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I want to model volatility of a stock index with 1460 observations. The specification of the GARCH model is planned to be EGARCH-X where X is the external regressor, with no mean specification. Unfortunately the result of LB test on raw series of diff-in-log (daily return) proves that autocorrelation exists

Box.test(na.omit(data_xts[,1]), lag = 10, type = "Ljung")
Box-Ljung test

data:  na.omit(data_xts[, 1])
X-squared = 19, df = 10, p-value = 0.04026

while the result of ARCH-LM test from FinTS library:

ArchTest(na.omit(data_xts[,1]), lag=10)
ARCH LM-test; Null hypothesis: no ARCH effects

data:  na.omit(data_xts[, 1])
Chi-squared = 314.92, df = 10, p-value < 2.2e-16

proves that the ARCH effect exists.

Should I use the mean specification such as ARMA to model the autocorrelation which in turn leads to GARCH-in-mean model specification? I prefer not to use GARCH-in-mean because my objective is "to forecast volatility and back-testing Value-at-Risk measures" especially without a mean specification.

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1 Answer 1

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To deal with autocorrelation it is common to use ARMA. To deal with conditional heteroskedasticity it is common to use GARCH. If both patterns are present in your data, you can use ARMA-GARCH. This is not the same as GARCH-in-mean.

An ARMA(p,q)-GARCH(r,s) model looks like this: \begin{aligned} x_t &= \mu_t + u_t, \\ \mu_t &= c + \varphi_1 x_{t-1} + \dots + \varphi_p x_{t-p} + \theta_1 u_{t-1} + \dots + \theta_q u_{t-q} \quad \text{("ARMA")}, \\ u_t &= \sigma_t \varepsilon_t, \\ \sigma_t^2 &= \omega + \alpha_1 u_{t-1}^2 + \dots + \alpha_s u_{t-s}^2 + \beta_1 \sigma_{t-1}^2 + \dots + \beta_r \sigma_{t-r}^2 \quad \text{("GARCH")}, \\ \varepsilon_t &\sim i.i.D(0,1), \end{aligned} where $D$ is some probability distribution with zero mean and unit variance.

A GARCH(r,s)-in-mean looks like this: \begin{aligned} x_t &= \mu_t + u_t, \\ \mu_t &= \dots + c\sigma_t^2 \quad \text{("GARCH in mean")}, \\ u_t &= \sigma_t \varepsilon_t, \\ \sigma_t^2 &= \omega + \alpha_1 u_{t-1}^2 + \dots + \alpha_s u_{t-s}^2 + \beta_1 \sigma_{t-1}^2 + \dots + \beta_r \sigma_{t-r}^2, \\ \varepsilon_t &\sim i.i.D(0,1), \end{aligned} where $D$ is some probability distribution with zero mean and unit variance. There could be some additional terms in the equation for $\mu_t$, thus the dots before $c\sigma_t^2$.

Also, note that it is impossible not to have a mean specification. The specification need not be complicated, though; it could be just a constant: $\mu_t\equiv\mu$.

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  • $\begingroup$ thank you so much, I got it $\endgroup$
    – Najib Noer
    Nov 25, 2022 at 12:20
  • $\begingroup$ @NajibNoer, you are welcome! I am glad I could help! $\endgroup$ Nov 25, 2022 at 12:24

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