How can I test that the sum of the $\alpha_1$ and $\beta_1$ parameters in a GARCH(1,1) model is significantly different from 1?

  • Hi. I am also trying to do the same thing here using R's ugarchfit. So do you mean "sGARCH" and "iGARCH" provide restricted and unrestricted GARCH models? Why is this? – Eric Mar 5 at 20:30
  • I really need to solve this problem. I posted the actual question I wanted to ask directly in the following:… – Eric Mar 14 at 10:23
up vote 4 down vote accepted

The restriction you are interested in is a simple linear restriction. The testing principle is the same as testing for a linear restriction in a regression setting, for example. You may use likelihood ratio (LR), Wald or Lagrange multiplier tests.

Consider the LR test:

  1. Estimate the unrestricted GARCH(1,1). Obtain the model log-likelihood $L_{unres.}$.
  2. Estimate the restricted GARCH(1,1) subject to the restriction $\alpha_1+\beta_1=1$. Obtain the model likelihood $L_{res.}$.
  3. Calculate the likelihood ratio statistic $LR=-2 \operatorname{log} \left( \frac{L_{res.}}{L_{unres.}} \right)$. Under the null hypothesis that the restriction holds in population, the $LR$ statistic will follow a $\chi^2(1)$ distribution (since there is one linear restriction that distinguishes the restricted model from the unrestricted model). Given that, you can obtain the p-value and compare it to the chosen significance level; then you will see whether you can or cannot reject the null hypothesis.

Since you added an R tag, I suppose you are also interested in implementing the test in R. There are three steps corresponding the three steps above:

  1. Specify the restricted model using ugarchspec with option variance.model = list(model = "sGARCH") and estimate it using ugarchfit. Obtain the log-likelihood from the slot fit sub-slot likelihood.
  2. Specify the restricted model using ugarchspec with option variance.model = list(model = "iGARCH") and estimate it using ugarchfit. Obtain the log-likelihood as above.
  3. Calculate $LR=2(\text{logLik}_{unres.}-\text{logLik}_{res.})$. Obtain the p-value as pchisq(q = LR, df = 1).

On a second thought, the case where $\alpha_1+\beta_1=1$ is a boundary case. When $\alpha_1+\beta_1<1$ the conditional variance is stationary; but when $\alpha_1+\beta_1=1$ the conditional variance becomes integrated.
Could this issue somehow invalidate the LR test? I am not quite sure. I hope someone could post an argument in the comments or as another answer.

  • 1
    Re first comment: Yes, I mean that. Why is this? Because sometimes you know what you want, either an integrated or a stationary equation, and you can specify this by selecting the appropriate option. Re second comment: A low p-value suggests rejection of the null hypothesis while a high one suggests the opposite. The null hypothesis is that the restriction holds. So a low p-value suggests it does not hold, hence, $\alpha_1+\beta_1<0$. – Richard Hardy Mar 6 at 8:10
  • 1
    @Eric, that's right. – Richard Hardy Mar 6 at 10:52
  • 1
    sGARCH is unrestricted because its coefficients do not need to sum to one. iGARCH is restricted because they must (which is a linear restriction: $\alpha_1+\beta_1=1$). – Richard Hardy Mar 9 at 9:17
  • 1
    Low p-value means low probability to have observed what you have observed given that the null hypothesis is correct, which is a motivation to reject the null hypothesis. Technically the likelihood of sGARCH cannot be lower than that of iGARCH; but there might be some bug in the calculation or oherwise poor definition in the function. – Richard Hardy Mar 9 at 12:37
  • 1
    Due to the linear restriction, you are estimating one coefficient less than in an sGARCH. That might be the reason for why you have those NA values. Knowing $\alpha_1$ and the fact that $\alpha_1+\beta_1=1$ you know $\beta_1$, so you are not estimating it. Then you have no standard error etc. – Richard Hardy Mar 10 at 8:02

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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