Consider the linear mixed effects model:

\begin{equation} X_i(t_{ij}) = \eta + Z_i(t_{ij})w_i + \epsilon_{ij}, \end{equation}

where $\eta$ is the mean, $Z_i(t_{ij}) = [1, \log(t_{ij})]$, $w_i = (w_{0i}, w_{1i})' \sim N(0,\Sigma_{w})$, $\epsilon_{ij} \sim N(0, \sigma^2)$, and

\begin{equation} \Sigma_{w} = \begin{pmatrix} \sigma^2_1& \rho\sigma_1\sigma_2 \\ \rho\sigma_1\sigma_2 & \sigma^2_2 \end{pmatrix}. \end{equation}

The conditional response is given by

\begin{equation} X_i(t_{ij}) \mid w_i, \eta, \theta, \sigma \sim N(\eta + Z_i(t_{ij})w_i, \sigma^2), \end{equation}

where $\theta = c(\sigma_1, \sigma_2, \rho)$.

But, the variance of $X_i$ is

\begin{equation} \begin{aligned} Cov(X_i(t_i), X_i(t_i)) & = Cov(\eta + Z_i(t_i)w_i + \epsilon_i, \eta + Z_i(t_i)w_i + \epsilon_i) \\ & = Var(Z_i(t_i)w_i) + Var(\epsilon_i) = Z_i(t_i)\Sigma_{w}Z'_i(t_i) + \sigma^2. \end{aligned} \end{equation}

So, why is the conditional response not given by

\begin{equation} X_i(t_{ij}) \mid w_i, \eta, \theta, \sigma \sim N(\eta, Z_i(t_i)\Sigma_{w}Z'_i(t_i) + \sigma^2). \end{equation}


1 Answer 1


Okay, I misunderstood.

The conditional response is given by

\begin{equation} X_i(t_{ij}) \mid w_i \sim N(\eta + Z_i(t_{ij})w_i, \sigma^2). \end{equation}

And, the marginal distribution is given by:

\begin{equation} X_i(t_{ij}) \sim N(\eta, Z_i(t_i)\Sigma_{w}Z'_i(t_i) + \sigma^2). \end{equation}

Ref: https://www.sciencedirect.com/science/article/pii/S002437951100320X


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