Large Sample Theory: Freedom-Equation Hypotheses

Abstract

In this chapter we assume once again that $\boldsymbol{\theta }\in W$. However our hypothesis H now takes the form of freedom equations, namely $\boldsymbol{\theta }=\boldsymbol{\theta } (\boldsymbol{\alpha })$, where $\boldsymbol{\alpha }= (\alpha _{1},\alpha _{2},\ldots,\alpha _{p-q})'$. We require the following additional notation. Let $\boldsymbol{\Theta }_{\boldsymbol{\alpha }}$ be the p × p − q matrix with (i, j)th element $\partial \theta _{i}/\partial \alpha _{j}$, which we assume to have rank p − q. As before, $L(\boldsymbol{\theta }) =\log \prod _{ i=1}^{n}f(x_{i},\boldsymbol{\theta })$ is the log likelihood function. Let $\mathbf{D}_{\boldsymbol{\theta }}L(\boldsymbol{\theta })$ and $\mathbf{D}_{\boldsymbol{\alpha }}L(\boldsymbol{\theta }(\boldsymbol{\alpha }))$ be the column vectors whose ith elements are $\partial L(\boldsymbol{\theta })/\partial \theta _{i}$ and $\partial L(\boldsymbol{\theta })/\partial \alpha _{i}$ respectively. As before, $\mathbf{B}_{\boldsymbol{\theta }}$ is the p × p information matrix with i, jth element

$$\displaystyle{-n^{-1}E_{\boldsymbol{\theta }}\left [\frac{\partial ^{2}L(\boldsymbol{\theta })} {\partial \theta _{i}\partial \theta _{j}} \right ] = -E\left [\frac{\partial ^{2}\log \,f(x,\boldsymbol{\theta })} {\partial \theta _{i}\partial \theta _{j}} \right ],}$$

and we add $\mathbf{B}_{\boldsymbol{\alpha }}$, the $p - q \times p - q$ information matrix with i, jth element $-E[\partial ^{2}\log \,f(x,\boldsymbol{\theta }(\boldsymbol{\alpha }))/\partial \alpha _{i}\partial \alpha _{j}]$. To simplify the notation we use $[\cdot ]_{\boldsymbol{\alpha }}$ to denote that the matrix in square brackets is evaluated at $\boldsymbol{\alpha }$, for example

$$\displaystyle{\mathbf{B}_{\boldsymbol{\alpha }} = [\boldsymbol{\varTheta }'\mathbf{B}_{\boldsymbol{\theta }}\boldsymbol{\varTheta }]_{\alpha } =\boldsymbol{\varTheta } _{\boldsymbol{\alpha }}'\mathbf{B}_{\boldsymbol{\theta }(\boldsymbol{\alpha })}\boldsymbol{\varTheta }_{\boldsymbol{\alpha }}.}$$

We note that

$$\displaystyle{\mathbf{D}_{\alpha }L(\boldsymbol{\theta }) =\boldsymbol{\varTheta } _{\alpha }'\mathbf{D}_{\boldsymbol{\theta }}L(\boldsymbol{\theta }(\boldsymbol{\alpha })).}$$