In Fumio Hayashi's Econometrics(2000), there're two types of F-ratio when testing linear hypotheses: the null hypothesis is $H0: R\hat \beta = \gamma$, and the alternative hypothesis is $H1: R\hat \beta \not = \gamma$, where $\gamma$ is an $r$-order vector. There're $K$ parameters and $N$ observations.
(1) $F_1 \equiv (R\hat \beta - \gamma)^T [\hat \sigma^2 R(X^TX)^{-1}R^T]^{-1} (R\hat \beta - \gamma) / r$
As can be seen at pp.41 (Equation 1.4.9), this F-ratio only needs one OLS estimation without any restriction.
(2) $F_2 = \frac{(SSR_r - SSR_u)/r}{SSR_u/(N-K)}$
As can be seen at pp.43 (Equation 1.4.11), this F-ratio needs an OLS without any restriction($SSR_u$), and an OLS with restriction of $R\hat \beta = \gamma$. $SSR_r$ is the sum square of residual of OLS with restriction.
How to prove that $F_1$ and $F_2$ are equivalent?
What I have done is shown below
For OLS with no restriction, the sum square of resid (SSR) is: $$SSR_U = (Y - X\hat \beta)^T (Y - X\hat \beta) = [Y - X(X^TX)^{-1} X^TY]^T [Y - X(X^TX)^{-1} X^TY]\\ = (MY)^T (MY) = [M(X\beta + \epsilon)]^T[M(X\beta + \epsilon)] = \epsilon^T M^TM\epsilon = \epsilon^T M \epsilon$$
And then estimate $\hat \beta$ using OLS with restriction:$\lim_{\beta} \epsilon^T \epsilon$, subject to $R\beta = \gamma$. Make a Lagrangian function: $$ L = (Y - X\beta)^T (Y - X\beta) - \lambda (R\beta - \gamma) $$ And then calculate the First-Order Criteria(FOC): $$\text{(1)}\frac{\partial L}{\partial \beta} = -2X^TY + 2X^TX\hat \beta - R^T \lambda^T = 0\\ \hat \beta = (X^TX)^{-1}[\frac{1}{2}R^T\lambda^T + X^TY]$$ $$\text{(2)} \frac{\partial L}{\partial \lambda} = R\beta - \gamma = 0\\\gamma = R\beta$$ How should I do then?
I've solved this question, my proof is shown below.
F2's numerator
For OLS with no restriction, the sum square of resid (SSR) is: $$SSR_U = (Y - X\hat \beta)^T (Y - X\hat \beta) = [Y - X(X^TX)^{-1} X^TY]^T [Y - X(X^TX)^{-1} X^TY]\\ = (MY)^T (MY) = [M(X\beta + \epsilon)]^T[M(X\beta + \epsilon)] = \epsilon^T M^TM\epsilon = \epsilon^T M \epsilon$$
Where $M = I - X(X^TX)^{-1}X^T$. And then estimate $\hat \beta$ using OLS with restriction:$\min_{\beta} \epsilon^T \epsilon$, subject to $R\beta = \gamma$. Make a Lagrangian function: $$ L = (Y - X\beta)^T (Y - X\beta) - \lambda (R\beta - \gamma) $$ And then calculate the First-Order Criteria(FOC): $$\text{(1)}\frac{\partial L}{\partial \beta} = -2X^TY + 2X^TX\hat \beta - R^T \lambda^T = 0\\ \hat \beta = (X^TX)^{-1}[\frac{1}{2}R^T\lambda^T + X^TY]$$ $$\text{(2)} \frac{\partial L}{\partial \lambda} = R\beta - \gamma = 0\\\gamma = R\beta$$
Put FOCs into a matrix equation, and solve it: $$ \left[\begin{matrix} 2X^TX & R^T \\ R & 0 \end{matrix}\right] \left[\begin{matrix} \bar\beta \\ \lambda^T \end{matrix}\right] = \left[\begin{matrix} 2X^TY \\ \gamma \end{matrix}\right]\\ \left[\begin{matrix} \bar\beta \\ \lambda^T \end{matrix}\right] = \left[\begin{matrix} 2X^TX & R^T \\ R & 0 \end{matrix}\right]^{-1}\left[\begin{matrix} 2X^TY \\ \gamma \end{matrix}\right] $$ Thus the restricted estimator $\bar \beta$ is: $$\bar \beta = \hat \beta - (X^TX)^{-1}R^TW(R\hat\beta - \gamma)$$
Where $W = [R(X^TX)^{-1}R^T]^{-1}$. Therefore, the SSR of restricted OLS is:
$$SSR_R = (Y - X\bar \beta)^T (Y - X\bar \beta) \\ Y-X\bar\beta = MY + X(X^TX)^{-1}R^TW[R(X^TX)^{-1}X^TY - \gamma]\\ Y - X\bar\beta = (M+S)Y - X(X^TX)^{-1}R^TW\gamma$$
Where $S = X(X^TX)^{-1}R^TWR(X^TX)^{-1}X^T$, it's easy to prove that both $S$ and $(M+S)$ are symmetric and idempotent.
Thus, the $SSR_R$ is: $$SSR_R = Y^T(M+S)Y - Y^T(M+S)^TX(X^TX)^{-1}R^TW\gamma\\ -\gamma^TWR(X^TX)^{-1}X^T(M+S)Y + \gamma^TW\gamma$$
(1)The second term of the equation above is: $$Y^TM^TX(X^TX)^{-1}R^TW\gamma = 0\\ Y^TS^TX(X^TX)^{-1}R^TW\gamma = Y^T X(X^TX)^{-1}R^TW\gamma$$
(2) The third term is: $$\gamma^TWR(X^TX)^{-1}X^TMY = \gamma^TWR(X^TX)^{-1}X^T[I-X(X^TX)^{-1}X^T]Y = 0\\ \gamma^TWR(X^TX)^{-1}X^TSY = \gamma^TWR(X^TX)^{-1}X^TY$$
Therefore, $SSR_R - SSR_U$ is: $$SSR_R - SSR_U = Y^TSY - \gamma^TWR(X^TX)^{-1}X^TY- Y^T X(X^TX)^{-1}R^TW\gamma + \gamma^TW\gamma\\ =Y^TSY - \gamma^TWR\hat \beta - \hat\beta^TR^T W\gamma + \gamma^TW\gamma$$ Where $Y^TSY = \hat \beta^T R^TWR\hat\beta$,so F2's numerator can be written as: $$\frac{SSR_R - SSR_U}{r} = (R\hat \beta - \gamma)^T [R(X^TX)^{-1}R^T]^{-1} (R\hat \beta - \gamma)/r$$
F2's denominator
Because $M$ is idempotent, $\epsilon^T M \epsilon = (M\epsilon)^T(M\epsilon) = [M(X\beta + \epsilon)]^T [M(X\beta + \epsilon)] = (MY)^T(MY) = \hat \epsilon^T \hat \epsilon$. $$\frac{SSR_U}{N-K} = \frac{\epsilon^T M\epsilon}{N-K} = \frac{\hat\epsilon^T \hat \epsilon}{N-K} = \hat \sigma^2$$ It's clear that $\hat \sigma^2$ is scalar.
In summary
$$F_2 = \frac{(SSR_R - SSR_U)/r}{SSR_U/(N-K)} = (R\hat \beta - \gamma)^T [R(X^TX)^{-1}R^T]^{-1} (R\hat \beta - \gamma)/(r \hat \sigma^2)\\ = (R\hat \beta - \gamma)^T [\boldsymbol{\hat \sigma^2} R(X^TX)^{-1}R^T]^{-1} (R\hat \beta - \gamma) / \boldsymbol{r} = F_1$$
Q.E.D.