I am trying to find log likelihood function for ar(1)-garch(1) model

I tried below links for solution:

Maximum likelihood in the GJR-GARCH(1,1) model

How to derive the conditional likelihood for a AR-GARCH model?

I understood that in this case we have to put value of mu=0zero +0one*rt-1 and value of variance as sigma^2

After that I don't know how to proceed?

Could anyone help me out?

Thank You


1 Answer 1


Suppose $r_t=\phi_0+\phi_1r_{t-1}+a_t$



where {$\epsilon_t$} is Gaussian white noise series with mean 0 and variance 1. Let's $\theta = (\phi_0,\phi_1,\alpha_0,\alpha_1,\beta_1)$. We have

$$r_t\mid r_{t-1},\sigma_{t-1},a_{t-1},\theta \sim \mathcal{N}(\phi_0+\phi_1r_{t-1}, \alpha_0+\alpha_1a_{t-1}^2+\beta_1\sigma_{t-1}^2)\,.$$ Then $$\mathcal{L} = p(r_1,r_2,\dots,r_T\mid\theta) = p(r_1\mid\theta)p(r_2\mid r_1,\sigma_1,a_1,\theta)p(r_3\mid r_2,\sigma_2,a_2,\theta)\dots p(r_T\mid r_{T-1},\sigma_{T-1},a_{T-1},\theta)$$

You can then compute the log likelihood recursively by supposing $r_1\sim\mathcal{N}(\frac{\phi_0}{1-\phi_1},\frac{\alpha_0}{1-\alpha_1-\beta_1})$.

Those mean and variance are obtained as follows :

Suppose the mean of $r_t$ is constant : $\mu = \mathbb{E}[r_t]$ then $$\mu = \mathbb{E}[r_t] = \phi_0 + \phi_1\mathbb{E}[r_{t-1}] + \mathbb{E}[a_t] = \phi_0+\phi_1\mu\,.$$ So $\mu = \frac{\phi_0}{1-\phi_1}$.

The same analyze for $\sigma_t^2$. First remark that $\mathbb{E}[\sigma_t^2] = \mathbb{E}[a_t^2]$ then $$\mathbb{E}[\sigma_t^2] = \alpha_0 + \alpha_1\mathbb{E}[\sigma_{t-1}^2] + \beta_1\mathbb{E}[\sigma_{t-1}^2]\,.$$ So $\mathbb{E}[\sigma_t^2] = \frac{\alpha_0}{1-\alpha_1-\beta_1}$.

It optimization can be done numerically. Suppose $T=2$, then $$log\mathcal{L} = \log p(r_1\mid\theta) + \log p(r_2\mid r_1,\sigma_1,a_1,\theta)\,.$$

As $r_1\sim\mathcal{N}(\frac{\phi_0}{1-\phi_1},\frac{\alpha_0}{1-\alpha_1-\beta_1})$ and $r_2\sim\mathcal{N}(\phi_0+\phi_1\,r_1,\alpha_0+\alpha_1\,a_1^2+\beta_1\sigma_1^2)$, for each $\theta$ the loglikelihood can be computed and maximized.

  • $\begingroup$ Thanks for your detailed response.Could you guide me how did you arrive at value of mean and variance.Also could you guide how to proceed forward as I am not familiar with the topic?You can also guide me relevant resource?Based on the above links I thought only value of mean has to be changed ? My apologies for following up? $\endgroup$
    – Wolf Gupta
    Aug 3, 2021 at 0:22
  • $\begingroup$ For the computation, at each date, you have to compute the mean and the variance base on the past value. $\endgroup$ Aug 3, 2021 at 1:22
  • $\begingroup$ I can understood the part that you beautifully explained above .Could you explain how to find value of individual term p(r1|phi)p(r2|r1,sigma1,a1,phi). I tried but unfortunately not able to come up with anything or point me to the right direction? $\endgroup$
    – Wolf Gupta
    Aug 5, 2021 at 22:51
  • $\begingroup$ This is Bayes formula. $$p(r_2,r_1\mid theta) = p(r_2\mid r_1,\theta)p(r_1\mid \theta)\,.$$ Also $p(r_2\mid r_1,\theta) = p(r_2\mid \sigma_1,a_1,r_1,\theta)$ as when you know $r_1$, you can compute $\sigma_1$ and $a_1$ $\endgroup$ Aug 5, 2021 at 23:24
  • $\begingroup$ So above equation mentioned as L is likelihood function for AR(1)-GARCH(1,1) model and it cannot be solved further as we do not have values for sigma,a1,r1.phi. Am I correct? $\endgroup$
    – Wolf Gupta
    Aug 5, 2021 at 23:44

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