0
$\begingroup$

This is a Gaussian model of spatial correlation: \begin{align} \boldsymbol y &\sim N(\mu, \boldsymbol V), \text{ where} \nonumber \\ \mu_i &= \ldots \text{ (depends on fixed effects of the model)}, \text{ and} \nonumber \\ V_{i_1, i_2} &= \begin{cases} 0, & \text{if } \texttt{individual}_{i_1} \neq \texttt{individual}_{i_2} \text{ and } i_1 \neq i_2 \\ \nu^2 + \tau^2 \exp\left\{\frac{-(\texttt{t}_{i_1} - \texttt{t}_{i_2})^2}{\rho^2}\right\}, & \text{if } \texttt{individual}_{i_1} = \texttt{individual}_{i_2} \text{ and } i_1 \neq i_2 \\ \nu^2 + \tau^2 + \sigma^2, & \text{if } i_1 = i_2 \\ \end{cases} \end{align} The matrix $\boldsymbol V$ is the covariance matrix for the observations $\boldsymbol y$, so $\boldsymbol V = \text{Cov}(\boldsymbol y)$.

In this structure, two observations from different individuals are independent, two observations "very close" together have covariance $\nu^2 + \tau^2$ and two observations "very far" apart have covariance $\nu^2$ (how fast this decline in covariance from $\nu^2 + \tau^2$ to $\nu^2$ occur depends on the parameter $\rho$).

The total variance of a single observation is $\nu^2 + \tau^2 + \sigma^2$.

Note that its covariance structure is an extension of the compound symmetry structure: \begin{align} \boldsymbol y &\sim N(\mu, \boldsymbol V), \text{ where} \nonumber \\ \mu_i &= \ldots \text{ (depends on fixed effects of the model)}, \text{ and} \nonumber \\ V_{i_1, i_2} &= \begin{cases} 0, & \text{if } \texttt{individual}_{i_1} \neq \texttt{individual}_{i_2} \text{ and } i_1 \neq i_2 \\ \sigma^2_\texttt{individual}, & \text{if } \texttt{individual}_{i_1} = \texttt{individual}_{i_2} \text{ and } i_1 \neq i_2 \\ \sigma^2_\texttt{individual} + \sigma^2, & \text{if } i_1 = i_2 \\ \end{cases} \end{align} Here, we only have one random intercept. The “individual” factor in this model is considered a random effect allowing two observations from the same individual to be positively correlated.

My questions are:

  1. What would the spatial Gaussian correlation look like for two nested random intercepts?

  2. And how about the exponential correlation structure with two nested random intercepts?

The exponential correlation structure has correlation term $\tau^2 \exp\left\{\frac{-|\texttt{t}_{i_1} - \texttt{t}_{i_2}|}{\rho}\right\}$.

$\endgroup$
  • $\begingroup$ Not clear enough. Dimension of $V$ = ?. $\endgroup$ – user158565 Dec 15 '18 at 2:27
  • $\begingroup$ The matrix $V$ is the covariance matrix for the observations $y$, so $V = \text{Cov}(y)$. $\endgroup$ – user3419936 Dec 15 '18 at 8:40
  • $\begingroup$ I tried to add some more details. I hope it is more clear now. $\endgroup$ – user3419936 Dec 15 '18 at 9:07
2
$\begingroup$

As far as I can see, the model you are looking for is

$$\left\{ \begin{array}{l} Y_{ijk} = \mu_{ijk} + b_i + b_{ij} + \varepsilon_{ijk},\\\\ b_i \sim \mathcal N(0, \tau^2), \quad b_{ij} \sim \mathcal N( 0, \phi^2),\\\\ \varepsilon_{ij} \sim \mathcal N(0, \sigma^2h_{k, k’}), \end{array} \right.$$

where $Y_{ijk}$ denotes the $k$-th measurement in the $j$-th level of the outcome variable for the $i$-th subject, $\mu_{ijk}$ is the corresponding mean, $b_i$ is the random effect for the $i$-th subject, $b_{ij}$ is the random effect for the $j$-th level for subject $i$, and the error terms $\varepsilon_{ij}$ have a serial correlation structure defined by the $h_{k,k’}$ function (e.g., Gaussian or exponential or another one).

If you integrate out the random effects, the elements of the marginal covariance matrix will look as follows:

$$\mbox{var}(Y_{ijk}) = \tau^2 + \phi^2 + \sigma^2,$$ $$\mbox{cov}(Y_{ijk}, Y_{ijk'}) = \tau^2 + \phi^2 + \sigma^2h_{k,k’}, \quad \mbox{for } k' \neq k,$$ $$\mbox{cov}(Y_{ijk}, Y_{ij'k}) = \tau^2, \quad \mbox{for } j' \neq j,$$ $$\mbox{cov}(Y_{ijk}, Y_{ij'k'}) = \tau^2 + \sigma^2h_{k,k’}, \quad \mbox{for } j' \neq j, \; \mbox{and} \; k \neq k'.$$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.