A Continuous-Time Approach to Intensive Longitudinal Data: What, Why, and How?

Abstract

The aim of this chapter is to (a) provide a broad didactical treatment of the first-order stochastic differential equation model—also known as the continuous-time (CT) first-order vector autoregressive (VAR(1)) model—and (b) argue for and illustrate the potential of this model for the study of psychological processes using intensive longitudinal data. We begin by describing what the CT-VAR(1) model is and how it relates to the more commonly used discrete-time VAR(1) model. Assuming no prior knowledge on the part of the reader, we introduce important concepts for the analysis of dynamic systems, such as stability and fixed points. In addition we examine why applied researchers should take a continuous-time approach to psychological phenomena, focusing on both the practical and conceptual benefits of this approach. Finally, we elucidate how researchers can interpret CT models, describing the direct interpretation of CT model parameters as well as tools such as impulse response functions, vector fields, and lagged parameter plots. To illustrate this methodology, we reanalyze a single-subject experience-sampling dataset with the R package ctsem; for didactical purposes, R code for this analysis is included, and the dataset itself is publicly available.

Appendix: Matrix Exponential

Similar to the scalar exponential, the matrix exponential can be defined as an infinite sum

$$\displaystyle \begin{aligned} \boldsymbol{e}^{\boldsymbol{A}}=\sum_{k=0}^{\infty}\frac{1}{k!}\boldsymbol{A}^k \end{aligned}$$

The exponential of a matrix is not equivalent to taking the scalar exponential of each element of the matrix, unless that matrix is diagonal. The exponential of a matrix can be found using an eigenvalue decomposition

$$\displaystyle \begin{aligned} \boldsymbol{A}=\boldsymbol{V}\boldsymbol{D}\boldsymbol{V^{-1}} \end{aligned}$$

where V is a matrix of eigenvectors of A and D is a diagonal matrix of the eigenvalues of A (cf. Moler and Van Loan 2003). The matrix exponential of A is given by

$$\displaystyle \begin{aligned} \boldsymbol{e}^{\boldsymbol{A}} = \boldsymbol{V} \boldsymbol{e}^{\boldsymbol{D}}\boldsymbol{V}^{-1} \end{aligned}$$

where e ^D is the diagonal matrix whose entries are the scalar exponential of the eigenvalues of A. When we want to solve for the matrix exponential of a matrix multiplied by some constant Δt, we get

$$\displaystyle \begin{aligned} \boldsymbol{e}^{\boldsymbol{A}\varDelta t} = \boldsymbol{V}\boldsymbol{e}^{\boldsymbol{D}\varDelta t}\boldsymbol{V}^{-1} \end{aligned} $$

(2.12)

Take it that we have a 2 × 2 square matrix given by

$$\displaystyle \begin{aligned} \boldsymbol{A} = \left[ \begin{array}{cc} a & b \\ c & d \\ \end{array} \right] \end{aligned}$$

and we wish to solve for e ^AΔt. The eigenvalues of A are given by

$$\displaystyle \begin{aligned} \lambda_1 &= \frac{1}{2}\left(a+d-\sqrt{a^2 + 4bc - 2ad + d^2}\right)\\ \lambda_2 &= \frac{1}{2}\left(a+d+\sqrt{a^2 + 4bc - 2ad + d^2}\right) \end{aligned} $$

where we will from here on denote

$$\displaystyle \begin{aligned} R = \sqrt{a^2 + 4bc - 2ad + d^2} \end{aligned}$$

for notational simplicity. The exponential of the diagonal matrix made up of eigenvalues multiplied by the constant Δt is given by

$$\displaystyle \begin{aligned} \boldsymbol{e}^{\boldsymbol{D}\varDelta t} = \left[ \begin{array}{cc} e^{\frac{1}{2}(a+d-R) \varDelta t} & 0 \\ 0 &e^{\frac{1}{2}(a+d+R)\varDelta t} \\ \end{array} \right] \end{aligned}$$

The matrix of eigenvectors of A is given by

$$\displaystyle \begin{aligned} \boldsymbol{V} = \left[ \begin{array}{cc} \frac{a-d-R}{2c} & \frac{a-d+R}{2c} \\ 1 & 1 \\ \end{array} \right] \end{aligned}$$

assuming c ≠ 0, with inverse

$$\displaystyle \begin{aligned} \boldsymbol{V}^{-1} = \left[ \begin{array}{cc} \frac{-c}{R} & \frac{a-d+R}{2R} \\ \frac{c}{R} & \frac{-a+d+R}{2R} \\ \end{array} \right]. \end{aligned}$$

Multiplying V e ^D V ⁻¹ gives us

$$\displaystyle \begin{aligned} \boldsymbol{e}^{\boldsymbol{A}\varDelta t} = \left[ \begin{array}{cc} \frac{R-a+d}{2R}e^{\lambda_1 \varDelta t} + \frac{R+a-d}{2R}e^{\lambda_2 \varDelta t} & \frac{b(-e^{\lambda_1 \varDelta t} + e^{\lambda_2 \varDelta t})}{R} \\ \frac{c(-e^{\lambda_1 \varDelta t} + e^{\lambda_2 \varDelta t})}{R} & \frac{R+a-d}{2R}e^{\lambda_1 \varDelta t} + \frac{R-a+d}{2R}e^{\lambda_2 \varDelta t} \\ \end{array} \right] {} \end{aligned} $$

(2.13)

Note that we present here only a worked out example for a 2 × 2 square matrix. For larger square matrices (representing models with more variables), the eigenvalue decomposition remains the same although the terms for the eigenvalues, eigenvectors, and determinants become much less feasible to present.