Solved – GARCH(1,1) fails to converge in “rugarch” in R

garchr

My code is very simpe

require(rugarch)
require(quantmod)
    #Daily GARCH(1,1)
    date_from = c("1996-01-01", "2000-01-02", "2004-01-03", "2008-01-04", "2012-01-05")
    date_to = c("2000-01-01", "2004-01-02", "2008-01-03", "2012-01-04", "2016-08-20")
    forex = vector(mode = 'list', length = 5)
    for (i in 1:5) {
      getSymbols("EUR/AUD", src="oanda", from = date_from[i], to = date_to[i])
      forex[[i]] = EURAUD
    }
    EURAUD = Reduce(rbind,forex)
    EURAUD$DEURAUD <- diff(log(EURAUD$EUR.AUD))
    spec7 = ugarchspec(variance.model = list(model = "sGARCH", garchOrder = c(1,1)),
                       mean.model = list(armaOrder = c(0,0), include.mean = TRUE))
    roll_7 = ugarchroll(spec7, EURAUD[,2], forecast.length = 1500, refit.every = 50, windows.size = 1500, refit.window = 'moving', solver = 'hybrid', calculate.VaR = FALSE, keep.coef = FALSE)

Unfortunately it fails to converge. What are my alternatives? The standard GARCH(1,1) is crucial as it is proven by Hanen and Lunde (2005) that provides the most accurate forecasts and I nee as a benchmark model for my thesis.

Best Answer

Assuming a GARCH model is sensible for your data (daily log-returns on currency exchange rates), here are some possible solutions:

Try different starting values. Most of the times this should do the job.
Add a negligible amount of noise to the original data (enough to get the solver unstuck but still not affecting the parameter estimates noticeably). Either do this once or perhaps multiple times and average over the outcomes.
Try a different model: GARCH(1,1) with variance targetting, GARCH(1,1) with a different conditional distribution (e.g. Student-$t$ in place of Normal) -- if this is still a satisfactory benchmark for you.

Related Solutions

Solved – GARCH(1,1) with insignificant constant

I can't comment yet so I try a short answer:

You could impose a parameter restriction and set $a_0=0$ to gain efficiency.

E.g. in R using the rugarch package you might use the option fixed.pars = list(omega=0) in the specification.

The model is still meaningful. The long term volatility in this case would be zero, and as long as $a_1$ (and $a_2$) is significant the model exhibits a mean reversion to the long term volatility of zero. Of course this long-term volatility would never be reached in practice since mean reversion is "slow" and new information will come in and push you away from the long-term vola.

GARCH Models – Resolving Contradictory Results in Estimating GJR-GARCH(1,1) with rugarch

There are two different parametrizations of the GJR-GARCH model in rugarch, and you're applying the formula for the persistence from one parametrization to the other.

For GJR-GARCH(1,1), the first one is the one you've shown, which in the documentation is written like this:

$$\sigma_t^2 = w + (\alpha + \gamma I_{t-1}) \varepsilon_{t-1}^2 + \beta \sigma_{t-1}^2$$

where $I_t = 1$ when $\varepsilon_t \leq 0$, and 0 otherwise.

As you say, when the standardized residuals are drawn from a symmetric distribution, the persistence is given by:

$$\alpha + \beta + \frac{1}{2}\gamma$$

There is an alternate, equivalent parametrization, which arises as a special case of the family-GARCH model, which goes like this:

$$\sigma_t^2 = w + \alpha' \sigma^2_{t-1} (|z_{t-1}|-\eta z_{t-1})^2 + \beta \sigma_{t-1}^2$$

Since the sign of $z_t$ and $\varepsilon_t$ is the same and since $\varepsilon_t = \sigma_t z_t$, you can show that this is the same as:

$$\sigma_t^2 = w + \left(\alpha'(1-\eta)^2 + \alpha'\left[(-1-\eta)^2-(1-\eta)^2\right]I_{t-1} \right)\varepsilon^2_{t-1} + \beta \sigma_{t-1}^2$$

Reading off the coefficients, we have the relationship between the two parametrizations:

$$\alpha = \alpha'(1-\eta)^2$$ $$\gamma = \alpha'\left[(-1-\eta)^2-(1-\eta)^2\right]$$

Therefore, in the second parametrization the persistence is given by:

$$\alpha'(1-\eta)^2 + \beta + \frac{1}{2}\alpha'\left[(-1-\eta)^2-(1-\eta)^2\right]$$

The more general version of this appears in the documentation you've linked to as equations 27-28.

In rugarch, the first parametrization is obtained from rugarch::ugarchspec by passing :

variance.model = list(model = "gjrGARCH", garchOrder = c(1,1))

The second, which is the one you have, by passing:

variance.model = list(model = "fGARCH", garchOrder = c(1,1), submodel = "GJRGARCH")

Plugging in your coefficient estimates (to the precision that you've displayed them), I get that the persistence of your GJR-GARCH model is $0.9575652$, which is very close to what rugarch::persistence gave you.