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I have two questions regarding the proof of the following theorem: Consider the regression model $Y=X\cdot\beta+\epsilon$, where $\epsilon\sim N(0,\sigma^2\cdot I_n)$ and X has full rank. Then, the least square estimator $\hat{\beta}$ is given by $\hat{\beta}=(X^TX)^{-1}X^TY$. In order to test $H_0: \hat{\beta}=\beta_0 $ vs. $H_1:\hat{\beta}\neq \beta_0$ the following test statistic is used: $$T=\frac{(\hat{\beta}-\beta_0)^T(X^TX)(\hat{\beta}-\beta_0)}{m\cdot S^2} $$ where $S^2=\frac{1}{n-m}(Y-X\hat{\beta})^T(Y-X\hat{\beta})$ is an unbiased estimator for $\sigma^2$. Then, under $H_0$ it holds that $$T\sim F_{m,n-m}$$ Question 1: In the lecture we used a theorem that if $Z\sim N(\mu, K)$ and A is a symmetric and idempotent matrix, then $Z^TAZ$ follows non-central $\chi^2$-distribution. But I do not think it is necessary to use the theorem here to follow that under $H_0$ it holds that $$\sigma^{-2}(\hat{\beta}-\beta_0)^T(X^TX)(\hat{\beta}-\beta_0)\sim\chi^2_m$$ Due to the fact that $$\hat{\beta}-\beta_0\sim N(0, \sigma^2(X^TX)^{-1}) $$ it holds that $$\sigma^{-1}X(\hat{\beta}-\beta_0)\sim N(0,I_n)$$ Thus, by the definition of the (central) $\chi^2$-distribution it holds that $$ \sigma^{-2}(\hat{\beta}-\beta_0)^TX^TX(\hat{\beta}-\beta_0)\sim \chi^2_m$$. So why would one use the theorem mentioned above?$$\ $$ Question 2: If we consider that $$\frac{n-m}{\sigma^2}S^2\sim\chi^2_{n-m} $$, then under $H_0$ it should hold that $$T*=\frac{\frac{(\hat{\beta}-\beta_0)^TX^TX(\hat{\beta}-\beta_0)}{\sigma^2}}{\frac{n-m}{\sigma^2}S^2}=\frac{(\hat{\beta}-\beta_0)^TX^TX(\hat{\beta}-\beta_0)}{(n-m)S^2}\sim F_{m,n-m} $$ But $T*\neq T$. So my question is how did $n-m$ got transformed to simply $m$ in the denominator?

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  • $\begingroup$ You need to divide the numerator and denominator in $T*$ by the respective degrees of freedom to get an F distribution under $H_0$. As for the first question, you are right. That $Z\sim N_p(\mu,K)\implies (Z-\mu)^TK^{-1}(Z-\mu)\sim \chi^2_p$ doesn't require any theorem. $\endgroup$ Feb 4, 2023 at 12:42

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For Q 2: recall that you can define the F distribution by $$ \frac{X_1/a_1}{X_2/a_2}, $$ where $X_1$ and $X_2$ are two independent chi-squared r.v. with $a_1$ and $a_2$ degrees of freedom, respectively. Thus, both for $T*$ and $T$, you can arrange the fraction accordingly.

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