From what I understand, a correlogram plots the autocorrelation of a stochastic process as a function of the delay. However, that is only possible in the case of wide-sense stationary processes, since the autocorrelation otherwise depends on the two precise moments in time, not just on the delay...

Am I correct in stating that correlograms can only be plotted for wide-sense stationary processes?


1 Answer 1


Your reasoning makes sense.

For a wide-sense stationary process, a correlogram can be interpreted in a straightforward way, since the population counterparts of the sample autocorrelations you see in it do actually exist.

For a process that is not wide-sense stationary, it is still technically possible to plot a correlogram. However, it loses its straightforward interpretation, as the population counterparts of sample autocorrelations do not exist. The correlogram may still be informative, though, as e.g. we can identify (or at least get a sense of) a unit-root, a long-memory or a trending process by looking at a correlogram.

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    $\begingroup$ To see this, play with simple examples, such as pure linear trend or trend plus some noise. Long lag autocorrelations have meaning as the correlations of $y_i$ and $y_{i+k}$ but they tell you mostly that there is a trend, and aren't informative otherwise. People often seem to forget that autocorrelations are essentially correlations so that the corresponding scatter plot underlines and allows interpretation of what is being calculated. $\endgroup$
    – Nick Cox
    Jan 4, 2022 at 12:46
  • $\begingroup$ +1: The phrasing in the second sentence is a bit awkward. Maybe rephrase along the lines of "For a wide-sense stationary process, a correlogram can be interpreted in a straightforward way, as a plot of sample estimates of the corresponding population autocorrelations." $\endgroup$ Jan 4, 2022 at 15:10
  • $\begingroup$ @ColorStatistics, mind the comma. I would agree with your suggestion if it did not contain a comma before as. I think my phrasing is OK (though certaintly not perfect), and the comma before as plays an important role there. I have now rephrased a bit nevertheless, hopefully for the better. $\endgroup$ Jan 4, 2022 at 17:43

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