Particle filter for estimation of static parameters I am considering particle filtering methods for the estimation of static and dynamic parameters. For the static parameters $\theta$, Liu and West (page 7, equation 3.1) describe an "artificial" perturbation:
$$
\theta_{t+1} = \theta_t + \zeta_{t+1}
$$
$$
\zeta_{t+1} \sim N(0, W_{t+1})
$$
in order to generate a sample from $p(\theta_{t+1} | D_t)$ before updating the sample based on the likelihood of the new data at time $t+1$ to obtain a sample from $p(\theta_{t+1} | D_{t+1})$. Note that the notation $\theta_t$ is used to mean that it denote the time $t$ posterior of $\theta$ and not that $\theta$ is a time-varying parameter.
Now adding this artificial perturbation increases the variance of the sample and so the estimate of the true posterior is over-dispersed. They describe a method in which you may perturb the samples without increasing the sample variance. That is, so:
$$
V(\theta_{t+1} | D_t) = V(\theta_t | D_t) + W_{t+1} + 2C(\theta_t, \zeta_{t+1} | D_t) = V(\theta_t | D_t)
$$
which implies:
$$
C(\theta_t, \zeta_{t+1} | D) = -W_{t+1}/2
$$
(this is all from page 9 of Liu and West). They then state that approximating joint normality of $(\theta_t, \zeta_{t+1} | D_t)$ then implies:
$$
\theta_{t+1} | \theta_t \sim N(A_{t+1} \theta_t + (1-A_{t+1}) \bar{\theta_t}, (1-A_{t+1}^2)V_t)
$$
where:
$$
A_{t+1} = I - W_{t+1}V_t^{-1}/2
$$
and the idea is that if $p(\theta_t | D_t)$ has finite mean $\bar{\theta}_t$ and variance matrix $V_t$, then the mean and variance matrix of $p(\theta_{t+1} | D_t)$ is also $\bar{\theta}_t$ and $V_t$.
My question:
How do you derive this form of $\theta_{t+1} | \theta_t$? It makes sense but is not obvious (to me) how to get there, any advice whatsoever would be appreciated.
 A: Answering my own question based on the help from @Taylor. A very nice proof of the results used is here, from which I have taken some notation.
For a multivariate normal vector $\boldsymbol{Y} \sim N(\boldsymbol{\mu}, \boldsymbol{\Sigma})$, consider partitioning
$\boldsymbol{\mu}$ and $\boldsymbol{Y}$ into:
\begin{equation}
\begin{aligned}
\boldsymbol\mu =
\begin{bmatrix}
 \boldsymbol\mu_1 \\
 \boldsymbol\mu_2
\end{bmatrix}
\end{aligned}
\qquad \text{and} \qquad
\begin{aligned}
\boldsymbol{Y}=
\begin{bmatrix}
\boldsymbol{y}_1 \\ 
\boldsymbol{y}_2 
\end{bmatrix}
\end{aligned}
\end{equation}
with a similar partition of $\boldsymbol{\Sigma}$ into:
$$ 
\begin{bmatrix}
\boldsymbol{\Sigma}_{11} & \boldsymbol{\Sigma}_{12} \\
\boldsymbol{\Sigma}_{21} & \boldsymbol{\Sigma}_{22}
\end{bmatrix}.
$$
The conditional distribution of the first partition given the second is then:
$$ 
\boldsymbol{y}_1 | \boldsymbol{y}_2 \sim
N(\boldsymbol{\mu}_1 + \boldsymbol{\Sigma}_{12} \boldsymbol{\Sigma}_{22}^{-1} (\boldsymbol{y}_2 - \boldsymbol{\mu}_2), 
\boldsymbol{\Sigma}_{11} - \boldsymbol{\Sigma}_{12}{\boldsymbol{\Sigma}_{22}}^{-1} \boldsymbol{\Sigma}_{21}).
$$
In this problem:
$$
\begin{aligned}
C(\boldsymbol{\theta}_{t+1}, \boldsymbol{\theta}_t | D_t)
&= E(\boldsymbol{\theta}_{t+1} \boldsymbol{\theta}_t^\intercal | D_t) - E(\boldsymbol{\theta}_{t+1} | D_t)E(\boldsymbol{\theta}_t | D_t)^\intercal \\
&= E((\boldsymbol{\theta}_t + \boldsymbol{\zeta}_{t+1}) \boldsymbol{\theta}_t^\intercal | D_t) - \bar{\boldsymbol{\theta}_t}\bar{\boldsymbol{\theta}_t}^\intercal \\
&= E(\bar{\boldsymbol{\theta}_t}\bar{\boldsymbol{\theta}_t}^\intercal + \boldsymbol{\zeta}_{t+1} \boldsymbol{\theta}_t^\intercal | D_t) - \bar{\boldsymbol{\theta}_t}\bar{\boldsymbol{\theta}_t}^\intercal \\
&= E(\bar{\boldsymbol{\theta}_t}\bar{\boldsymbol{\theta}_t}^\intercal | D_t) + E(\boldsymbol{\zeta}_{t+1} \boldsymbol{\theta}_t^\intercal | D_t) - \bar{\boldsymbol{\theta}_t}\bar{\boldsymbol{\theta}_t}^\intercal \\
&= (\mathbf{V}_t + \bar{\boldsymbol{\theta}_t}\bar{\boldsymbol{\theta}_t}^\intercal ) + (-\frac{1}{2} \mathbf{W}_{t+1} + 0) - \bar{\boldsymbol{\theta}_t}\bar{\boldsymbol{\theta}_t}^\intercal \\
&= \mathbf{V}_t -\frac{1}{2} \mathbf{W}_{t+1} \\
\boldsymbol{\mu}_1 &= \boldsymbol{\mu}_2 = \bar{\boldsymbol{\theta}}_t \\
\boldsymbol{\Sigma}_{11} &= \boldsymbol{\Sigma}_{22} = \mathbf{V}_t \\
\boldsymbol{\Sigma}_{12} &= \boldsymbol{\Sigma}_{21} = \mathbf{V}_t -\frac{1}{2} \mathbf{W}_{t+1}
\end{aligned}
$$
and so the conditional normal evolution is:
$$ 
\boldsymbol{\theta}_{t+1} | \boldsymbol{\theta}_t \sim N(\bar{\boldsymbol{\theta}}_t + (\mathbf{V}_t -\frac{1}{2} \mathbf{W}_{t+1}) \mathbf{V}_t^{-1} (\boldsymbol{\theta}_t -\bar{\boldsymbol{\theta}}_t), 
\mathbf{V}_t - (\mathbf{V}_t -\frac{1}{2} \mathbf{W}_{t+1}) \mathbf{V}_t^{-1} (\mathbf{V}_t -\frac{1}{2} \mathbf{W}_{t+1}))
$$
which simplifies to:
$$
\boldsymbol{\theta}_{t+1} | \boldsymbol{\theta}_t \sim N(\mathbf{A}_t \boldsymbol{\theta}_t + (\mathbf{I} - \mathbf{A}_t) \bar{\boldsymbol{\theta}}_t, 
(\mathbf{I} - \mathbf{A}_t^2) \mathbf{V}_t)
$$
where:
$$ 
\mathbf{A}_t = \mathbf{I} - \frac{1}{2} \mathbf{W}_{t+1} \mathbf{V}_t^{-1}.
$$
A: This follows from a property of multivariate normal random vectors. Here's a link:
https://en.wikipedia.org/wiki/Multivariate_normal_distribution#Conditional_distributions
