2
$\begingroup$

I attempt to fit the following probit model to a time series where we observe the binary variable $R_{t}$ and another variable $X_{t}$, a latent unobserved variable $y^{*}_{t}$ and a state variable $s_{t}$: $$ y^{*}_{t} = -c_{0t} -c_{1t}\,X_{t-k} + u_{t} $$ where u_{t} is (0,1) normally distributed and $R_{t} = 1$ if $y^{*}_{t} \le 0$ and $R_{t} = 0$ otherwise. $c_{it}$ depends on the current state $s_{t}$ which can take either 0 or 1 values, such that $$ P(R_{t} = 1 | s_{t}, X_{t-k}) = N(c_{0t} +c_{1t}\,X_{t-k}) $$ See for example this paper page 47. Now my problem is that I do not find any derivation of how to estimate the parameters. It should be straightforward and easy to solve, but I do not find anything related to probit but only (vector) autoregressions with Markov switching. Any ideas?

$\endgroup$
1
$\begingroup$

Estimating a hidden Markov model is not straightforward. Look for "particle filters" and "SMC methods" within the Bayesian literature.

This is a probit model with hidden Markov dependence: the distribution of $R_t$ conditional on $X_{1:t}$ is Bernoulli with probability $$\mathbb{P}(R_t=1|X_{1:t})= 1-\mathbb{P}(R_t=0|X_{1:t})=p(X_{1:t})=\Phi(c_{0t}+c_{1t}X_{t-k})$$

What you omitted in the definition of your model is the dependence on the hidden state $s_t$: it should read $$y^{*}_{t} = -c_{0t}(s_t) -c_{1t}(s_t)\,X_{t-k} + u_{t}$$resulting into the probit model$$\mathbb{P}(R_t=1|X_{1:t},s_t)= 1-\mathbb{P}(R_t=0|X_{1:t},s_t)=p(X_{1:t},s_t)=\Phi(c_{0t}(s_t)+c_{1t}(s_t)X_{t-k})$$ Once again, this is far from a trivial problem.

$\endgroup$
  • $\begingroup$ My problem is that I need to somehow back out unconditional transition probabilities for each point in time. The paper just says to choose as transition value which is "appropriate" though. $\endgroup$ – user3673486 Apr 26 '15 at 14:26

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.