I am modelling an univariate GARCH and I would like to know if I am operating in the correct way.

First of all, I suppose that the performance of my asset can be described as:

$Y_t=\mu+\epsilon_t $

in other words, I suppose the conditional mean of the process is constant over time, thus $\text{Var}(Y_t)=\text{Var}(\epsilon_t)=\sigma^2_t$. Later I develop the following GARCH(1,1) model:

$\sigma^2_t=\omega+\alpha Y_{t-1}^2+\beta\sigma^2_{t-1} $

At the implementation level, therefore, I fix $\mu_0 = \bar{Y}, \omega_0,\alpha_0,\beta_0$, I get the value of

$\epsilon_t =Y_t - \mu_0$

and finally I set the log likelihood function of the GARCH model


My question is: is it correct that the parameter $\mu$ is estimated together as the parameters $\omega,\alpha,\beta$?

  • $\begingroup$ Thanks Richard, I completely agree with you! Observing the formulation of the log likelihood, I think that in the equation for $\sigma^2_t$, $Y^2_{t-1}$ should be replaced by $\epsilon^2_{t-1}$ since, subsequently, the standardized errors are defined as $\xi_{t}=\frac{\epsilon_t}{\sigma_t}$ $\endgroup$ Aug 29, 2019 at 6:29
  • $\begingroup$ Sounds correct. $\endgroup$ Aug 29, 2019 at 7:03

1 Answer 1


Yes it is correct, indeed you are performing MLE where, given a postulated model like yours (with constant conditional mean and GARCH error term), all the unknown parameters are estimated at once (you simultaneously optimize the log likelihood to get the cond mean and the GARCH parameters).

A two-stage approach where you first estimate $\mu$ and then you use the series $y_{t}-\mu$ in a second-stage MLE estimation to estimate the GARCH parameters is also possible. Very often it is used to reduce the computational complexity of a single-stage estimation in complex models. However, notice that in a two-stage approach, at the second stage you assume that the first-stage estimate of $\mu$ is "true" value of the parameter and estimate the other parameters for the conditional variance. Therefore, as usual in MLE, if you make the wrong assumption about the true value of a parameter (and/or you mispecify the other parts of the postulated model), you fall in QMLE (where the Q means Quasi, see the link), which is still consistent but may be less efficient than MLE.

What is wrong in your question is the notation used for Variance $\text{Var}(y_{t})$, which is usually used to indicate the unconditional variance of $y$, not the conditional variance $\text{Var}_{t}(y_{t+1})=\mathbb{E}_{t}(y_{t+1}^{2})-\mathbb{E}_{t}(y_{t+1})^{2}$

and the log likelihood should be:

$$\sum_{t=1}^{T} -\frac{1}{2} \big{[} \ln(2\pi)+\ln(\sigma^2_t)+\frac{\epsilon^2_t}{\sigma^2_t}\big{]}$$

With $\sigma_{t}^{2}= \omega +\alpha \epsilon_{t-1}^{2} + \beta \sigma_{t-1}^{2}$

  • $\begingroup$ Thanks for the answer! I think also that the formulation of the log likelihood is wrong: I think that it is: $-\frac{1}{2}\sum_{t}[ln(\sigma^2_t)+\frac{\epsilon^2_t}{\sigma^2_t}]$ $\endgroup$ Aug 28, 2019 at 11:32
  • $\begingroup$ Your correction is fine but you need an additional one, I will put in in the edit in a few minutes $\endgroup$
    – Fr1
    Aug 28, 2019 at 11:39
  • $\begingroup$ Thanks for editing the log likelihood. I was not interested in the constant term $ln(2\pi)$ since i was looking at the maximum problem! But I totally agree with you! $\endgroup$ Aug 28, 2019 at 11:57
  • $\begingroup$ Sure, for max/min wrt parameters, the constant can be dropped $\endgroup$
    – Fr1
    Aug 28, 2019 at 12:30
  • $\begingroup$ Thank you so much, now I can proceed without problems!! $\endgroup$ Aug 28, 2019 at 12:34

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