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. \(H_A: C\beta \neq d\).