\[ln(\sigma_t^2)=\omega+\beta ln(\sigma_{t-1}^2)+\alpha \left| \frac{u_{t-1}}{\sigma_{t-1}}\right|+Y(\frac{u_{t-1}}{\sigma_{t-1}})\]

take expectation on both sides:

\[E(ln(\sigma_t^2))=\omega+\beta E(ln(\sigma_{t-1}^2))+\alpha \sqrt{\frac{2}{\pi}}\] (note: \[z=\frac{u_{t-1}}{\sigma_{t-1}} is N(0,1), E(|z|)=\sqrt{\frac{2}{\pi}}\])


let long term log variance \[E(ln(\sigma^2))=x\]
\[x=\omega+\beta x+\alpha \sqrt{\frac{2}{\pi}}\]

\[x=\frac{\omega+\alpha \sqrt{\frac{2}{\pi}}}{1-\beta}\]

because E(f(y)) <>f(E(y)) (if f is not a linear function. in this case f=ln()), therefore we can use an approximation. If you really want a variance not log long term variance you can use approximation, given y is not too large. Usually variance is a small number.

\[ln(y) \approx y-1, E(ln(y))\approx E(y)-1, E(y)\approx E(ln(y))+1\]

\[E(\sigma^2) \approx \frac{\omega+\alpha \sqrt{\frac{2}{\pi}}}{1-\beta}+1\]

请问t分布的egarch无条件方差是什么

EGARCH的无条方差和模型所假设的残差分布类型没有关系，不管是norm, std, 还是ged, 永远都是同样的一个公式，因为无条件方差(long-term variance)只是一个站在EGARCH这个异方差结构设定上得出的无条件方差的结论。

谢谢，是\[e^{(\omega + \alpha \sqrt{2/\pi })/(1-\beta )}\],对吧