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If a stochastic process is generated by a vector autoregressive process of order $d$, can it be strictly stationary. I know that under the stability condition, that this is weakly stationary process.

$\begin{align} \mathbf{y}_t = A_1\mathbf{y}_{t-1} + \dots + A_d\mathbf{y}_{t-d} + \boldsymbol{\epsilon}_t, \quad t \in \mathbb{Z} \end{align}$

I was also wondering, if anyone had examples of stochastic process that are strictly stationary, alpha-mixing, and have heavy tails?

Richard Hardy
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Dylan Dijk
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  • Intuitively, what should prevent a weakly stationary VAR process from being strictly stationary? For simplicity, let us assume the error term $\boldsymbol\epsilon_t$ is i.i.d. – Richard Hardy May 03 '23 at 18:56
  • So a process is strictly stationary if the joint distribution for any n-tuple is the same. Not sure how you would show that such a process is strictly stationary. – Dylan Dijk May 03 '23 at 20:26
  • Is it because it can be written with an MA representation, and then the shifts in joint distributions will always be the same as it will always be a sum of infinite independent wn r.s with the same coefficients? – Dylan Dijk May 03 '23 at 21:01
  • Well, if we reduce the the joint distribution for any $n$-tuple to just the first couple of moments of that distribution (provided that they exist), do we not get weak stationarity? Then try this in reverse and see whether there is a necessary reason for the higher-order moments to vary when the first two moments are fixed. – Richard Hardy May 04 '23 at 07:42
  • Well we already know its weakly stationary. Are you saying that if we show that all the moments are the same, for any shift then we have the same joint distribution, because that does not work? You mean show that the MGFs are the same? – Dylan Dijk May 04 '23 at 12:47
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    if we show that all the moments are the same, for any shift then we have the same joint distribution Yes. And you probably do not need to resort to MGFs for that, since you only need one example (existence), not all of them (necessity). – Richard Hardy May 04 '23 at 13:10
  • Ok, thank you for your help. – Dylan Dijk May 04 '23 at 13:42
  • In the end I think it is easiest to show that the process is strictly stationary, by using that if the stability condition holds then we have a MA($\infty$) representation. And hence, using this answer the process is strictly stationary. – Dylan Dijk Jul 28 '23 at 07:48
  • @RichardHardy I have asked a question about geometric ergodicity and was wondering if you could help? – Dylan Dijk Aug 02 '23 at 12:35
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    That one is too tough for me, but +1 for the question. – Richard Hardy Aug 02 '23 at 16:27

1 Answers1

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The answer to your first question is yes, as can be found in a paper by Bougerol and Picard, dated back to 1992 and titled "Strict Stationarity of Generalized Autoregressive Processes", available at this link

https://projecteuclid.org/journals/annals-of-probability/volume-20/issue-4/Strict-Stationarity-of-Generalized-Autoregressive-Processes/10.1214/aop/1176989526.full

See in particular Section 4 therein, dealing with multivariate ARMA, and note that nothing prevents innovation vectors to be heavy tailed, as long as the logarithm of their norm has finite expectation.

Jack London
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