Spherical radial approximation for nested mixed effects models

Abstract

We consider a likelihood approximation in generalized linear mixed-effects models (GLMM) with multilevel nested random effects. Likelihood evaluation in such models is difficult, hindered by the need for high dimensional integration, where the dimension is proportional to the number of units per level and the number of random effects per unit. Various integration approaches have been proposed, including the penalized quasi-likelihood method, Laplace approximation, quadrature approximation, simulation, and MCMC algorithms. We propose a new quadrature approximation method, which is based on the spherical radial integration approach of Monahan and Genz (J Am Stat Assoc 92:664–674 1997), and at the same time takes advantage of the hierarchical structure of the integration. Our new hierarchical spherical radial method has a time complexity that is linear in the number of units per level and the number of random effects per unit, in contrast to the exponential complexity of the adaptive Gaussian quadrature method of Pinheiro and Chao (J Comput Graph Stat 15:58–81 2006) for the same problem. Using a spline approximation, the generalized additive mixed models (GAMM) are GLMMs with two levels of nested random effects. We apply our method to estimation of GAMMs. We compare it with competing methods through simulations and apply our method to analyze virologic and immunologic responses in an AIDS clinical trial. An R package is written and available at http://users.wpi.edu/~jgagnon/software.html.

Appendix

Lemma 1

If $h(\mathbf {u})=h_0(\mathbf {x}_3)h_1(\mathbf {x}_1,\mathbf {x}_3)h_2(\mathbf {x}_2,\mathbf {x}_3)$ is spherically symmetric in $\mathbf {u}=(\mathbf {x}_1,\mathbf {x}_2,\mathbf {x}_3)$, then so is: $g(\mathbf {x}_3) = h_0(\mathbf {x}_3) \int h_1(\mathbf {x}_1,\mathbf {x}_3)h_2(\mathbf {x}_2,\mathbf {x}_3) d\mathbf {x}_1 d\mathbf {x}_2$ in $\mathbf {x}_3$.

Proof

Let R be a rotation matrix (orthogonal and determinant=1) we then have:

$$\begin{aligned} g(R\mathbf {x}_3)&= \int h(\mathbf {x}_1,\mathbf {x}_2,R\mathbf {x}_3) d\mathbf {x}_1 d\mathbf {x}_2 \nonumber \\&= \int h(\left( \begin{array}{ccc} \mathbf {I} &{} \mathbf {0} &{}\mathbf {0} \\ \mathbf {0} &{} \mathbf {I} &{} \mathbf {0} \\ \mathbf {0} &{} \mathbf {0} &{} \mathbf {R} \end{array} \right) \left( \begin{array}{c} \mathbf {x}_1 \\ \mathbf {x}_2 \\ \mathbf {x}_3 \end{array} \right) ) d\mathbf {x}_1 d\mathbf {x}_2 . \end{aligned}$$

(30)

Since $\left( \begin{array}{ccc} \mathbf {I} &{} \mathbf {0} &{}\mathbf {0} \\ \mathbf {0} &{} \mathbf {I} &{} \mathbf {0} \\ \mathbf {0} &{} \mathbf {0} &{} \mathbf {R} \end{array} \right)$ is orthogonal and has determinant one, it is a rotation matrix, so:

$$\begin{aligned} \int h(\left( \begin{array}{ccc} \mathbf {I} &{} \mathbf {0} &{}\mathbf {0} \\ \mathbf {0} &{} \mathbf {I} &{} \mathbf {0} \\ \mathbf {0} &{} \mathbf {0} &{} \mathbf {R} \end{array} \right) \left( \begin{array}{c} \mathbf {x}_1 \\ \mathbf {x}_2 \\ \mathbf {x}_3 \end{array} \right) ) d\mathbf {x}_1 d\mathbf {x}_2&= \int h(\mathbf {x}_1,\mathbf {x}_2,\mathbf {x}_3) d\mathbf {x}_1 d\mathbf {x}_2 \nonumber \\&= g(\mathbf {x}_3) \end{aligned}$$

(31)

which proves that $g(\mathbf {x}_3)$ is spherically symmetric. $\square$

Lemma 2

Let $h(\mathbf {u})$ be the same as in lemma 1. If $h(\mathbf {u})$ is spherically symmetric with respect to $\mathbf {u}$, then so is $h_1(\mathbf {x}_1,\mathbf {x}_3)$ in $\mathbf {x}_1$ and $h_2(\mathbf {x}_2,\mathbf {x}_3)$ in $\mathbf {x}_2$.

Proof

We show it for $h_1(\mathbf {x}_1,\mathbf {x}_3)$ in $\mathbf {x}_1$ only. Let R be a rotation matrix of the same size of the length of $\mathbf {x}_1$. Then, $G = \left( \begin{array}{ccc} \mathbf {R} &{} \mathbf {0} &{} \mathbf {0} \\ \mathbf {0} &{} {\mathbf {I}_{x_2}} &{} \mathbf {0} \\ \mathbf {0} &{} \mathbf {0} &{} {\mathbf {I}_{x_3}} \end{array} \right)$ is a rotation matrix as well with ${\mathbf {I}_{x_2}}$ being an identity matrix whose size is the length of $\mathbf {x}_2$ and ${\mathbf {I}_{x_3}}$ is an identity matrix whose size is the length of $\mathbf {x}_3$. Because $h(\mathbf {u})$ is spherically symmetric, we have:

$$\begin{aligned} h(\mathbf {u}) = h(G\mathbf {u}) = h_0(\mathbf {x}_3)h_1(R\mathbf {x}_1,\mathbf {x}_3)h_2(\mathbf {x}_2,\mathbf {x}_3) \end{aligned}$$

(32)

which implies:

$$\begin{aligned} h_1(\mathbf {x}_1,\mathbf {x}_3) = h_1(R\mathbf {x}_1,\mathbf {x}_3) \end{aligned}$$

(33)

$\square$

Lemma 3

Let $h(\mathbf {u})$ be the same as in lemma 1 and $\hat{\mathbf {u}} \!= argmax (ln (h(\mathbf {u})))$. Let H be its Hessian where:

$$\begin{aligned} \mathbf {H} = -\frac{\partial ^2 ln(h(\mathbf {u}))}{\partial \mathbf {u}^2}\bigg |_{\mathbf {u}=\hat{\mathbf {u}}} \end{aligned}$$

(34)

and let of the Cholesky decomposition of $\mathbf {H}$ be $\mathbf {L}^{T}\mathbf {L}$. Furthermore, define $f^{*}(\tilde{\mathbf {u}})$ as $|\mathbf {L}^{-1}| f(\hat{\mathbf {u}} + \mathbf {L}^{-1}\tilde{\mathbf {u}})$ with $\tilde{\mathbf {u}}$ being $(\tilde{\mathbf {x}_1},\tilde{\mathbf {x}_2},\tilde{\mathbf {x}_3})^{T}$. Then, we have: $h^*(\tilde{\mathbf {u}})=h_0^{*}(\tilde{\mathbf {x}_3})h_1^{*}(\tilde{\mathbf {x}_1},\tilde{\mathbf {x}_3})h_2^{*}(\tilde{\mathbf {x}_2},\tilde{\mathbf {x}_3})$ for some $h_0^*$, $h_1^*$, and $h_2^*$.

Proof

Pinheiro and Chao (2006) showed that the Hessian has the following block structure:

$$\begin{aligned} \mathbf {H} = \left( \begin{array}{ccc} \triangle &{} 0 &{} \triangle \\ 0 &{} \triangle &{} \triangle \\ \triangle &{} \triangle &{} \triangle \end{array} \right) \end{aligned}$$

(35)

Using the Cholesky decomposition of $\mathbf {H}$, we can show that $\mathbf {L}$ has the following block structure:

$$\begin{aligned} \mathbf {L} = \left( \begin{array}{ccc} \triangle &{} 0 &{} \triangle \\ 0 &{} \triangle &{} \triangle \\ 0 &{} 0 &{} \triangle \end{array} \right) \end{aligned}$$

(36)

Consequently, its inverse has the same block structure by the block matrix inverse formula. Define

$$\begin{aligned} \mathbf {L^{-1}} = \left( \begin{array}{ccc} a_{11} &{} 0 &{} a_{13} \\ 0 &{} a_{22} &{} a_{23} \\ 0 &{} 0 &{} a_{33} \end{array} \right) \end{aligned}$$

(37)

Under this transformation, we have:

$$\begin{aligned}&h^{*} (\tilde{\mathbf {u}}) = |\mathbf {L}^{-1}| h(\hat{\mathbf {u}} + \mathbf {L}^{-1} \tilde{\mathbf {u}}) = |\mathbf {L}^{-1}| h(\hat{\mathbf {x}}_1+a_{11}\tilde{\mathbf {x}}_1+a_{13}\tilde{\mathbf {x}}_3,\nonumber \\&\hat{\mathbf {x}}_2+a_{22}\tilde{\mathbf {x}}_2+a_{23}\tilde{\mathbf {x}}_3,\hat{\mathbf {x}}_3+a_{33}\tilde{\mathbf {x}}_3) = |\mathbf {L}^{-1}|h_0(\hat{\mathbf {x}}_3+a_{33}\tilde{\mathbf {x}}_3)\nonumber \\&h_1(\hat{\mathbf {x}}_1+a_{11}\tilde{\mathbf {x}}_1+a_{13}\tilde{\mathbf {x}}_3,\hat{\mathbf {x}}_3+a_{33}\tilde{\mathbf {x}}_3)h_2(\hat{\mathbf {x}}_2+a_{22}\tilde{\mathbf {x}}_2+a_{23}\tilde{\mathbf {x}}_3,\nonumber \\&\hat{\mathbf {x}}_3+a_{33}\tilde{\mathbf {x}}_3)= h_0^*(\tilde{\mathbf {x}}_3)h_1^*(\tilde{\mathbf {x}}_1,\tilde{\mathbf {x}}_3)h_2^*(\tilde{\mathbf {x}}_2,\tilde{\mathbf {x}}_3). \end{aligned}$$

(38)

$\square$