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For the simplest ARCH model $x_t = \sigma_t \epsilon_t$ where $\sigma_t^2 = a_0 + a_1x_{t-1}^2$, $x_t$ is stationary when $a_1 <1$. However, isn't the whole point of using an ARCH model is to model unstationary process (the variance of $x_t$ changes)? How could we end up with modeling a stationary process with an ARCH model?

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  • $\begingroup$ When they say stationary they are referring to the unconditional distribution of the process. These models are conditional models (autoregressive "conditional" heteroscedasticity), and it is permitted to have short term divergences as long as the long run behavior is preserved. $\endgroup$ – Cowboy Trader Jul 13 '17 at 19:22
  • $\begingroup$ @CagdasOzgenc Thanks, and that makes sense. If you move this comment to an answer, I'm happy to accept it, or let me know if you think this question is trivial, and I'll just delete it $\endgroup$ – Probably Jul 13 '17 at 19:29
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    $\begingroup$ This may be useful stats.stackexchange.com/a/72628/28746 $\endgroup$ – Alecos Papadopoulos Aug 5 '17 at 15:41
  • $\begingroup$ @AlecosPapadopoulos Indeed! Upvoted your answer since I had the same doubt as the OP of that question did $\endgroup$ – Probably Aug 9 '17 at 13:07
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No, the point of ARCH is to model time-varying conditional variance which does not have to be nonstationary. Stationarity is not the same as constancy. A time series can be stationary without being constant. E.g. the variance $\sigma_t$ of a variable $x_t$ can be stationary without being constant, i.e. without $\sigma_t\equiv \sigma$ for some fixed value $\sigma$.

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Although you refer to the weak stationarity requirement of the simple ARCH(1) model. Actually strict stationarity holds up to something like a(1)<3,56, which also suggests that the unconditional variance of the process is undefined above 1 while the process itself is still strictly stationary.

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