I was wondering why in practical terms (aka when modeling and so on), why it's more useful to work a weakly stationary time series than a strict stationary one?


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    $\begingroup$ It is less restrictive but it may not be the most appropriate depending on your series. $\endgroup$ Commented Apr 25, 2017 at 15:32

1 Answer 1


With the stochastic process model for a time series, there is usually just one sample path or realization of the process to work with, and a weakly stationary model is much easier to fit than a a strictly stationary model. Remember that all we need to fit a weakly stationary model is the value of the (constant) mean, and this is easily estimated as $$E[X_i] \approx\mu = \frac 1n \sum_{k=0}^{n-1} x_k$$ from the single available sample path $x_0, x_1, \cdots, x_{n-1}$, Similarly, the autocorrelation function $R_X(\ell) = E[X_iX_{i+\ell}]$ can be estimated as $$R_X(\ell) = E[X_iX_{i+\ell}] \approx \frac 1n \sum_{k=0}^{n-1-\ell} x_kx_{k+\ell}, ~ \ell = 0, 1, 2, \ldots$$ with the caveat that the estimate is likely to be suspect for values of $\ell$ close to $n-1$. In contrast, the fitting of a strictly stationary model requires estimation of the distribution of the $X_i$, the estimation of the joint distribution of $X_i$ and $X_{i+\ell}$ for each $\ell$, the joint distribution of $X_i$, $X_{i+\ell}$ and $X_{i+\ell + m}$, and so on, all of which estimations should be viewed with a great deal of skepticism.

  • $\begingroup$ Aren't all marginal distributions at each time identical in a strictly stationary process? Estimation of marginals should be easy. No? $\endgroup$ Commented Apr 25, 2017 at 17:17
  • $\begingroup$ @CagdasOzgenc With a single sample path, we can estimate the distribution by, for example, ooking at a histogram of $x_0, x_1, \ldots$ but this very act presupposes that the process is stationary. If we had several sample paths $\mathbf x^{(i)}$, we could look at the histogram of $x_i^{(1)}, x_i^{(2)}, x_i^{(3)},\ldots$ to get an empirical distribution for $X_i$ and compare it to the empirical distribution for $X_k$ to see whether the assumption of stationarity is even reasonably defendable, and if so, we can proceed. With a single sample path, this is not possible. $\endgroup$ Commented Apr 25, 2017 at 18:25
  • $\begingroup$ Yes but aren't we in the same situation by assuming weak stationarity? I mean if assumption doesn't hold the results will be not so meaningful in any case, no? $\endgroup$ Commented Apr 25, 2017 at 18:28
  • $\begingroup$ @CagdasOzgenc The point is that weak stationarity requires fewer assumptions than strict stationarity, and the estimation is more robust. As George Cox famously said, All models are incorrect, but some are useful. In this case, weak stationarity is (or can be) just as incorrect as strict stationarity, but it is often more useful, and easier to use, than strict stationarity. $\endgroup$ Commented Apr 25, 2017 at 23:05

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