Stochastic Volatility Model In Kim et al. (1998) stochastic volatility model is specified as:
$y_t=\beta\exp({\frac{h_t}{2}})\varepsilon_t,\quad t\geqslant1$ 
$h_{t+1}=\mu+\phi(h_t-\mu)+\sigma_\eta\eta_t$
$h_1\sim N(\mu,\frac{\sigma^2}{1-\phi^2})$
and $\varepsilon_t, \eta_t$ are independent standard normal.
Later on they write (page 365) that it can be shown that
$f(h_t|h_{\backslash t},\theta)=f(h_t|h_{t+1}, h_{t-1}, \theta)=f_N(h_t|h^*_t, \nu^2)$
where $h^*_t=\mu+\frac{\phi( (h_{t-1}-\mu)+(h_{t+1}-\mu) )}{1+\phi^2},\quad\nu^2=\frac{\sigma^2_\eta}{1+\phi^2}$
Also, $\theta=(\mu,\sigma^2,\phi)$, $f_N$ - normal pdf, $h_{\backslash t}$ - values of the volatility vector without the value at $t$
I do not understand how they derive this $f_N(h_t|h^*_t, \nu^2)$. Somehow they play around with AR(1) equations for $h_{t+1}$ and $h_t$, but how? They plug what into what?
 A: Some fairly simple algebra will get you there.
From the forward equation, you have:
$$
\begin{align}
h_{t+1} &= \mu + \phi(h_t - \mu)  + \sigma_\eta \eta_t \\
\phi^2(h_t - \mu) &= \phi(h_{t+1} - \mu) -  \phi\sigma_\eta \eta_t \\
\end{align}
$$
Similarly, from the backward equation, you have:
$$
\begin{align}
h_t - \mu &=  \phi(h_{t-1} - \mu)  + \sigma_\eta \eta_{t-1} \\
\end{align}
$$
Adding the two expressions together gives
$$
\begin{align}
(1+\phi^2)(h_t - \mu) &= \phi((h_{t-1} - \mu) + (h_{t+1} - \mu)) + \sigma_\eta (\eta_{t-1}-  \phi \eta_t) \\
(h_t - \mu) &= \frac{\phi}{1+\phi^2}((h_{t-1} - \mu) + (h_{t+1} - \mu)) + \frac{\sigma_\eta}{1+\phi^2} (\eta_{t-1}-  \phi \eta_t) \\
h_t &= \mu + \frac{\phi}{1+\phi^2}((h_{t-1} - \mu) + (h_{t+1} - \mu)) + \frac{\sigma_\eta}{1+\phi^2} (\eta_{t-1}-  \phi \eta_t)
\end{align}
$$
It is easy to see that if the $\{\eta_t\}$ are standard normal, then $h_t$ is (conditionally on $h_{t-1}$ and $h_{t+1}$) normally distributed, and
$$
\begin{align}
\mathbb{E}(h_t\mid h_{t-1}, h_{t+1}) &= \mu + \frac{\phi}{1+\phi^2}((h_{t-1} - \mu) + (h_{t+1} - \mu)) \\
&\equiv h_t^* \\
\mathbb{V}(h_t\mid h_{t-1}, h_{t+1}) &= \frac{\sigma_\eta^2}{1+\phi^2}\\
&\equiv \nu^2
\end{align}
$$
