Multiplying Gaussian distributions of different dimensions The multiplication of multivariate Gaussian distributions defined over some parameter vector of a given dimension can be achieved by the following. Assuming that the Gaussian is parametrized by the precision matrix ($\Lambda$) and mean ($\mu$). We can compute the new precision as:
$$
\Lambda = \Lambda_1 + \Lambda_2 
$$
Also, the new mean can be computed using the identity:
$$
\Lambda \mu = \Lambda_1 \mu_1 + \Lambda_2 \mu_2
$$
So, this is very convenient. However, I have a situation where I need to perform such a multiplication where one of the Gaussian distribution depends on $n$ number of variables. However, the second distribution depends on a subset of these $n$ variables. So, to perform this multiplication, I need to somehow make the second distribution also depend on all the $n$ variables, though in reality only a subset of these $n$ variables are affecting the underlying distribution. 
The exact situation is that I have a posterior distribution $q(\theta)$ where $\theta$ is n-dimensional and I have the likelihood of the following form:
$$
L(\theta) = \prod_{i=1}^{k}L(\theta_i)
$$
Here, $L(\theta_i)$ is a subset of $n$ variables and $L(\theta_i)$ is independent of $L(\theta_j)$ for $i \neq j$. I need to multiply $q(\theta)$ with a $L(\theta_i)$. So I somehow need to express this $L(\theta_i)$ as a function of $\theta$. Each $L(\theta_i)$ is a 3-dimensional normal distribution i.e. $\theta_i$ is 3-dimensional.
Does anyone know how this can be done?
 A: You might want to add into the question the comment which says that your likelihood $L(\theta) = \prod_{i=1}^n L(\theta_i)$ where each $\theta_i$ is 3-dimensional and $L(\theta_i)$ is 3D Gaussian.
So, if I understood correctly, your observation model is 
$y_i|\theta\sim N(\theta_i, \Sigma_i)$, where $y_i,\theta_i\in R^3$ and $\Sigma_i\in R^{3\times3}$ are, and $y_i$:s are conditionally independent. This can be equivalently written as a $3n$-dimensional normal distribution
\begin{equation}
y \sim N(\theta,\Sigma)
\end{equation}
where $y,\theta$ are 3n-dimensional vectors (just the $y_i$s and $\theta_i$s stacked together), and the joint covariance matrix $\Sigma$ is a block diagonal matrix where the elements corresponding to components belonging to the same $i$ are taken from the $\Sigma_i$s, and other elements of $\Sigma$ are 0.
To prove this, note that any linear combination of $y$ is a sum of linear combinations of $y_i$s, i.e., a sum of independent Gaussians, thus Gaussian. By definition, this implies that $y$ is $3n$-dimensional multivariate Gaussian. Then, to deduce the parameters, note that the means of $y_{k,i}$, and covariances of $y_{k,i},y_{l,i}$ are not impacted by taking the other variables into account, and based on the independence, covariances $Cov(y_{k,i},y_{l,j})$ must be zero for $i\neq j$.
A: ok, I am slightly confused. So, imagine for simplicity we have 4 parameters divided into 2 bivariate Gaussians. So, imagine the precision matrix for the first likelihood term is a 2x2 matrix with the terms 
$$
\begin{pmatrix}
  a_{1,1} & a_{1,2} \\
  a_{2,1} & a_{2,2}
\end{pmatrix}
$$
Similarly, the precision matrix for the second likelihood term is
$$
\begin{pmatrix}
  b_{1,1} & b_{1,2} \\
  b_{2,1} & b_{2,2}
\end{pmatrix}
$$
Now, are you saying that if I want to represent either of the likelihood term as a 4-dimensional Gaussian, they will have the precision matrix as:
$$
\begin{pmatrix}
  a_{1,1} & a_{1,2} & 0 & 0 \\
  a_{2,1} & a_{2,2} & 0 & 0 \\
  0 & 0 & b_{1,1} & b_{1,2} \\
  0 & 0 & b_{2,1} & b_{2,2}
\end{pmatrix}
$$
So I have the same precision matrix regardless of whether I want to represent likelihood term 1 or likelihood term 2. However, I am confused as to how both the terms can have the same precision matrix? Because if I just multiply this by itself, all the precision components which are non zero would change, even though it is equivalent to either multiplying by just one of the likelihood term.
A: Ok, it seems it is very simple. The precision matrix can be set to zeros for rows (i.e. uniform distribution) which do not affect the given likelihood term. Similarly, the $\Lambda \mu$ variable will have zero entries for variables not affecting the likelihood term.
