Testing the sum of GARCH(1,1) parameters

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: stats.stackexchange.com/questions/333256/… – Eric Mar 14 at 10:23

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.

• 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
• @Eric, that's right. – Richard Hardy Mar 6 at 10:52
• 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
• 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
• 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