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I read that "strictly stationary implies that long-run variance is finite and positive", which means that long-run variance can be 0 or negative. However, according to this, it seems that long-run variance can not be negative.

So is it possible for long-run variance to be negative or 0?

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    $\begingroup$ Are you sure that it does not imply that the long-run variance is infinite and positive? $\endgroup$
    – Alexis
    Feb 20, 2018 at 22:04
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    $\begingroup$ A constant time series is strictly stationary with long-run variance zero. Therefore--by pure logic--it is possible for it to be negative or zero. $\endgroup$
    – whuber
    Feb 20, 2018 at 22:27

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strictly stationary implies that long-run variance is finite and positive

No, that isn't true. Let $X_1, \ldots, X_t$ be iid Cauchy random variables. None of these even have a variance (or mean). Yet the process is stationary.

which means that long-run variance can be 0 or negative

No variance can ever be negative, because for any random variable $Y$, $(Y-E[Y])^2 \ge 0$ with probability $1$, then its expectation must be non-negative.

Under typical assumptions, it's strictly positive, too. If you look at the formula from the link you posted, $$ \lim_{T\to \infty}\text{Var}[\sqrt{T}(\bar{X}_T - \mu)] = \sum_{i=-\infty}^{\infty} \gamma(i) $$ you might not be able to tell right away because a lot of those terms in the sum might be negative. However, recall that autocovariance functions $\gamma(\cdot)$ are positive definite. So that term has to be positive. The only thing you have to worry about is whether it's $\infty$ or not. But that is usually taken care of by the assumption of absolute summability, or that $\sum_i |\gamma(i)| < \infty$. Finiteness of this implies finiteness of the other.

So is it possible for long-run variance to be negative or 0?

A stationary process can have a long-run variance of $0$, sure. Take a random sample from a distribution that only has probability on $\mu$. Then $\text{Var}(\sqrt{n}(\bar{X}_n - \mu) = 0$ which means its limit is $0$ as well.

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If you use the popular equation Var(X)=E[X²]-E[X]² then the well-known numerical instabilities can cause the result to be negative if E[x] is larger than the standard deviation. But then the result is clearly bad, because variance is never negative, and negative values cause the standard deviation to be undefined.

While this equation is found in many many textbooks and papers (and is undoubtedly mathematically correct), it is all but reliable with finite floating point numbers as used in computers. So it is okay to use this in proofs, but not in code.

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  • $\begingroup$ If numerical instabilities can cause the result to be negative, when why "the result is clearly bad"? $\endgroup$
    – Aqqqq
    Feb 21, 2018 at 9:36
  • $\begingroup$ @Aqqq because it’s negative. $\endgroup$
    – Taylor
    Feb 21, 2018 at 16:47
  • $\begingroup$ What exactly is numerical instabilities? Is it something opposite to this en.wikipedia.org/wiki/Numerical_stability? If yes: how does that result in the variance to be negative? $\endgroup$
    – Aqqqq
    Feb 21, 2018 at 19:02
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    $\begingroup$ The exact reason is catastrophic cancellation. You subtract two values of similar magnitude, and all the first bits are the same, and rounding errors become relatively big. If you round down the first, round up the second, it can become negative. $\endgroup$ Feb 22, 2018 at 7:02

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