We have $Y_i = \beta_0 +\beta_1(X_i -\bar X )+\epsilon_i$ for i=1,...,n
$$\epsilon_i \sim N(0,\sigma^2)$$
We know that $SSR= Y^T P_xY - n\bar Y^2=Y^T (P_x -n^{-1} J_nJ_n^T)Y$
$$J_n=\big[\begin{matrix}1 \\ ... \\ 1\end{matrix}\big]_{nxn}$$
$$MSR=SSR/(r-1)$$
How can I find the distribution of MSR?
Suppose $Y\sim N_p(\mu, V)$ and Rank(V)=p
If AV is idempotent of ranks then $Y^TAY \sim \chi_s^2(\lambda)$ where $\lambda =1/2 (\mu^TA\mu)$
Will I use this result?
I have done for $SSR=Y^TP_xY$ Uncorrected model.
By using this result and find
$$\frac{Y^TP_xY}{\sigma^2}=Y^TBY\sim \chi_s^2(\lambda)$$ where $\lambda =1/2 (\mu^TB\mu)$ and$B= \sigma^{-2 }P_x$
But I couldn't find for corrected model MSR.