Consider the adapted ARCH model: $X_t^2 = \alpha_0 + \alpha_1X_{t-1}^2+\eta_t$, where $\eta_t=(\epsilon_t^2-1)(\alpha_0+\alpha_1X_{t-1}^2)$ and $\epsilon_t \sim \mathcal{N}(0,1)$.
My question is, how can I calculate $\mathbb{E}(\eta_tX_{t-s}^2)?$ I have tried simplifying but can't get to the desired result which should be zero... e.g.
$$ \mathbb{E}(\eta_tX^2_{t-s})=\mathbb{E}(\eta_t(\alpha_0 + \alpha_1X_{t-1-s}^2+\eta_{t-s}))=\mathbb{E}(((\epsilon_t^2-1)(\alpha_0+\alpha_1X_{t-1}^2))(\alpha_0 + \alpha_1X_{t-1-s}^2+\eta_{t-s}))$$
Not sure if i'm tackling this the wrong way, because this becomes quite messy, am i missing something ? Note I have calculated $\mathbb{E}(\eta_t)=0$.
Assume that $(\epsilon_t)_{t\geqslant1}$ is i.i.d. and independent of $X_0$. Then, for every $t\geqslant1$, $X_{t-1}$ is measurable with respect to $(X_0;(\epsilon_s)_{1\leqslant s\leqslant t-1})$, which is independent of $\epsilon_t$, hence, for every $t\geqslant1$, $\epsilon_t$ is independent of $(X_s)_{1\leqslant s\leqslant t-1}$. In particular, for every $t\geqslant s\geqslant1$, $\epsilon_t$ is independent of $(X_{t-1},X_{t-s})$ hence $$E(\eta_t\cdot X_{t-s}^2)=E(\epsilon^2_t-1)\cdot E((\alpha_0+\alpha_1X_{t-1}^2)\cdot X_{t-s}^2)=0$$