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I'm going through "GARCH models" by Francq and Zakoian (2010). They define the TGARCH(1,1) as $$\sigma_t = \omega + \beta_1 \sigma_{t-1} + \alpha_{1,+}\epsilon_{t-1}^+ - \alpha_{1,-}\epsilon_{t-1}^- $$ where $\epsilon_{t-1}^+$ equals $\epsilon_{t-1}$ if positive and 0 if negative (opposite holds for $\epsilon_{t-1}^-$).

On p.252 they write the following TGARCH(1,1) stationarity condition $$E[(\alpha_{1,+}z_t^+ - \alpha_{1,-}z_t^- + \beta_1)^2]<1$$ which, assuming $N(0,1)$ innovations $z_t$, simplifies to $$\frac{1}{2}(\alpha_{1,+}^2 + \alpha_{1,-}^2)+\frac{2\beta_1}{\sqrt{2 \pi}}(\alpha_{1,+}+\alpha_{1,-}) + \beta_1^2<1$$ I've been trying to derive the second formula from the first one. Whereas I've obtained the last two terms of the equation, I struggle with the first one. More precisely, solving the brackets from the first equation I'm stuck at: $$\alpha_{1,+}^2 E[z_t^2]+\alpha_{1,-}^2E[z_t^2]-2\alpha_{1,+}\alpha_{1,-}E[z_t^+z_t^-]$$ How do you go from this expression to $$\frac{1}{2}(\alpha_{1,+}^2 + \alpha_{1,-}^2)$$ PS: They also mention that $$E[z_t^+]=-E[z_t^-]=\sqrt{\frac{1}{2\pi}}$$How do you derive this result knowing that $z_t$ is $N(0,1)$ distributed?

Kondo
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First of all, we note that

\begin{align} \alpha_{1,+}^2 E[z_t^2]+\alpha_{1,-}^2E[z_t^2]-2\alpha_{1,+}\alpha_{1,-}E[z_t^+z_t^-] \end{align}

should be

\begin{align} \alpha_{1,+}^2 E[(z_t^+)^2]+\alpha_{1,-}^2E[(z_t^-)^2]-2\alpha_{1,+}\alpha_{1,-}E[z_t^+z_t^-] \end{align}

Secondly, we note that $E[z_t^+z_t^-] = 0$ since either $z_t^+$ or $z_t^-$ will be zero (cannot both be non zero at the same time). Now,the only thing we need to show is

$$E[(z_t^+)^2] = E[(z_t^-)^2]=\frac{1}{2}$$

Intuitively, this must be the case since $N(0,1)$ is a symmetric distribution with $E[z_t^2] = 1$. To formally show it, one has to evaluate

$$E[(z_t^+)^2] = \frac{1}{\sqrt{2\pi}}\intop_{0}^{\infty} z^2\exp\left(-\frac{z^2}{2}\right) dz $$ and $$E[(z_t^-)^2] = \frac{1}{\sqrt{2\pi}}\intop_{-\infty}^{0} z^2\exp\left(-\frac{z^2}{2}\right) dz $$ In the same way, you can show that e.g. $$E[z_t^-] = \frac{1}{\sqrt{2\pi}}\intop_{-\infty}^{0} z\exp\left(-\frac{z^2}{2}\right) dz = -\frac{1}{\sqrt{2\pi}}$$

See e.g. this question for the actual steps in relation to evaluating the integrals

Johan Stax Jakobsen
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