In Kim et al. (1998) stochastic volatility model is specified as:

$y_t=\beta\exp({\frac{h_t}{2}})\varepsilon_t,\quad t\geqslant1$


$h_1\sim N(\mu,\frac{\sigma^2}{1-\phi^2})$

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?

  • $\begingroup$ Let me know if you are not able to follow the answer below and I can make it a little more detailed. $\endgroup$ – tchakravarty Oct 19 '14 at 18:21

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 $$ \begin{align} \mathbb{E}(h_t\mid h_{t-1}, h_{t+1}) &= \mu + \frac{\phi}{1+\phi^2}((h_{t-1} - \mu) + (h_{t+1} - \mu)) \\ &\equiv h_t^* \\ \mathbb{V}(h_t\mid h_{t-1}, h_{t+1}) &= \frac{\sigma_\eta^2}{1+\phi^2}\\ &\equiv \nu^2 \end{align} $$


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