# Covariance matrix of CAR model

Does anyone know how to obtain the covariance matrix of the conditional autoregressive (CAR) model?

The general idea is that a random variable $Y_{i}$ may depend on other values $Y_{j}$ where $j \neq i$. $Y_{i}$ can be a measurement at location $i$ that is correlated with measurement $Y_{j}$ at location $j$. In a CAR model, these variables follow the conditional mean:

$$E(Y_{i}|\text{all }Y_{j\neq i}) = \mu_{i} + \rho\Sigma_{j\neq i}w_{ij}(Y_{j}-\mu_{j})$$

where $\mu_{i}$ is the mean of $Y_{i}$, $w_{ij}$ are weights that determine the influence of location $j$ on $i$ and $\rho$ describes a positive or negative direction. We also know that the conditional variance is:

$$Var(Y_{i}|\text{all }Y_{j\neq i})=\sigma^{2}$$

I want to know how to derive the covariance matrix:

$$V = (I - \rho W)^{-1}M$$

in which $M$ is a $n\times n$ matrix of conditional variances $(\sigma_{1}^{2},\dots, \sigma_{n}^{2})$ and $W$ is a $n\times n$ matrix of weights $w_{ij}$ with zeros on the diagonal.

I'm guessing it should be a somewhat straightforward calculation. Mainly, $$\Sigma_{ij} = cov(Y_{i},Y_{j}) = E[(Y_{i}-E[Y_{i}])(Y_{j}-E[Y_{j}])]$$

However, I can't see how to obtain the first term from this operation. I think and correct me if I'm wrong, $E(Y_{i},Y_{j})$ under this model should be zero. That should simplify the derivation.

This is the description I have been reading, in case it may be helpful. The original paper by Besag assumes the existence of such a covariance matrix, but as far as I understand, he doesn't derive it.

Any help would be appreciated.

UPDATE:

There is a very simple way to prove the covariance matrix.

If we use the vector of all $Y_{i}$ as $Y$ and define the CAR model in terms of $Y$:

$$Y = \rho W Y + M$$

(assuming $\mu_{i}=0$ and with a diagonal $n\times n$ matrix sharing the same $\sigma^{2}$)

Then, $$Y = (I - \rho W)^{-1}M$$ and it is quite obvious that a distribution $Y = N(0, (I - \rho W)^{-1}M)$ gives the desired result. However, I'm not quite satisfied with this result since I had to redefine the CAR model, which no longer has the expected value as above.

The answer was a bit more complicated than I thought. In order to prove the covariance matrix of the CAR model, we have to use Brook's lemma to write the joint probability $p(Y)$ using the conditional distributions $$p(Y_{i}|\text{all }Y_{j\neq i})=(2\tau_{i}^{2})^{-1/2}\exp{\displaystyle(\frac{1}{2\tau_{i}^{2}}[y_{i}-\mu_{i}-\rho\sum_{j=1}^{n}w_{ij}(y_j-\mu_{j})])}$$
In Brooke's lemma, we choose $\mu_{j}$ as the point usually written as $y_{j0}$ in:
$$p(Y) = C\prod_{i}^{n}\frac{p_{i}(y_{i}|Y_{j}=y_{j0}, j< i; Y_{j}=y_{j},j>i)}{{p_{i0}(y_{i}|Y_{j}=y_{j0}, j<i; Y_{j}=y_{j},j>i)}}$$
The result of this product is compared to the exponent of the multivariate normal with covariance $(I - \rho W)^{-1}M$.