It is well known that if a generic autoregressive process of $n$ order, $AR(n)$, has a gaussian white noise error term ("innovations"), then it is gaussian too.

So I presume that if the error term is not gaussian, neither the process necessary is.

Anyway, in a paper I'm reading, it is said that $AR(n)$ must be gaussian, because only in this case the process would have the same distribution at every moment - and that's a property of autoregressive processes.

Is there a point where I'm wrong?

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    $\begingroup$ Please give a reference to the paper $\endgroup$ – Juho Kokkala Apr 23 '17 at 14:42
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    $\begingroup$ Maybe you are thinking about a form of stationarity. $\endgroup$ – Michael R. Chernick Apr 23 '17 at 14:53
  • $\begingroup$ @JuhoKokkala Unfortunately it's in Italian - I'll translate the whole interesting part: The autoregressive model of n order, $y_t=\sum_{i=1}^n a_iy_{t-i} + e_t$, could describe only gaussian processes because only when $e_t$ and $y$ are gaussian, the distribution of $y$ would be of the same type for every time step. $\endgroup$ – Lo Scrondo Apr 23 '17 at 14:54
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    $\begingroup$ Well, using that argument it could also have some other stable (sum-stable) distribution. $\endgroup$ – kjetil b halvorsen Apr 23 '17 at 15:54
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    $\begingroup$ The quote is either out of context or plain wrong. Look at Levy processes $\endgroup$ – Aksakal Apr 24 '17 at 13:06

No. Consider:

$$ X_{t} = \rho X_{t-1} + Z_{t} $$

where $|\rho|<1$ and $Z \sim NIG(\alpha,\beta) $, a Normal-inverse Gaussian distribution (which is not a Normal distribution; it has non-zero Skewness). Assume all parameters are positive, and than $|\beta|<\alpha$.

Then, it can be shown that:

$$ X_{\infty} \sim NIG\left(\alpha,\beta\right) $$

In other words, in this example, the AR(1) inherits the distribution of the disturbance, which is not Gaussian.

The proof of the above result, and other examples like this one can be found in this article.


An AR(n) process with Gaussian errors is Gaussian if and only if all the parameters are known. Otherwise, it depends upon whether you are using a Frequentist or Bayesian methodology. For a Frequentist methodology, with a random shock with a finite, positive second moment and a first moment centered on zero (including the Gaussian), then the sampling distribution of the estimator for a parameter will either be Gaussian, Cauchy or the mixture of the two.

See Rao, M M., Consistency and Limit Distributions of Estimators of Parameters in Explosive Stochastic Difference Equations, The Annals of Mathematical Statistics, volume 32, number 1, March 1961, pp 195-218.

Mann, H B and Wald, A; On the Statistical Treatment of Linear Stochastic Difference Equations, Econometrica, volume 11, 1943, pp 173-200

For the Bayesian, the answer depends upon the form of the error term.


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