Negative signs in GARCH equation

When one try to fit a GARCH on a time series it may happen that one or more coefficients in the estimation output have negative sign. In these cases:

1. all the negative coefficients (and relative orders) must necessary be removed from the equation, even if they are significant?

2. Is this eventually true for both ARCH and GARCH components?

3. The fact that variance equation can't contain negative coefficients is even true for the other garch extensions (EGARCH, TARCH, ...)?

Thank you

A negative coefficient does not necessarily entail a negative $\sigma^{2}$. Usually we do not impose positivity constraints during estimation, then we check if $\sigma^{2}$ takes some negative values or if coefficients respect some known positivity constraints (when these constraints are known).

Regarding the standard Garch model, you can force all the coefficients to be positive during maximisation of the likelihood but it is not a necessary condition (just a sufficient condition), the original formulation of Bollerslev (86) imposes this sufficient constraint. If you employ this constraint, you should not "remove" negative coefficients as they should never appear as a plausible result during estimation.

For each Garch-type of model, sufficient conditions (i.e parameter restrictions) are different, as an example see Tsai, H., & Chan, K. S. (2008) for Garch model. Egarch, GJR... have different constraints.

Tsai, H., & Chan, K. S. (2008). A note on inequality constraints in the garch model. Econometric Theory, 24(3), 823–828. http://doi.org/10.1017/S0266466608080432

Bollerslev, T. (1986). Generalized Autoregressive Conditional Heteroskedasticity. Journal of Econometrics, 31, 307–327.

• thank you. So, if I understood, in a standard GARCH model it is sufficient that the sigma_sq itself is positive, while it isn't necessary that every single coefficient is positive, right? Commented Sep 2, 2018 at 13:07
• @LeoAn Yes, the paper of Tsai and Al. argue that imposing all coefficients to be nonnegative is too restrictive. If $\sigma^{2}$ is positive you don't need to bother you with positivity constraints. Commented Sep 2, 2018 at 13:15