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 $\alpha+\beta$.

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

$$ \sigma^2_t = \omega + (\alpha+\gamma \mathbb{I}_{\varepsilon_{t-1} < 0})\varepsilon^2_{t-1} + \beta\sigma^2_{t-1} $$

The persistence parameter equals $\alpha+\gamma/2+\beta$.

Now consider the TGARCH model by Zakoian (1994):

$$ \sigma_t = \omega + (\alpha\mathbb{I}_{\varepsilon_{t-1} \geq 0}-\gamma \mathbb{I}_{\varepsilon_{t-1} < 0})\varepsilon_{t-1} + \beta\sigma_{t-1}. $$

Question: What is the persistence parameter in the TGARCH model?


1 Answer 1


The first important point is to understand where the "persistence parameter" comes from in the GJR-GARCH model. The specification depends on a normality assumption for the innovations $\varepsilon_t$:

$$ E[\gamma \mathbb{I}_{\varepsilon < 0} \varepsilon^2] = \intop_{-\infty}^\infty \gamma \mathbb{I}_{\varepsilon < 0} \frac{\varepsilon^2 }{ \sqrt{2 \sigma^2 \pi}} \exp\left(-\frac{\varepsilon^2 }{ 2\sigma^2}\right) d\varepsilon = \intop_{-\infty}^0 \gamma \frac{\varepsilon^2 }{ \sqrt{2 \sigma^2 \pi}} \exp\left(-\frac{\varepsilon^2 }{ 2\sigma^2}\right) d\varepsilon = \frac{\gamma}{2}\sigma^2 $$

The TGARCH model, as you present it, is specified in terms of $\sigma_t$ and not $\sigma_t^2$. We can perform similar calculations to obtain the persistance of $\sigma_t$

$$ E[-\gamma \mathbb{I}_{\varepsilon < 0} \varepsilon] = \intop_{-\infty}^\infty -\gamma \mathbb{I}_{\varepsilon < 0} \frac{\varepsilon }{ \sqrt{2 \sigma^2 \pi}} \exp\left(-\frac{\varepsilon^2 }{ 2\sigma^2}\right) d\varepsilon = \intop_{-\infty}^0 -\gamma \frac{\varepsilon }{ \sqrt{2 \sigma^2 \pi}} \exp\left(-\frac{\varepsilon^2 }{ 2\sigma^2}\right) d\varepsilon = \frac{\gamma}{\sqrt{2 \pi}}\sigma $$


$$ E[\alpha \mathbb{I}_{\varepsilon \geq 0} \varepsilon] = \intop_{-\infty}^\infty \alpha \mathbb{I}_{\varepsilon \geq 0} \frac{\varepsilon }{ \sqrt{2 \sigma^2 \pi}} \exp\left(-\frac{\varepsilon^2 }{ 2\sigma^2}\right) d\varepsilon = \intop_{0}^\infty \alpha \frac{\varepsilon }{ \sqrt{2 \sigma^2 \pi}} \exp\left(-\frac{\varepsilon^2 }{ 2\sigma^2}\right) d\varepsilon = \frac{\alpha}{\sqrt{2 \pi}}\sigma $$

Thus, we can say that the persistance of $\sigma_t$ is given by

$$ \frac{\alpha}{\sqrt{2 \pi}} + \frac{\gamma}{\sqrt{2 \pi}} + \beta $$

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    $\begingroup$ Is $\sigma$ the unconditional variance of $\varepsilon$? $\endgroup$ Jul 20, 2021 at 6:13
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    $\begingroup$ Good answer ! But do we need the normal assumption for the GJR-GARCH ? In my opinion, if we assume that the density of $\epsilon_t$ is symmetric around zero, we should get the same result. The same should be valid for TGARCH as well. $\endgroup$
    – Count
    Jul 20, 2021 at 7:08
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    $\begingroup$ Also what do we exactly mean if we talk about persistence in GARCH models ? I usually calculate the ACF of $\epsilon_t^2$ and then look at which rate the ACF converges to zero and call this the persistence of volatility. However, depending on the model this can be quite difficult. I know that your results for T-GARCH are correct, but how can we infer that the ACF of $\epsilon_t^2$ converges to zero at this rate without calculating the ACF ? Did I misunderstood something ? $\endgroup$
    – Count
    Jul 20, 2021 at 7:24

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