# Chapter 3 The F test for Comparing Reduced vs. Full Models

• Assume GMMNE and suppose $$\mathcal{C}(X_0) \subset \mathcal{C}(X)$$. We wish to test $$H_0: E(y)\in C(X_0)$$ vs. $$H_A: E(y)\in C(X)\backslash C(X_0)$$

• For the general case, consider the test statistics

\begin{aligned} F & = \frac{y'(P_X-P_{X_0})y/[rank(X) - rank(X_0)]}{y'(I-P_X)y/[n-rank(X)]} \\ & \sim F_{rank(X)-rank(X_0), n-rank(X)}\left(\frac{\beta X'(P_X-P_{X_0})X\beta}{2\sigma^2}\right) \end{aligned}

• Because $$(\frac{P_X-P_{X_0}}{\sigma^2})(\sigma^2I) = P_X - P_X$$ is idempotent and $$rank(P_X - P_{X_0}) = rank(P_X)-rank(P_{X_0})$$, $y'(\frac{P_X-P_{X_0}}{\sigma^2})y \sim \chi_{rank(X)-rank(X_0)}^2(\frac{1}{2}\beta'X'(\frac{P_X-P_{X_0}}{\sigma^2})X\beta)$

• $$y'(\frac{I-P_X}{\sigma^2})y \sim \chi_{n-rank(X)}^2$$

• $$y'(\frac{P_X-P_{X_0}}{\sigma^2})y$$ is independent of $$y'(\frac{I-P_X}{\sigma^2})y$$ because $$(\frac{P_X-P_{X_0}}{\sigma^2})(\sigma^2I)(\frac{I-P_X}{\sigma^2}) = 0$$

• If $$H_0$$ is true, then $$(P_X-P_{X_0})X\beta = 0$$

• $$\beta'X(P_X-P_{X_0})X\beta = \|(P_X-P_{X_0})X\beta\|^2 = \|X\beta-P_{X_0}X\beta\|^2 = \|E(y) - P_{X_0}E(y)\|^2$$

• $$y'(P_X-P_{X_0})y = y'(I-P_{X_0})y - y'(I-P_X)y = SSE_{REDUCED} - SSE_{FULL}$$

• Thus, $$F = \frac{(SSE_{REDUCED} - SSE_{FULL})/(DFE_{REDUCED} - DEF_{FULL})}{SSE_{FULL}/DFE_{FULL}}$$

• Equivalence of $$F$$ test: this reduced vs. full model F test is equivalent to the F test for testing $$H_0: C\beta = d$$ vs. $$H_A: C\beta \neq d$$.