2
$\begingroup$

I am currently working on the MS GARCH model proposed by Haas paper given below

$\epsilon_n=Z_n \sigma_{\Delta_n,n}$ where $\epsilon_n$ is a time-series of residuals, $\left\{Z_n ,n\in \mathbb{Z}\right\}$ is a sequence of normal independent and identically distributed random variables with mean zero and unit variance, and $\left\{ \Delta_n,n\in \mathbb{Z}\right\}$ is a Markov chain.

The $k \times 1$ vector $\boldsymbol{\sigma}_n^2=[\sigma_{1n}^2,\sigma_{2n}^2,\ldots,\sigma_{kn}^2]'$ of regime variances follows the multivariate GARCH(1,1) equation \begin{equation} \label{2.2} \boldsymbol{\sigma}_n^2=\boldsymbol{\alpha_0}+\boldsymbol{\alpha_1}\epsilon_{n-1}^2+\boldsymbol{\beta}\boldsymbol{\sigma}_{n-1}^2\,, \end{equation} where $\boldsymbol{\alpha_i}=[\alpha_{i1},\alpha_{i2},\ldots,\alpha_{ik}]'$, $i=0,1$; $\boldsymbol{\beta}=diag(\beta_1,\beta_2,\ldots,\beta_k)$.

Now I am trying to link the underlying latent Markov chain to the theory of hiddem markov models. In the theory of HMM I am finding that the observations sequence should be independent. In my case the observation sequence is the residuals $\epsilon_n$ which even though they are uncorrelated, they are not independent.

Can this assumption of independence be eliminated in some way? Or am I making some mistake by joining HMM and MS GARCH?

$\endgroup$
  • $\begingroup$ you haven't described what $\sigma_{\Delta_n,n}$ is $\endgroup$ – Taylor Mar 21 '18 at 21:37
  • $\begingroup$ Yes, it is the square root of one of the elements of the vector $\boldsymbol{\sigma}_n$. We have k elements which can take the value of $\sigma_{\Delta_n,n}$, and it is chosen from these k elements depending on which regime we are in at time $n$. $\endgroup$ – Anna Mar 21 '18 at 21:43
  • $\begingroup$ okay, I understand now $\endgroup$ – Taylor Mar 21 '18 at 21:46
2
$\begingroup$

In the theory of HMM I am finding that the observations sequence should be independent.

Typically for HMMs, conditioning on the states, the observations are conditionally independent. In the case of your model, it is not true, as $$ p(\epsilon_1, \ldots, \epsilon_n \mid \Delta_1, \ldots, \Delta_n) \neq p(\epsilon_1 \mid \Delta_1) \times \cdots \times p(\epsilon_n \mid \Delta_n). $$

For your model, the distribution of $\epsilon_t$ depends on the previous values, as well as the contemporaneous hidden state. So this isn't a typical example of a HMM, although I wouldn't object to it being called that.

Can this assumption of independence be eliminated in some way?

If you changed the GARCH part of your model to an ARCH part, and if you treated the $\sigma^2_t$ process as being part of the state, there would be conditional independence. The $k+1 \times 1$ state vector could be $$ \left[ \begin{array}{c} \Delta_t \\ \boldsymbol{\sigma}_t^2 \end{array}\right]. $$ This would give you a latent first-order Markov process, although its state space is not finite, which is another typical assumption of what are called Hidden Markov Models. Also note that the $\boldsymbol{\sigma}_t^2$ is totally deterministic, which isn't typical either, for these sorts of models.

$\endgroup$
  • $\begingroup$ I do not understand why removing the GARCH part helps, since actually there is still the dependence on $\epsilon_{n-1}$ which depends on the regime at time $n-1$. and what do you mean treat the $\sigma_t^2$ as being part of the state? $\endgroup$ – Anna Mar 22 '18 at 6:34
  • $\begingroup$ @Anna see my edits. Overall I'd say if you're interested in specifically this model, it won't help much in the short run to go exploring HMMs thinking they're somehow related. $\endgroup$ – Taylor Mar 22 '18 at 15:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.