Persistence in TGARCH 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?

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 $\sigma_t$
$$
E[-\gamma \mathbb{I}_{\varepsilon < 0} \varepsilon] = \intop_{-\infty}^\infty -\gamma \mathbb{I}_{\varepsilon < 0} \frac{\varepsilon }{ \sqrt{2 \sigma^2 \pi}} \exp\left(-\frac{\varepsilon^2 }{ 2\sigma^2}\right) d\varepsilon = \intop_{-\infty}^0 -\gamma  \frac{\varepsilon }{ \sqrt{2 \sigma^2 \pi}} \exp\left(-\frac{\varepsilon^2 }{ 2\sigma^2}\right) d\varepsilon = \frac{\gamma}{\sqrt{2 \pi}}\sigma
$$
and
$$
E[\alpha \mathbb{I}_{\varepsilon \geq 0} \varepsilon] = \intop_{-\infty}^\infty \alpha \mathbb{I}_{\varepsilon \geq 0} \frac{\varepsilon }{ \sqrt{2 \sigma^2 \pi}} \exp\left(-\frac{\varepsilon^2 }{ 2\sigma^2}\right) d\varepsilon = \intop_{0}^\infty \alpha  \frac{\varepsilon }{ \sqrt{2 \sigma^2 \pi}} \exp\left(-\frac{\varepsilon^2 }{ 2\sigma^2}\right) d\varepsilon = \frac{\alpha}{\sqrt{2 \pi}}\sigma
$$
Thus, we can say that the persistance of $\sigma_t$ is given by
$$
\frac{\alpha}{\sqrt{2 \pi}} + \frac{\gamma}{\sqrt{2 \pi}} + \beta
$$
