# Portion of Derivation of F-test

In Greene's Econometric Analysis there is a derivation regarding the F stat. The setup is a null hypothesis of the form: $$H_0: R\beta =q$$ where $$\beta$$ is a $$k\times 1$$ vector of parameters, $$R$$ is a $$J\times k$$ matrix and $$q$$ is a $$J\times 1$$ vector. $$R$$ and $$q$$ together define restrictions that are hypothesized. $$e_*$$ are residuals that result from OLS when imposing the restrictions and $$e$$ are residuals when estimating OLS without restrictions. $$b_*$$ and $$b$$ are respective parameter estimates. (Note that $$Rb_*=q$$).

Greene's derivation has a portion in which this line is shown,

$$(1) \hspace{3.2cm} e_*'e_*=e'e +(b-b_*)'X'X(b-b_*)$$ The next line shows,

$$(2) \hspace{0.5cm}e_*'e_*-e'e =(Rb-q)'[R(X'X)^{-1}R']^{-1}(Rb-q)$$

I am curious how to go between these lines. If $$R$$ were an invertible square matrix, it is immediate because $$e'e +(b-b_*)'X'X(b-b_*) =e'e +(b-b_*)'R'(R')^{-1}X'XR^{-1}R(b-b_*)$$ then we simplify further.

$$R$$ is not an invertible square matrix however, so this can't be the method. Futhermore, appealing to the pseudoinverse does not help, because $$R$$ does not have full column rank, thus $$R^+R\ne I$$ and we can't proceed with $$e'e +(b-b_*)'X'X(b-b_*) =e'e +(b-b_*)'R'(R')^{+}X'XR^{+}R(b-b_*)$$.

In short, there must be an algebraic trick to go from equation (1) to (2). I am curious what that trick is.

• This looks like a good candidate for Cross Validated Stack Exchange. Commented Oct 18, 2023 at 16:14
• Michael: Unfortunately, the proof at the link below uses different notation that you used but I'm quite confident that you can follow it and map it to your case. I did find a textbook derivation that uses your notation ( a introduction to econometric theory by james davidson ) but it's a fairly new text and probably not worth purchasing for one derivation. web.vu.lt/mif/a.buteikis/wp-content/uploads/PE_Book/… Commented Oct 19, 2023 at 5:30
• Mark Leeds this is great. Thank you. You could consider making this an answer rather than a comment. Commented Oct 20, 2023 at 8:54
• Hi Michael: I'm glad it helps. I'm running out now but I'll try to remember later. Note that I re-read my comment in above and I don't mean to denigrate Davidson's book. It's quite a nice intro book. If someone is interested in that type of book, I highly recommend it. Commented Oct 20, 2023 at 16:23

Consider the restricted least squares problem: $$\min \underbrace{(Y - Xb_\ast)'}_{(1\times n)}\underbrace{(Y - Xb_\ast)}_{n\times 1} \text{ s.t. } \underbrace{R}_{(j \times k)}\underbrace{b_\ast}_{(k \times 1)} - \underbrace{q}_{(j \times 1)}$$ We set up the Lagrangian $$L(b_\ast,\lambda) = (Y - Xb_\ast)'(Y - Xb_\ast) - 2 \underbrace{\lambda'}_{(1 \times j)}(Rb_\ast-q).$$ The first order conditions give: $$\underbrace{X'(Y - X b_\ast)}_{(k \times k)} + \underbrace{R'\lambda}_{(k \times k)} = 0.$$ This gives: $$b_\ast = \underbrace{(X'X)^{-1}X'Y}_{b} - (X'X)^{-1}R'\lambda.$$ The first term is the usual (unrestricted) OLS estimator, say $$b$$. Multiplying both sides with $$R$$ gives: $$\underbrace{Rb_\ast}_{q} = \underbrace{R b}_{(j \times 1)} - \underbrace{R(X'X)^{-1}R'}_{(j \times j)} \underbrace{\lambda}_{j \times j}$$ If $$R(X'X)^{-1}R'$$ is of full rank, this can be solved for $$\lambda$$: $$\lambda = (R(X'X)^{-1}R')^{-1}(Rb - q).$$ Subsituting into the expression for $$b_\ast$$ gives: $$b_\ast - b = \underbrace{(X'X)^{-1}}_{(k \times k)} \underbrace{R'}_{(k \times j)}\underbrace{(R(X'X)^{-1}R')^{-1}}_{(j \times j)} \underbrace{(q - Rb)}_{(j \times 1)}$$
Now, substitute this into the expression $$(b-b_\ast)'X'X(b - b_ast)$$
$$(q - Rb)' (R(X'X)^{-1}R')^{-1}R(X'X)^{-1} X'X (X'X)^{-1}R'(R(X'X)^{-1}R')^{-1}(q - Rb)$$
We can simplify the middle part as $$X'X (X'X)^{-1} = I$$. $$(q - Rb)' (R(X'X)^{-1}R')^{-1}R(X'X)^{-1} R'(R(X'X)^{-1}R')^{-1}(q - Rb)$$ Next, also $$R(X'X)^{-1}R'(R(X'X)^{-1}R')^{-1} = I$$ so we get that: $$(b- b_\ast) X'X (b - b_\ast) = (q - Rb)' (R(X'X)^{-1}R')^{-1}(q - Rb)$$ This is expression (2).