Suppose I have jointly normal random vectors $[\bf{v_1}, \ldots, \bf{v_K}]$' with mean $ \bf{M}$ and joint block tridiagonal precision matrix $ \bf{P}$:
$$ \bf{M}= \begin{bmatrix}\bf{\mu_1} \\\ldots \\ \bf{\mu_K} \end{bmatrix} $$
$$ \bf{P}= \begin{bmatrix}\bf{P}_{11} & \bf{P}_{12} & \bf{0} & \bf{0}\\ \bf{P}_{21} & \bf{P}_{22} & \ddots & \bf{0} \\\bf{0} & \ddots & \ddots & \bf{P}_{KK-1} \\ \bf{0} & \bf{0} & \bf{P}_{K-1K} & \bf{P}_{KK} \\\\ \end{bmatrix} $$
$\bf{v_1}, \ldots,\bf{v_K}$ are all of dimension $N \times 1$. I want to derive the generic conditional distribution $ \bf{v_n}|\bf{v_{n-1}}$ for $1<n<K$ as a function of the elements of $\bf{M}$ and of the elements of the precision matrix $\bf{P}$.
I am following the approach outlined here link (equations 2.73 2.75), that in a two random vectors set up show that the conditional mean and the conditional variance can be written as a function of the elements of the vector of means and the precision matrix.
I am proceeding as follows:
First, I marginalize all the $\bf{v}_i$'s for $i = 1,\ldots,n-2$ and for $i = n+1,\ldots,K$. Then I derive the mean and the variance of $\bf{v_n}|\bf{v_{n-1}}$ as the article suggests. Using this approach I am struggling in finding a general formula for the conditional mean and conditional variance as a function of the elements of the precision matrix.
I think that the fact that this precision matrix implies a Markov structure such that conditionally on its predecessor each random vector is independent of the other random vectors preceding it should help me, but still the fact that I have to marginalize all the other random vectors gives me other extra terms.
Is it the way to procede?