# 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.

For a multivariate normal vector $$\boldsymbol{Y} \sim N(\boldsymbol{\mu}, \boldsymbol{\Sigma})$$, consider partitioning $$\boldsymbol{\mu}$$ and $$\boldsymbol{Y}$$ into: \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} 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: $$\mathbf{A}_t = \mathbf{I} - \frac{1}{2} \mathbf{W}_{t+1} \mathbf{V}_t^{-1}.$$