For a Markov process given by $$x_t = \mu +\kappa(x_{t-1} - \mu) + \sigma \cdot \varepsilon_t $$ where $\varepsilon_t \sim N(0,1)$ and $\mu$, $\kappa$, $\sigma^2$ are the parameters, how would I find the conditional distribution $p\left(x_t \vert x_{t-1}, x_{t+1},\kappa ,\mu,\sigma^2\right)$?
The paper I'm reading says it is given by: $$N\left(\mu + \kappa[(x_{t-1}-\mu)+(x_{t+1}-\mu)] / [1+\kappa^2], \quad\sigma^2/(1-\kappa^2)\right)$$ however I can't work out how this is derived.
Many thanks in advance!
First notice that: $$ \begin{align} x_{t+1} &= \mu+\kappa(x_t-\mu)+\sigma \epsilon_{t+1}\\ & = \mu+\kappa( \mu+\kappa(x_{t-1}-\mu)+\sigma \epsilon_t -\mu)+\sigma \epsilon_{t+1}\\ & = \mu+\kappa( \kappa(x_{t-1}-\mu)+\sigma \epsilon_t)+\sigma \epsilon_{t+1}\\ & = \mu+\kappa( \kappa(x_{t-1}-\mu))+ \kappa \sigma \epsilon_t +\sigma \epsilon_{t+1} \end{align} $$
Using conditional expectation: $$ \begin{align} \mathbb{E}(x_t | x_{t-1}, x_{t+1}) & = \mathbb{E}( \mu+\kappa(x_{t-1}-\mu)+\sigma \epsilon_{t} | x_{t-1}, x_{t+1})\\ & = \mu+\kappa(x_{t-1}-\mu)+\mathbb{E}(\sigma\epsilon_{t} | x_{t-1}, x_{t+1})\\ \end{align} $$
I imagine that the noises $\epsilon_t$ are all independent, right? Then: $$ \begin{align} \mathbb{E}(x_t | x_{t-1}, x_{t+1}) & = \mu+\kappa(x_{t-1}-\mu)+\mathbb{E}(\sigma\epsilon_{t} | \kappa \sigma \epsilon_t +\sigma \epsilon_{t+1} = x_{t+1} - (\mu+\kappa( \kappa(x_{t-1}-\mu))) )\\ \end{align} $$
Now all you need to do is to understand $\mathbb{E}( X | \kappa X+Y )$ where $X,Y \sim N(0,\sigma^2)$ with $X,Y$ independent. But the sum of independent gaussians is a gaussian and you know that: $$ f_{X|X+Y}(x|y) = \frac{f_{X,Y}(x,y)}{f_X(x)} $$
Can you continue from here?