})^{-1}\mathbf{X_1}'(\mathbf{y} - \mathbf{X_2}\mathbf{b_2}) \end{equation} Define $\mathbf{G}$ to be the $N \times G$ matrix of groups identifiers with the element $G_{ig} = 1$ if the $i$th observation is in group $g$ and 0 otherwise. Then we can write $\mathbf{X_1} = \mathbf{G}\mathbf{\overline{X}_1}$ where $\mathbf{\overline{X}_1}$ is the $G \times K_1$ matrix of group-level regressors with duplicate rows removed. Then by simple manipulation we have: \begin{equation} \mathbf{X_1} = \mathbf{G}(\mathbf{G}'\mathbf{G})^{-1}\mathbf{G}'\mathbf{X_1} \end{equation} Note that in this $(\mathbf{G}'\mathbf{G})^{-1}\mathbf{G}'$ is a matrix that finds group-level averages of the matrix that post-multiplies it. Thus, if there is no group-level variation in $\mathbf{X_2}$, we will have $(\mathbf{G}'\mathbf{G})^{-1}\mathbf{G}'\mathbf{X_2} = \mathbf{0}$
The result that if there's no group-level variation in $\mathbf{X_2}$, then the coefficients on $\mathbf{X_1}$ do not depend on $\mathbf{X_2}$ is straightforward. We can write: \begin{align} \mathbf{X_1}'\mathbf{X_2} &= [\mathbf{G}(\mathbf{G}'\mathbf{G})^{-1}\mathbf{G}'\mathbf{X_1}]'\mathbf{X_2} \\ &= \mathbf{X_1}'\mathbf{G}(\mathbf{G}'\mathbf{G})^{-1}\mathbf{G}'\mathbf{X_2} \\ &= \mathbf{0} \end{align} where the last line follows because $(\mathbf{G}'\mathbf{G})^{-1}\mathbf{G}'\mathbf{X_2} = \mathbf{0}$. Thus the formula for $\mathbf{b_1}$ becomes: \begin{equation} \mathbf{b_1} = (\mathbf{X_1}'\mathbf{X_1})^{-1}\mathbf{X_1}'\mathbf{y} \end{equation} and so does not depend on $\mathbf{X_2}$.
The outstanding problem
But here's the problem: if there is group-level variation in $\mathbf{X_2}$ I have only shown that $\mathbf{X_1}'\mathbf{X_2}$ only depends on group-level variation in $\mathbf{X_2}$. It seems to me that it is still possible for individual-level variation in $\mathbf{X_2}$ to matter through its effect on $\mathbf{b_2}$.
The full formula for $\mathbf{b_1}$ with $\mathbf{b_2}$ eliminated is: \begin{equation} \mathbf{b_1} = (\mathbf{X_1}'\mathbf{M_2}\mathbf{X_1})^{-1}\mathbf{X_1}'\mathbf{M_2}\mathbf{y} \end{equation} where $\mathbf{M_2} = \mathbf{I} - \mathbf{X_2}(\mathbf{X_2}'\mathbf{X_2})^{-1}\mathbf{X_2}'$. I can't see how I can substitute in $\mathbf{X_1} = \mathbf{G}(\mathbf{G}'\mathbf{G})^{-1}\mathbf{G}'\mathbf{X_1}$ and rearrange to end up only with group-level averages of $\mathbf{X_2}$.
