[proofplan] We express the unrestricted and restricted least-squares estimators explicitly and identify the loss caused by imposing the affine constraint $Rb=r$. The excess residual sum of squares is the quadratic form in the constraint discrepancy $R\hat{\beta}-r$, while the unrestricted residual sum of squares is the squared length of the Gaussian residual projection. Under $H_0$, these two quantities are independent chi-squared variables after division by $\sigma^2$, with degrees of freedom $q$ and $n-p$ respectively. The stated statistic is therefore their ratio divided by degrees of freedom, which is the defining form of the $F_{q,n-p}$ distribution. [/proofplan] [step:Write the unrestricted fit as an orthogonal projection] Define \begin{align*} C &:= (X^\top X)^{-1} \in \mathbb{R}^{p \times p}, \\ P_X &:= XCX^\top \in \mathbb{R}^{n \times n}. \end{align*} Since $X$ has rank $p$, $X^\top X$ is invertible, and $P_X$ is the [orthogonal projection](/theorems/437) of $\mathbb{R}^n$ onto $\operatorname{Range}(X)$. The unrestricted least-squares estimator is the map \begin{align*} \hat{\beta}: \mathbb{R}^n &\to \mathbb{R}^p \\ y &\mapsto C X^\top y. \end{align*} Thus the unrestricted fitted value is $X\hat{\beta}=P_XY$, and \begin{align*} \operatorname{RSS}_1 = |Y-X\hat{\beta}|^2 = |(I_n-P_X)Y|^2. \end{align*} [guided] We first separate the part of $Y$ explained by the design matrix from the residual part. Define \begin{align*} C &:= (X^\top X)^{-1} \in \mathbb{R}^{p \times p}, \\ P_X &:= XCX^\top \in \mathbb{R}^{n \times n}. \end{align*} The rank assumption $\operatorname{rank}(X)=p$ implies that $X^\top X$ is invertible. The matrix $P_X$ is symmetric and idempotent: \begin{align*} P_X^\top = P_X, \qquad P_X^2 = XCX^\top XCX^\top = XC(X^\top X)CX^\top = XCX^\top = P_X. \end{align*} Therefore $P_X$ is the orthogonal projection onto $\operatorname{Range}(X)$. The unrestricted least-squares estimator is the [linear map](/page/Linear%20Map) \begin{align*} \hat{\beta}: \mathbb{R}^n &\to \mathbb{R}^p \\ y &\mapsto C X^\top y. \end{align*} For the observed random vector $Y$, the fitted value is $X\hat{\beta}=XCX^\top Y=P_XY$. Since least squares residuals are orthogonal to the column space of $X$, the unrestricted residual vector is \begin{align*} Y-X\hat{\beta}=(I_n-P_X)Y. \end{align*} Hence \begin{align*} \operatorname{RSS}_1 = |Y-X\hat{\beta}|^2 = |(I_n-P_X)Y|^2. \end{align*} [/guided] [/step] [step:Compute the excess sum of squares from the affine restriction] Define \begin{align*} A := RCR^\top \in \mathbb{R}^{q \times q}. \end{align*} The matrix $A$ is positive definite because $R$ has rank $q$ and $C$ is positive definite. The restricted estimator is \begin{align*} \tilde{\beta} := \hat{\beta} - CR^\top A^{-1}(R\hat{\beta}-r). \end{align*} Indeed, \begin{align*} R\tilde{\beta} = R\hat{\beta} - RCR^\top A^{-1}(R\hat{\beta}-r) = R\hat{\beta} - (R\hat{\beta}-r) = r. \end{align*} For every $b \in \mathbb{R}^p$ with $Rb=r$, write $h:=b-\tilde{\beta}$. Then $Rh=0$, and \begin{align*} (Y-Xb) = (Y-X\hat{\beta}) + X(\hat{\beta}-\tilde{\beta}) - Xh. \end{align*} The three terms are pairwise orthogonal: $Y-X\hat{\beta}$ is orthogonal to $\operatorname{Range}(X)$, and \begin{align*} \langle X(\hat{\beta}-\tilde{\beta}),Xh\rangle_{\mathbb{R}^n} &= (\hat{\beta}-\tilde{\beta})^\top X^\top Xh \\ &= (R\hat{\beta}-r)^\top A^{-1}RCX^\top Xh \\ &= (R\hat{\beta}-r)^\top A^{-1}Rh \\ &= 0. \end{align*} Therefore $\tilde{\beta}$ minimizes the restricted least-squares problem, and \begin{align*} \operatorname{RSS}_0-\operatorname{RSS}_1 &= |X(\hat{\beta}-\tilde{\beta})|^2 \\ &= (\hat{\beta}-\tilde{\beta})^\top X^\top X(\hat{\beta}-\tilde{\beta}) \\ &= (R\hat{\beta}-r)^\top A^{-1}(R\hat{\beta}-r). \end{align*} [guided] The restricted problem is an affine least-squares problem because the constraint is $Rb=r$, not necessarily $Rb=0$. Define \begin{align*} A := RCR^\top \in \mathbb{R}^{q \times q}. \end{align*} We need $A^{-1}$ to exist. Since $C=(X^\top X)^{-1}$ is positive definite and $R$ has rank $q$, every nonzero vector $u \in \mathbb{R}^q$ satisfies $R^\top u \ne 0$, and hence \begin{align*} u^\top A u = u^\top RCR^\top u = (R^\top u)^\top C(R^\top u) > 0. \end{align*} Thus $A$ is positive definite and invertible. Define \begin{align*} \tilde{\beta} := \hat{\beta} - CR^\top A^{-1}(R\hat{\beta}-r). \end{align*} This is the unrestricted estimator corrected in the directions needed to enforce the constraint. We verify feasibility: \begin{align*} R\tilde{\beta} = R\hat{\beta} - RCR^\top A^{-1}(R\hat{\beta}-r) = R\hat{\beta} - A A^{-1}(R\hat{\beta}-r) = r. \end{align*} Now let $b \in \mathbb{R}^p$ satisfy $Rb=r$, and define $h:=b-\tilde{\beta} \in \mathbb{R}^p$. Since both $b$ and $\tilde{\beta}$ satisfy the restriction, $Rh=0$. Decompose the residual: \begin{align*} Y-Xb = (Y-X\hat{\beta}) + X(\hat{\beta}-\tilde{\beta}) - Xh. \end{align*} The first term is orthogonal to $\operatorname{Range}(X)$ by the normal equations for unrestricted least squares, so it is orthogonal to both $X(\hat{\beta}-\tilde{\beta})$ and $Xh$. It remains to check the orthogonality of the two fitted-space terms: \begin{align*} \langle X(\hat{\beta}-\tilde{\beta}),Xh\rangle_{\mathbb{R}^n} &= (\hat{\beta}-\tilde{\beta})^\top X^\top Xh \\ &= \bigl(CR^\top A^{-1}(R\hat{\beta}-r)\bigr)^\top X^\top Xh \\ &= (R\hat{\beta}-r)^\top A^{-1}RCX^\top Xh \\ &= (R\hat{\beta}-r)^\top A^{-1}Rh \\ &= 0. \end{align*} The [Pythagorean theorem](/theorems/3266) gives \begin{align*} |Y-Xb|^2 = |Y-X\hat{\beta}|^2 + |X(\hat{\beta}-\tilde{\beta})|^2 + |Xh|^2. \end{align*} This is minimized exactly when $h=0$, so $\tilde{\beta}$ is the restricted least-squares estimator. Therefore \begin{align*} \operatorname{RSS}_0-\operatorname{RSS}_1 &= |X(\hat{\beta}-\tilde{\beta})|^2 \\ &= (\hat{\beta}-\tilde{\beta})^\top X^\top X(\hat{\beta}-\tilde{\beta}) \\ &= (R\hat{\beta}-r)^\top A^{-1}RCX^\top XCR^\top A^{-1}(R\hat{\beta}-r) \\ &= (R\hat{\beta}-r)^\top A^{-1}RCR^\top A^{-1}(R\hat{\beta}-r) \\ &= (R\hat{\beta}-r)^\top A^{-1}(R\hat{\beta}-r). \end{align*} [/guided] [/step] [step:Identify the numerator as a chi-squared variable with $q$ degrees of freedom] Under $H_0$, $R\beta=r$. Since \begin{align*} \hat{\beta} = C X^\top Y = \beta + C X^\top \varepsilon, \end{align*} we have \begin{align*} R\hat{\beta}-r = RCX^\top \varepsilon. \end{align*} Thus $R\hat{\beta}-r$ is Gaussian with mean $0$ and covariance matrix \begin{align*} \operatorname{Cov}(R\hat{\beta}-r) = \sigma^2 RCX^\top XCR^\top = \sigma^2 RCR^\top = \sigma^2 A. \end{align*} Let $A^{1/2} \in \mathbb{R}^{q \times q}$ denote the unique symmetric positive definite square root of $A$, and let $A^{-1/2}:=(A^{1/2})^{-1}$. Define the random vector \begin{align*} Z := \sigma^{-1} A^{-1/2}(R\hat{\beta}-r) \in \mathbb{R}^q. \end{align*} Then $Z$ has the $q$-dimensional normal distribution with mean $0$ and covariance $I_q$, denoted $Z \sim \mathcal{N}_q(0,I_q)$. Therefore $Z^\top Z$ has the chi-squared distribution with $q$ degrees of freedom, denoted $\chi^2_q$, and \begin{align*} \frac{\operatorname{RSS}_0-\operatorname{RSS}_1}{\sigma^2} = Z^\top Z \sim \chi^2_q. \end{align*} [guided] Under the null hypothesis $H_0$, the true parameter satisfies $R\beta=r$. The normal linear model writes the observed random vector as $Y=X\beta+\varepsilon$, where $\varepsilon \sim \mathcal{N}_n(0,\sigma^2 I_n)$. Substituting this into the unrestricted estimator gives \begin{align*} \hat{\beta} = C X^\top Y = C X^\top(X\beta+\varepsilon) = C(X^\top X)\beta + C X^\top \varepsilon = \beta + C X^\top \varepsilon. \end{align*} Applying $R$ and using $R\beta=r$ yields \begin{align*} R\hat{\beta}-r = RCX^\top \varepsilon. \end{align*} Thus $R\hat{\beta}-r$ is a linear transformation of the Gaussian vector $\varepsilon$, so it is Gaussian. Its mean is zero, and its covariance matrix is \begin{align*} \operatorname{Cov}(R\hat{\beta}-r) &= \operatorname{Cov}(RCX^\top \varepsilon) \\ &= RCX^\top \operatorname{Cov}(\varepsilon)XCR^\top \\ &= \sigma^2 RCX^\top XCR^\top \\ &= \sigma^2 RCR^\top \\ &= \sigma^2 A. \end{align*} We now standardize this Gaussian vector. Since $A$ is symmetric positive definite, it has a unique symmetric positive definite square root $A^{1/2} \in \mathbb{R}^{q \times q}$. Define $A^{-1/2}:=(A^{1/2})^{-1}$ and define \begin{align*} Z := \sigma^{-1} A^{-1/2}(R\hat{\beta}-r) \in \mathbb{R}^q. \end{align*} The covariance of $Z$ is \begin{align*} \operatorname{Cov}(Z) &= \sigma^{-2}A^{-1/2}\operatorname{Cov}(R\hat{\beta}-r)A^{-1/2} \\ &= \sigma^{-2}A^{-1/2}(\sigma^2 A)A^{-1/2} \\ &= I_q. \end{align*} Hence $Z$ has the $q$-dimensional normal distribution with mean $0$ and covariance $I_q$, denoted $Z \sim \mathcal{N}_q(0,I_q)$. By the definition of the chi-squared distribution, $Z^\top Z$ has the chi-squared distribution with $q$ degrees of freedom, denoted $\chi^2_q$. Finally, \begin{align*} Z^\top Z &= \sigma^{-2}(R\hat{\beta}-r)^\top A^{-1/2}A^{-1/2}(R\hat{\beta}-r) \\ &= \sigma^{-2}(R\hat{\beta}-r)^\top A^{-1}(R\hat{\beta}-r) \\ &= \frac{\operatorname{RSS}_0-\operatorname{RSS}_1}{\sigma^2}. \end{align*} Therefore \begin{align*} \frac{\operatorname{RSS}_0-\operatorname{RSS}_1}{\sigma^2} \sim \chi^2_q. \end{align*} [/guided] [/step] [step:Identify the denominator as an independent chi-squared variable with $n-p$ degrees of freedom] Since $P_X$ is an orthogonal projection of rank $p$, the matrix $I_n-P_X$ is an orthogonal projection of rank $n-p$. Under $H_0$, \begin{align*} (I_n-P_X)Y = (I_n-P_X)(X\beta+\varepsilon) = (I_n-P_X)\varepsilon, \end{align*} because $X\beta \in \operatorname{Range}(X)$. Therefore \begin{align*} \frac{\operatorname{RSS}_1}{\sigma^2} = \frac{|(I_n-P_X)\varepsilon|^2}{\sigma^2} \sim \chi^2_{n-p}. \end{align*} Moreover, $R\hat{\beta}-r=RCX^\top\varepsilon$ and $(I_n-P_X)\varepsilon$ are jointly Gaussian linear transformations of $\varepsilon$, and their covariance is \begin{align*} \operatorname{Cov}\bigl(RCX^\top\varepsilon,(I_n-P_X)\varepsilon\bigr) = \sigma^2 RCX^\top(I_n-P_X) = 0. \end{align*} Joint Gaussian random vectors with zero covariance are independent, so \begin{align*} \frac{\operatorname{RSS}_0-\operatorname{RSS}_1}{\sigma^2} \quad\text{and}\quad \frac{\operatorname{RSS}_1}{\sigma^2} \end{align*} are independent. [guided] The unrestricted residual is the part of the noise lying orthogonally to the column space of $X$. Since $P_X$ is an orthogonal projection onto a $p$-dimensional subspace, $I_n-P_X$ is an orthogonal projection onto the orthogonal complement, whose dimension is $n-p$. Under $H_0$ we still have $Y=X\beta+\varepsilon$, and $X\beta \in \operatorname{Range}(X)$. Hence \begin{align*} (I_n-P_X)Y = (I_n-P_X)(X\beta+\varepsilon) = (I_n-P_X)X\beta + (I_n-P_X)\varepsilon = (I_n-P_X)\varepsilon. \end{align*} Because $\varepsilon \sim \mathcal{N}_n(0,\sigma^2 I_n)$, the squared length of its projection onto an $(n-p)$-dimensional subspace, divided by $\sigma^2$, has the $\chi^2_{n-p}$ distribution: \begin{align*} \frac{\operatorname{RSS}_1}{\sigma^2} = \frac{|(I_n-P_X)\varepsilon|^2}{\sigma^2} \sim \chi^2_{n-p}. \end{align*} We also need independence from the numerator. The numerator depends only on \begin{align*} R\hat{\beta}-r=RCX^\top\varepsilon, \end{align*} while the denominator depends only on $(I_n-P_X)\varepsilon$. These are jointly Gaussian because both are linear transformations of the Gaussian vector $\varepsilon$. Their covariance is \begin{align*} \operatorname{Cov}\bigl(RCX^\top\varepsilon,(I_n-P_X)\varepsilon\bigr) = \sigma^2 RCX^\top(I_n-P_X). \end{align*} Since $P_X=XCX^\top$, we have $X^\top P_X=X^\top$, and therefore \begin{align*} X^\top(I_n-P_X)=X^\top-X^\top P_X=0. \end{align*} Thus the covariance is zero. For jointly Gaussian random vectors, zero covariance implies independence. Consequently, \begin{align*} \frac{\operatorname{RSS}_0-\operatorname{RSS}_1}{\sigma^2} \quad\text{and}\quad \frac{\operatorname{RSS}_1}{\sigma^2} \end{align*} are independent. [/guided] [/step] [step:Form the ratio and conclude the $F$ distribution] Define \begin{align*} U := \frac{\operatorname{RSS}_0-\operatorname{RSS}_1}{\sigma^2}, \qquad V := \frac{\operatorname{RSS}_1}{\sigma^2}. \end{align*} The previous steps show that $U \sim \chi^2_q$, $V \sim \chi^2_{n-p}$, and $U$ and $V$ are independent. The distribution $F_{q,n-p}$ denotes the law of $(U/q)/(V/(n-p))$ when $U$ and $V$ are independent chi-squared random variables with $q$ and $n-p$ degrees of freedom. Hence \begin{align*} \frac{U/q}{V/(n-p)} \sim F_{q,n-p}. \end{align*} Substituting the definitions of $U$ and $V$ gives \begin{align*} \frac{U/q}{V/(n-p)} = \frac{(\operatorname{RSS}_0-\operatorname{RSS}_1)/q}{\operatorname{RSS}_1/(n-p)}. \end{align*} Therefore \begin{align*} F = \frac{(\operatorname{RSS}_0-\operatorname{RSS}_1)/q}{\operatorname{RSS}_1/(n-p)} \sim F_{q,n-p}, \end{align*} as required. [/step]