Consider the following GARCH(1,1) process: $$ \epsilon_t = \sigma_t \eta_t \quad \text{where} \quad (\eta_t) \overset{iid}{\sim} \mathcal{N} (0,1)$$ $$ \sigma_t^2 = \omega + \alpha \epsilon_{t-1}^2 + \beta \sigma_{t-1}^2 \quad \text{where} \quad \omega, \alpha, \beta \geq 0$$.
It is a special case of the stochastic difference equation $Y_t = A_t Y_{t-1} + B_t $ for which proof of the existence of the stationary solution was developed by Brandt (1986). In book GARCH Models: Structure, Statistical Inference and Financial Applications by Francq and Zakoian (2010) there is a thm and proof of the existence of strict stationary solution to GARCH(1,1).
Theorem If $- \infty \leq \gamma := \mathbb{E} [\ln (\alpha \eta_t^2 +\beta )] <0 $ then the infinite sum $$ h_t = \omega \left[ 1 + \sum_{i=1}^{\infty} a(\eta_{t-1}) \dots a(\eta_{t-i}) \right] $$ converges almost surely and the process $(\epsilon_t) $ defined by $\epsilon_t = \sqrt{h_t} \eta_t $ is the unique strictly stationary solution to GARCH(1,1).
I try to understand the case if $\gamma = 0 $ then strict stationary solution does not exist. Let's denote $a(\eta_t) = \alpha \eta_t^2 + \beta $. In the case of $\gamma =0$ authors give proof by contradiction as following:
My questions are:
How we deduce that $ \omega \prod_i^n a(\eta_{t-i}) $ converges to $0$ as $n$ goes to infinity? I can't see this :(
Regarding Chung-Fuchs thm (footnote below) it seems that in our case $ X_t = \ln (\alpha \eta_t^2 + \beta) $. Of course $X_1, \dots X_n $ are iid and $\mathbb{E} X = 0 $ by assumptions of the thm. But how does $ \mathbb{E} \lvert X \rvert = \mathbb{E} \lvert \alpha \eta_t^2 + \beta \rvert > 0 $ ?
