Original model you mentioned in the comments to the other answer:
$$
Y_{t+1} = Y_t + \mu \Delta + \sqrt{v_t \Delta} \epsilon_{t+1}^y + \xi_{t+1}^y N_{t+1}^y \\
v_{t+1} = v_t + \kappa(\theta - v_t)\Delta + \sigma_v \sqrt{v_t\Delta}\epsilon_{t+1}^v
$$
with $\text{corr}(\epsilon_{t+1}^y,\epsilon_{t+1}^v) = \rho$. The reference you linked to is also linked to at the end of this.
1
Let's call $\epsilon_{t+1}^y = e^1_{t+1}$ and $\epsilon_{t+1}^v = \rho e^1_{t+1} + \sqrt{(1-\rho^2)}e^2_{t+1}$ with $e^1_{t+1}$ and $e^2_{t+1}$ independent standard normals. Making the substitutions we get
$$
Y_{t+1} - Y_t = \mu \Delta + \sqrt{v_t \Delta} e^1_{t+1} + \xi_{t+1}^y N_{t+1}^y \\
v_{t+1} = v_t + \kappa(\theta - v_t)\Delta + \sigma_v \sqrt{v_t\Delta}\left[\rho e^1_{t+1} + \sqrt{(1-\rho^2)}e^2_{t+1} \right]
$$
Then let $\phi = \sigma_v \rho$ and $w_v = \sigma^2_v(1-\rho^2)$. They say to make this transformation on page 33.
$$
Y_{t+1} - Y_t = \mu \Delta + \sqrt{v_t \Delta} e^1_{t+1} + \xi_{t+1}^y N_{t+1}^y \\
v_{t+1} = v_t + \kappa(\theta - v_t)\Delta + \phi\sqrt{v_t\Delta} e^1_{t+1} + \sqrt{v_t\Delta}\sqrt{w_v}e^2_{t+1}
$$
2
They menton that $\Theta = \{\mu, \kappa, \theta, \sigma_v, \rho, \lambda_y, \mu_y, \sigma_y\}$. After the transformation it's actually $\Theta = \{\mu, \kappa, \theta, \phi, w_v, \lambda_y, \mu_y, \sigma_y\}$ for us now. They also describe posteriors for the following (and these must be part of the state vector at some time): $\xi^y_{t+1}$ $N_{t+1}^y$, $v_{t+1}$.
So we could define the state vector
$$
x_t = [v_{t+1}, v_t, \xi^y_{t+1} N_{t+1}^y]',
$$
and this would represent a state space model closer to what the other answer was talking about. But there are probably a lot of ways to do this. At the moment I can't tell if this paper does it that way.
3
Anyway, back to your question...I'm not sure why you relabelled everything becuase that makes it way harder to follow along, but you said in the comment that you're trying to get at the 'conditional posterior of $v_{t+1}$.' If you mean $p(v_{t+1}|y_{1:T}, \Theta)$, then that's a marginal of the smoothing distributon $p(x_{t+1}|y_{1:T}, \Theta)$ that the other answer was talking about.
On the other hand, if you were trying to sample from $p(x_{t}|y_{1:T},x_{1:t-1},x_{t+1:T})$ then
\begin{align*}
p(x_{t}|y_{1:T},x_{1:t-1},x_{t+1:T}) &\propto \prod_{t=2}^T p(y_t|x_t)p(x_t|x_{t-1})p(y_1|x_1)p(x_1) \\
&\propto p(x_t|x_{t-1})p(y_t|x_t)p(x_{t+1}|x_t) \\
&\propto p(x_t|x_{t-1},x_{t+1},y_t)
\end{align*}
which the other answer also mentioned. I think this is called a "single-site sampler," perhaps useful if you wanted to get at $p(x_{1:T}|y_{1:T},\Theta)$ Gibbs-style. I am guessing that this is what you want, actually. You would get this if you used the state vector in part 2, and used the log returns $Y_{t+1} -Y_t$ as the observations.
So I'm kind of echoing the other answer here: it's probably one of those two things. Hope this helps.
Reference: http://lib.dr.iastate.edu/cgi/viewcontent.cgi?article=1121&context=stat_las_preprints
