Loglinear model M 0 is a special case of loglinear model M 1 a Haberman (1974a) showed that when i satisfy any model that is a special case of M 0 , i 1i log i i 0i log i Thus, 0 is the orthogonal projection of 1 onto the linear manifold of log satisfying M 0 Using this, show that G 2 (M 0 ) G 2 (M...

a Using the Haberman result it follows that The first ...

Loglinear model M 0 is a special case of loglinear model M 1 . a. Haberman (1974a)

Question:

Loglinear model M₀ is a special case of loglinear model M₁.

a. Haberman (1974a) showed that when {µ̂_i} satisfy any model that is a special case of M₀, ∑_i µ̂_1i log µ̂_i = ∑_i µ̂_0i log µ̂_i. Thus, µ̂₀ is the orthogonal projection of µ̂₁ onto the linear manifold of {log µ} satisfying M₀. Using this, show that G²(M₀) – G²(M₁) = 2∑_i µ̂_1i log(µ̂_1i/µ̂_0i).