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My understanding is 'no', a stationary process does not imply a normal distribution of the data. However I haven't found a clear indication in my library or online. I am interested in other resources comparing stationary and heteroscedastic processes to data distributions.

I understand a stationary process joint probability distribution does not change over time. But does it necessarily mean that the shape of the distribution is normal?

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As far as I can see this answers your questions, as would any time series analysis text. – conjugateprior Apr 7 '12 at 13:51
@Conjugate Prior You clearly know this field better than i do, but I'm not finding the answer on that page, nor any mention of "normal" or "Gaussian." I also don't recall seeing a time series text that addresses it. Can you be more specific? – rolando2 Apr 7 '12 at 14:05
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Hint: Let $\{X_i\}$ be a sequence of iid random variables with distribution $F$. Is the process $\{X_i\}$ stationary? – cardinal Apr 7 '12 at 14:08
@ConjugatePrior That's one of the first resources I looked at. But all I read there is that a stationary process has constant mean and variance over time. It doesn't mean the distribution of the outcomes is normal. – Robert Kubrick Apr 7 '12 at 14:10
@cardinal Yes to your question. But let's reverse the question: the sequence of ${X_i}$ shows stable mean and variance over time. Does it mean the overall distribution is normal? – Robert Kubrick Apr 7 '12 at 14:13
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There was a discussion about this on dsp.SE some months ago. My answer there might help resolve some of the issues that the OP has.

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(+1) That is an excellent answer and addresses nearly all the main points that one would wish to know. It is fortuitous for the OP that you've specifically addressed his concern, as well. There is one point that has appeared here repeatedly that may, or may not, come up in the DSP community. Several times here, I have seen people make the statement to the effect that because the mean and variance are constant in time, the process is wide-sense stationary. While the converse is true, the one stated is not; ignoring the invariance of the autocorrelation seems to be a common misunderstanding. – cardinal Apr 8 '12 at 0:08

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