6
$\begingroup$

Consider the standard GARCH model:

$$ \sigma^2_t = \omega + \alpha\varepsilon^2_{t-1} + \beta\sigma^2_{t-1}.$$

The so-called persistence parameter is defined as the sum $\alpha+\beta$.

Now consider the GJR-GARCH model by Glosten et al. (1993):

$$ \sigma^2_t = \omega + (\alpha+\gamma \mathbb{I}_{t-1})\varepsilon^2_{t-1} + \beta\sigma^2_{t-1} $$

where $\mathbb{I}_{t-1}$ is the indicator function:

$\mathbb{I}_{t-1}(\varepsilon_{t-1})=\varepsilon_{t-1}$ for $\varepsilon_{t-1}>0$ and
$\mathbb{I}_{t-1}(\varepsilon_{t-1})=0$ otherwise.

Question: What is the persistence parameter in the GJR-GARCH model? Could someone provide some references where this is explained?

My guess is that the persistence parameter equals $\alpha+\gamma/2+\beta$, but I am not sure. The guess is based on the material in V-Lab and the similarities between the standard GARCH and the GJR-GARCH model.

References

  • Glosten, L. R., R. Jagannathan, and D. E. Runkle, 1993. On The Relation between The Expected Value and The Volatility of Nominal Excess Return on stocks. Journal of Finance 48: 1779-1801.

  • V-Lab

$\endgroup$
3
$\begingroup$

I found an answer in the "vignette" to the "rugarch" package in R. Here is a quote from pages 7-8 (emphasis is mine):

Because of the presence of the indicator function, the persistence of the model now crucially depends on the asymmetry of the conditional distribution used. The persistence of the model $\hat P$ is,

$$ \hat P = \sum_{j=1}^q \alpha_j + \sum_{j=1}^p \beta_j + \sum_{j=1}^q \gamma_j\kappa $$

where $\kappa$ is the expected value of the standardized residuals $z_t$ below zero (effectively the probability of being below zero),

$$ \kappa = \mathbb{E}(\mathbb{I}_{t-j} z_{t-j}^2) = \int_{-\infty}^0 f(z,1,0,\dotsc) dz $$

where $f$ is the standardized conditional density with any additional skew and shape parameters $(\dotsc)$. In the case of symmetric distributions the value of $\kappa$ is simply equal to 0.5.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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