Combining Euclidean and composite likelihood for binary spatial data estimation

Abstract

In this paper we propose a blockwise Euclidean likelihood method for the estimation of a spatial binary field obtained by thresholding a latent Gaussian random field. The moment conditions used in the Euclidean likelihood estimator derive from the score of the composite likelihood based on marginal pairs. A feature of this approach is that it is possible to obtain computational benefits with respect to the pairwise likelihood depending on the choice of the spatial blocks. A simulation study and an analysis on cancer mortality data compares the two methods in terms of statistical and computational efficiency. We also study the asymptotic properties of the proposed estimator.

Appendix

The appendix collects the proofs of the asymptotic results. First of all, we introduce the following notation: $\nabla _{\varvec{\theta }}$ and $\nabla _{\varvec{\lambda }}$ are the first derivative operators for ${\varvec{\theta }}$ and ${\varvec{\lambda }}$ respectively, while $\nabla _{{\varvec{\theta }}{\varvec{\theta }}}$, $\nabla _{{\varvec{\lambda }}{\varvec{\lambda }}}$ and $\nabla _{{\varvec{\theta }}{\varvec{\lambda }}}$ indicate second and cross derivatives and are defined accordingly. Similarly, for a certain function $\widehat{R}({\varvec{\theta }},{\varvec{\lambda }})$ defined below, $\widehat{R}_{\varvec{\theta }}({\varvec{\theta }},{\varvec{\lambda }})$ is its first derivative with respect to ${\varvec{\theta }}$. Derivatives with respect to ${\varvec{\lambda }}$, second derivatives and cross derivatives are defined in a similar manner. In addition to that, we use

$$\begin{aligned} {\varvec{\varSigma }}({\varvec{\theta }})=\frac{b^d}{N}\sum _{\varvec{\kappa }}{\varvec{m}}_{\varvec{\kappa }}({\varvec{\theta }}){\varvec{m}}_{\varvec{\kappa }}({\varvec{\theta }})^\intercal \end{aligned}$$

instead of the expression in Eq. (20). As noticed in Newey and Smith (2004) this choice does not change the estimator but it simplifies the asymptotic analysis.

Proof

(Proof of Theorem 1) We have to show that, for some $\delta >0$, $P(\Vert \widehat{\varvec{\theta }}-{\varvec{\theta }}_0\Vert >\delta )\rightarrow 0$ as $n\rightarrow \infty$. By continuity of $Q({\varvec{\theta }})$ and the assumption that ${\varvec{\theta }}_0$ is the unique minimizer, we have that, for some $\varepsilon >0$, $\{ \Vert \widehat{\varvec{\theta}}-{\varvec{\theta}}_0\Vert > \delta \} \Longrightarrow \{ \vert \ Q(\widehat{\varvec{\theta }})-Q({\varvec{\theta}}_0)\vert >\varepsilon \}$. This is, the latter set contains the former. Hence, $P(\Vert \widehat{\varvec{\theta }}-{\varvec{\theta }}_0\Vert >\delta )\le P(\vert \ Q(\widehat{\varvec{\theta }})-Q({\varvec{\theta }}_0)\vert >\varepsilon )$. By assumptions C1–C2 and arguments similar to those in Theorem 2 we have the following uniform convergence condition

$$\begin{aligned} \sup _{\varvec{\theta }\in \theta }\vert \widehat{Q}_n({\varvec{\theta }})-Q({\varvec{\theta }})\vert \rightarrow _p0. \end{aligned}$$

(23)

Therefore,

$$\begin{aligned} \varepsilon\,&<\,\vert \ Q(\widehat{\varvec{\theta }})-Q({\varvec{\theta }}_0)\vert =\vert \ Q(\widehat{{\varvec{\theta }}})-\widehat{Q}_n({\varvec{\theta }}_0)+\widehat{Q}_n({\varvec{\theta }}_0)-Q({\varvec{\theta }}_0)\vert \\&\le 2\sup _{{\varvec{\theta }}\in \theta }\vert \widehat{Q}_n({\varvec{\theta }})-Q({\varvec{\theta }})\vert \rightarrow _p0 \end{aligned}$$

where the latter inequality follows from the triangular inequality and the uniform convergence condition in (23). This implies $P(\Vert \widehat{\varvec{\theta }}-{\varvec{\theta }}_0\Vert >\delta )\le P(\vert \ Q(\widehat{\varvec{\theta }})-Q({\varvec{\theta }}_0)\vert >\varepsilon )\rightarrow 0$ as $n\rightarrow \infty$. Hence, $\widehat{\varvec{\theta }}\rightarrow _p{\varvec{\theta }}_0$. Before showing asymptotic normality we show that the estimate of the Lagrange multiplier $\widehat{\varvec{\lambda }}$ converges to zero in probability. Similarly to (7) we have that

$$\begin{aligned} \frac{\widehat{\varvec{\lambda }}}{b^d}=\widehat{\varvec{\varSigma }}(\widehat{\varvec{\theta }})^{-1}\widehat{\varvec{m}}(\widehat{\varvec{\theta }}). \end{aligned}$$

(24)

By a mean value argument, assumptions C1-C2, results in Theorem 2 and continuous mapping theorem we get

$$\begin{aligned} \frac{\widehat{\varvec{\lambda }}}{b^d}\rightarrow _p{\bf{0}}\end{aligned}$$

In analogy with (5) let us define

$$\begin{aligned} \pi _{\varvec{\kappa }}({\varvec{\theta }},{\varvec{\lambda }})=\frac{1}{N}\Big (1-{\varvec{\lambda }}^\intercal ({\varvec{m}}_{\varvec{\kappa }}({\varvec{\theta }})-\widehat{{\varvec{m}}}({\varvec{\theta }}))\Big ) \end{aligned}$$

and

$$\begin{aligned} \widehat{R}({\varvec{\theta }},{\varvec{\lambda }})=-\frac{1}{2N}\sum _{{\varvec{\kappa }}}(N\pi _{\varvec{\kappa }}({\varvec{\theta }},{\varvec{\lambda }})-1)^2. \end{aligned}$$

Then by simple calculations we can write

$$\begin{aligned} \widehat{R}({\varvec{\theta }},{\varvec{\lambda }})=-{\varvec{\lambda }}^\intercal \widehat{\varvec{m}}({\varvec{\theta }})+\frac{1}{2b^d_n}{\varvec{\lambda }}^\intercal \widehat{\varvec{\varSigma }}({\varvec{\theta }})^\intercal {\varvec{\lambda }}. \end{aligned}$$

The first order conditions of $\widehat{R}(\widehat{\varvec{\theta }},\widehat{\varvec{\lambda }})$ with respect to ${\varvec{\theta }}$ and ${\varvec{\lambda }}$ are

$$\begin{aligned} 0=\widehat{R}_{\varvec{\theta }}(\widehat{\varvec{\theta }},\widehat{\varvec{\lambda }})=-\nabla \widehat{\varvec{m}}(\widehat{\varvec{\theta }})\widehat{\varvec{\lambda }}+\frac{\varvec{\lambda }^\intercal }{Nb^d_n}\sum _{{\varvec{\kappa }}}{\varvec{m}}_{\varvec{\kappa }}(\widehat{\varvec{\theta }})\nabla {\varvec{m}}_{\varvec{\kappa }}(\widehat{\varvec{\theta }})\widehat{\varvec{\lambda }} \end{aligned}$$

(25)

$$\begin{aligned} 0=\widehat{R}_{\varvec{\lambda }}(\widehat{\varvec{\theta }},\widehat{\varvec{\lambda }})=-\widehat{\varvec{m}}(\widehat{\varvec{\theta }})+\frac{1}{b^d}\widehat{\varvec{\varSigma }}(\widehat{\varvec{\theta }})\widehat{\varvec{\lambda }}. \end{aligned}$$

(26)

Let us now take a mean value expansion of the first order conditions (25) and (26) about the true values $(\varvec{\theta }^\intercal ,{\varvec{\lambda }}^\intercal )^\intercal =({\varvec{\theta }}_0^\intercal ,{\bf{0}}^\intercal )^\intercal$

$$\begin{aligned} {\bf{0}}=\widehat{R}_{\varvec{\theta }}(\widehat{\varvec{\theta }},\widehat{\varvec{\lambda }})&=\widehat{R}_{\varvec{\theta }}({\varvec{\theta }}_0,{\bf{0}})+\widehat{R}_{\varvec{\theta }{\varvec{\lambda }}}(\dot{\varvec{\theta }},\dot{\varvec{\lambda }})\widehat{\varvec{\lambda }}+\widehat{R}_{\varvec{\theta }\varvec{\theta }}(\dot{\varvec{\theta }},\dot{\varvec{\lambda }})(\widehat{\varvec{\theta }}-{\varvec{\theta }}_0)\\\nonumber&=\widehat{R}_{\varvec{\theta }{\varvec{\lambda }}}(\dot{\varvec{\theta }},\dot{\varvec{\lambda }})\frac{\sqrt{n}}{b^d}\widehat{\varvec{\lambda }}+\frac{1}{b^d}\widehat{R}_{\varvec{\theta }{\varvec{\theta }}}(\dot{\varvec{\theta }},\dot{\varvec{\lambda }})\sqrt{n}(\widehat{\varvec{\theta }}-{\varvec{\theta }}_0) \end{aligned}$$

(27)

$$\begin{aligned} {\bf{0}}=\widehat{R}_{\varvec{\lambda }}(\widehat{\varvec{\theta }},\widehat{\varvec{\lambda }})&=\widehat{R}_{\varvec{\lambda }}({\varvec{\theta }}_0,{\bf{0}})+\widehat{R}_{\varvec{\lambda }{\varvec{\lambda }}}(\dot{\varvec{\theta }},\dot{\varvec{\lambda }})\widehat{\varvec{\lambda }}+\widehat{R}_{\varvec{\lambda }\varvec{\theta }}(\dot{\varvec{\theta }},\dot{\varvec{\lambda }})(\widehat{\varvec{\theta }}-{\varvec{\theta }}_0)\\\nonumber&=\sqrt{n}\widehat{R}_{\varvec{\lambda }}({\varvec{\theta }}_0,{\bf{0}})+b^d\widehat{R}_{\varvec{\lambda }{\varvec{\lambda }}}(\dot{\varvec{\theta }},\dot{\varvec{\lambda }})\frac{\sqrt{n}}{b^d}\widehat{\varvec{\lambda }}+\widehat{R}_{\varvec{\lambda }{\varvec{\theta }}}(\dot{\varvec{\theta }},\dot{\varvec{\lambda }})\sqrt{n}(\widehat{\varvec{\theta }}-{\varvec{\theta }}_0). \end{aligned}$$

(28)

More compactly,

$$\begin{aligned} \left( {\begin{array}{c} {\bf{0}}\\ \sqrt{n}\widehat{R}_{\varvec{\lambda }}({\varvec{\theta }}_0,{\bf{0}}) \end{array}}\right) =- \left( {\begin{array}{cc} \frac{1}{b^d}\widehat{R}_{\varvec{\theta }{\varvec{\theta }}}(\dot{\varvec{\theta }},\dot{\varvec{\lambda }}) &{} \widehat{R}_{{\varvec{\theta }}{\varvec{\lambda }}}(\dot{\varvec{\theta }},\dot{\varvec{\lambda }})\\ \widehat{R}_{\varvec{\lambda }{\varvec{\theta }}}(\dot{\varvec{\theta }},\dot{\varvec{\lambda }}) &{} b^d\widehat{R}_{\varvec{\lambda }{\varvec{\lambda }}}(\dot{\varvec{\theta }},\dot{\varvec{\lambda }}) \end{array}}\right) \left( {\begin{array}{c} \sqrt{n}(\widehat{\varvec{\theta }}-{\varvec{\theta }}_0)\\ \frac{\sqrt{n}}{b^d}\widehat{\varvec{\lambda }} \end{array}}\right) . \end{aligned}$$

By the UWLLN $\frac{1}{b^d}\widehat{R}_{\varvec{\theta }{\varvec{\theta }}}(\dot{\varvec{\theta }},\dot{\varvec{\lambda }})\rightarrow _p{\bf{0}}$, $b^d\widehat{R}_{\varvec{\lambda }{\varvec{\lambda }}}(\dot{\varvec{\theta }},\dot{\varvec{\lambda }})\rightarrow _p{\varvec{\varSigma }}({\varvec{\theta }}_0)$ and $\widehat{R}_{\varvec{\lambda }{\varvec{\theta }}}(\dot{\varvec{\theta }},\dot{\varvec{\lambda }})\rightarrow _p-\nabla _{\varvec{\theta }}{\varvec{m}}({\varvec{\theta }}_0)$. Hence,

$$\begin{aligned} \left( {\begin{array}{cll} \sqrt{n}(\widehat{\varvec{\theta }}-{\varvec{\theta }}_0)\\ \frac{\sqrt{n}}{b^d}\widehat{\varvec{\lambda }} \end{array}}\right) =- \left( {\begin{array}{cc} -{\varvec{\varOmega }}({\varvec{\theta }}_0) &{} {\varvec{\varOmega }}({\varvec{\theta }}_0)\nabla _{\varvec{\theta }}{\varvec{m}}({\varvec{\theta }}_0)^\intercal {\varvec{\varSigma }}({\varvec{\theta }}_0)^{-1}\\ {\varvec{\varSigma }}({\varvec{\theta }}_0)^{-1}\nabla _{\varvec{\theta }}{\varvec{m}}({\varvec{\theta }}_0){\varvec{\varOmega }}({\varvec{\theta }}_0) &{}{\varvec{\varLambda }}({\varvec{\theta }}_0) \end{array}}\right) \left( {\begin{array}{c} {\bf{0}}\\ \sqrt{n}\widehat{\varvec{m}}({\varvec{\theta }}_0) \end{array}}\right) \\ +o_p(1) \end{aligned}$$

where ${\varvec{\varOmega }}({\varvec{\theta }}_0)=(\nabla _{\varvec{\theta }}{\varvec{m}}({\varvec{\theta }}_0)^\intercal {\varvec{\varSigma }}({\varvec{\theta }}_0)^{-1}\nabla _{\varvec{\theta }}{\varvec{m}}({\varvec{\theta }}_0))^{-1}$ and ${\varvec{\varLambda }}({\varvec{\theta }}_0)={\varvec{\varSigma }}({\varvec{\theta }}_0)^{-1}-{\varvec{\varSigma }}({\varvec{\theta }}_0)^{-1}\nabla _{\varvec{\theta }}{\varvec{m}}({\varvec{\theta }}_0){\varvec{\varOmega }}({\varvec{\theta }}_0)\nabla _{\varvec{\theta }}{\varvec{m}}({\varvec{\theta }}_0)^\intercal {\varvec{\varSigma }}({\varvec{\theta }}_0)^{-1}$. The result follows from an application of the CLT and the continuous mapping theorem. $\square$

Proof

(Proof of Theorem 2) Assume $\widehat{\varvec{\theta }}$ consistent and define the following mean value expansion around ${\varvec{\theta }}_0$

$$\begin{aligned} \widehat{\varvec{\varSigma }}(\widehat{\varvec{\theta }})=\widehat{\varvec{\varSigma }}({\varvec{\theta }}_0)+\nabla _{\varvec{\theta }}\widehat{\varvec{\varSigma }}(\dot{\varvec{\theta }})(\widehat{\varvec{\theta }}-{\varvec{\theta }}_0). \end{aligned}$$

By rearranging, taking the Euclidean norm and the Cauchy-Schwarz inequality we get

$$\begin{aligned} \Vert \widehat{\varvec{\varSigma }}(\widehat{\varvec{\theta }})-\widehat{\varvec{\varSigma }}({\varvec{\theta }}_0)\Vert&\le \frac{2b^d}{N}\Big \Vert \sum _{\varvec{\kappa }}{\varvec{m}}_{\varvec{\kappa }}(\dot{\varvec{\theta }})\sum _{\varvec{\kappa }}\nabla {\varvec{m}}_{\varvec{\kappa }}(\dot{\varvec{\theta }})\Big \Vert \Vert \widehat{\varvec{\theta }}-{\varvec{\theta }}_0\Vert \\&\le \frac{2b^d}{N}\Big ( \sum _{\varvec{\kappa }}\sup _{\varvec{\theta }\in \theta }\Vert {\varvec{m}}_{\varvec{\kappa }}(\dot{\varvec{\theta }})\Vert ^2\Big )^{\frac{1}{2}}\Big (\sum _{\varvec{\kappa }}\sup _{\varvec{\theta }\in \theta }\Vert \nabla {\varvec{m}}_{\varvec{\kappa }}(\dot{\varvec{\theta }})\Vert ^2\Big )^{\frac{1}{2}}\Vert \widehat{\varvec{\theta }}-{\varvec{\theta }}_0\Vert \\&=o_p(1). \end{aligned}$$

The last line follows from the consistency of $\widehat{\varvec{\theta }}$ and from the fact that $b^d=o(\sqrt{n})$. Thus,

$$\begin{aligned} \Vert \widehat{\varvec{\varSigma }}(\widehat{\varvec{\theta }})-{\varvec{\varSigma }}({\varvec{\theta }}_0)\Vert&\le \Vert \widehat{\varvec{\varSigma }}(\widehat{\varvec{\theta }})-\widehat{\varvec{\varSigma }}({\varvec{\theta }}_0)+\widehat{\varvec{\varSigma }}({\varvec{\theta }}_0)-{\varvec{\varSigma }}({\varvec{\theta }}_0)\Vert \\&\le \Vert \widehat{\varvec{\varSigma }}(\widehat{\varvec{\theta }})-\widehat{\varvec{\varSigma }}({\varvec{\theta }}_0)\Vert +\Vert \widehat{\varvec{\varSigma }}({\varvec{\theta }}_0)-{\varvec{\varSigma }}({\varvec{\theta }}_0)\Vert \\&\le o_p(1)+\sup _{\varvec{\theta }\in \theta }\Vert \widehat{\varvec{\varSigma }}({\varvec{\theta }}_0)-{\varvec{\varSigma }}({\varvec{\theta }}_0)\Vert \\&\le o_p(1). \end{aligned}$$

The result follows from the application of the triangular inequality and UWLLN. $\square$