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I know for the additive autoregressive model the stationary distribution of $\{X_t\}$ can be found, if it exists, in the following way: \begin{align} X_t &= \alpha X_{t-1} + \epsilon_t\\ \Rightarrow X_t &= (1-\alpha B)^{-1} \epsilon_t \\ &= \epsilon_t + \alpha \epsilon_{t-1} + \cdots. [\text{if}~ -1<\alpha<1] \end{align} Now if $e_t \stackrel{\text{iid}}{\thicksim} N(0, 1)$, say, one can easily find the distribution of $X_t$.

My question: What will be the stationary distribution, if it exists, of the multiplicative autoregressive model $X_t = \alpha X_{t-1}\epsilon_t$ ? What will be the condition(s) of $\alpha$ in this case? It will also be helpful if some hints are suggested to how it can be generalized to the higher order models.

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  • $\begingroup$ Are you sure you want that form? This would look like a bilinear model if $\epsilon_t$ was lagged more, and you had a nonlagged $\epsilon_t$ by itself $\endgroup$
    – Taylor
    Commented Apr 30, 2018 at 16:56
  • $\begingroup$ Yes. That is my model of interest. I do not have any prior knowledge of the stationary distributions of such models. So the possibility of the non-existence of stationary distribution remains. You can prove, if possible, that the stationary distribution of the stated model does not exist for any choices of the distribution (normal, Cauchy) of errors. $\endgroup$
    – Shanks
    Commented Apr 30, 2018 at 17:48
  • $\begingroup$ yeah, it's strange. Under certain conditions $\operatorname{Var}(X_t) = \alpha^2 \operatorname{Var}(X_{t-1})$ which only holds if $\alpha=1$, getting rid of (possibly) the only parameter. $\endgroup$
    – Taylor
    Commented Apr 30, 2018 at 18:31

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