Independence of parameters in linear regression $\varepsilon \sim\operatorname{N}(0,\sigma^2I_n)$
Consider adding a covariate $ \mu \in \mathbb R^n$
to get the model, $y = X\beta + \gamma\mu + \varepsilon$, and assume $X^T \mu = 0, \gamma \in \mathbb R.$
Let ($\hat\beta_\text{new}, \hat\gamma$) be the LSE of parameters ($\beta, \gamma$) in this new model.
True or False: $\hat\beta_\text{new}$ and $\hat\gamma$ are independent.
I start by saying that since the model is multivariate Normal, the covariance of these two variables will determine independence. If covariance = 0, then they are independent. So I want to show:
$$\operatorname{cov}(\hat\beta_\text{new}, \hat\gamma)=0.$$
get $\operatorname{E}(\hat\beta_\text{new}, \hat\gamma) - \operatorname{E}(\hat\beta_\text{new})\operatorname{E}(\hat\gamma) = 0$
$\operatorname{E}(\hat\beta_\text{new}) = \beta_\text{new}$ and $\operatorname{E}(\hat\gamma) = \gamma$, so want to show:
$\operatorname{E}(\hat\beta_\text{new}, \hat\gamma) = \beta_\text{new}\gamma$
Am I doing this correct?
 A: One of the consequences of the orthogonality of $\mu$ and the columns of $X$ is that $\widehat\beta_\text{new} = \widehat\beta.$
\begin{align}
\left[ \begin{array}{cc} \widehat\beta_\text{new} \\ \widehat\gamma \end{array} \right] & = \left( \left[ \begin{array}{cc} X & \mu \end{array} \right]^\top \left[ \begin{array}{cc} X & \mu \end{array} \right] \right)^{-1} \left[ \begin{array}{cc} X & \mu \end{array} \right]^\top y \\[10pt]
& = \left[ \begin{array}{cc} X^\top X, & X^\top\mu \\ \mu^\top X, & \mu^\top\mu \end{array} \right]^{-1} \left[ \begin{array}{cc} X^\top \\ \mu^\top \end{array} \right] y = \left[ \begin{array}{cc} (X^\top X)^{-1}X^\top y \\ (\mu^\top \mu)^{-1} \mu y \end{array} \right] = \left[ \begin{array}{cc} \widehat\beta \\ \widehat\gamma \end{array} \right].
\end{align}
In the same way, the coefficient $\widehat\gamma$ is the same as what it would have been if $X$ had not been in the model.
\begin{align}
\operatorname{cov}(\widehat\beta,\widehat\gamma) & = \operatorname{cov}\left( (X^\top X)^{-1} X^\top y, (\mu^\top \mu)^{-1} \mu^\top y \right) \\[10pt]
& = (X^\top X)^{-1} X^\top y \Big( \operatorname{cov}(y,y) \Big) \mu(\mu^\top\mu)^{-1} \\[12pt]
\text{(since in general, } \operatorname{cov}(Au,Bv) & = A \Big(\operatorname{cov}(u,v) \Big) B^\top \text{)}
\end{align}
and the orthogonality of $\mu$ and the columns of $X$ shows that that is $0.$
