## 7.4 Four subpopulations with temporally uncorrelated errors

The model for one well-mixed population was not very good. Another reasonable assumption is that the different census regions are measuring four different temporally independent subpopulations. We write a model of the log subpopulation abundances for this case as: $$$\begin{gathered} \begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix}_t = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix}_{t-1} + \begin{bmatrix}u\\u\\u\\u\end{bmatrix} + \begin{bmatrix}w_1\\w_2\\w_3\\w_4\end{bmatrix}_t \\ \textrm{ where } \mathbf{w}_t \sim \,\text{MVN}\begin{pmatrix}0, \begin{bmatrix} q & 0 & 0 & 0 \\ 0 & q & 0 & 0\\ 0 & 0 & q & 0 \\ 0 & 0 & 0 & q \end{bmatrix}\end{pmatrix}\\ \begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix}_0 = \begin{bmatrix}\mu_1\\\mu_2\\\mu_3\\\mu_4\end{bmatrix}_t \end{gathered} \tag{7.7}$$$ The $$\mathbf{Q}$$ matrix is diagonal with one variance value. This means that the process variance (variance in year-to-year population growth rates) is independent (good and bad years are not correlated) but the level of variability is the same across regions. We made the $$\mathbf{u}$$ matrix with one $$u$$ value. This means that we assume the population growth rates are the same across regions.

Notice that we set the $$\mathbf{B}$$ matrix equal to a diagonal matrix with 1 on the diagonal. This is the “identity” matrix and it is like a 1 but for matrices. We do not need $$\mathbf{B}$$ for our model, but MARSS() requires a value.

### 7.4.1 The observation process

In this model, each survey is an observation of a different $$x$$: $$$\left[ \begin{array}{c} y_{1} \\ y_{2} \\ y_{3} \\ y_{4} \end{array} \right]_t = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix}_t + \left[ \begin{array}{c} 0 \\ 0 \\ 0 \\ 0 \end{array} \right] + \left[ \begin{array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \end{array} \right]_t \tag{7.8}$$$ No $$a$$’s can be estimated since we do not have multiple observations of a given $$x$$ time series. Our $$\mathbf{R}$$ matrix doesn’t change; the observation errors are still assumed to the independent with different variances.

Notice that our $$\mathbf{Z}$$ matrix changed. $$\mathbf{Z}$$ is specifying which $$y_i$$ goes to which $$x_j$$. The one we have specified means that $$y_1$$ is observing $$x_1$$, $$y_2$$ observes $$x_2$$, etc. We could have set up $$\mathbf{Z}$$ like so $$$\begin{bmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{bmatrix}$$$

This would mean that $$y_1$$ observes $$x_2$$, $$y_2$$ observes $$x_1$$, $$y_3$$ observes $$x_4$$, and $$y_4$$ observes $$x_3$$. Which $$x$$ goes to which $$y$$ is arbitrary; we need to make sure it is one-to-one. We will stay with $$\mathbf{Z}$$ as an identity matrix since $$y_i$$ observing $$x_i$$ makes it easier to remember which $$x$$ goes with which $$y$$.

### 7.4.2 Fitting the model

We set up the model list for MARSS() as:

mod.list.1 <- list(B = "identity", U = "equal", Q = "diagonal and equal",
Z = "identity", A = "scaling", R = "diagonal and unequal",
x0 = "unequal", tinitx = 0)

We introduced a few more short-cuts. "equal" means all the values in the matrix are the same. "diagonal and equal" means that the matrix is diagonal with one value on the diagonal. "unequal" means that all values in the matrix are different.

We can then fit our model for 4 subpopulations as:

fit.1 <- MARSS::MARSS(dat, model = mod.list.1)