# A question regarding the proof of a statistical test for a regression parameter

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?

• 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. Feb 4, 2023 at 12:42

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.