Can every non stationary time series be converted to a stationary time series by applying differencing? Also, how do you decide the order of the differencing to be applied?

Do you just difference with intervals 1,2...n, and perform unit root test of stationary each time to see if the resulting series is stationary?


2 Answers 2


No. As a counterexample, let $X$ be any random variable and let the time series have the value $\exp(t X)$ at time $t$. The $k^\text{th}$ difference at time $i=0, 1, 2, \ldots$ is a linear combination

$$\Delta^k(i) = \sum_{j=0}^k w_j \exp((i+j)X) = \exp(iX) \sum_{j=0}^k w_j \exp(jX) = \exp(iX) \Delta^k(0).$$

for coefficients $w_j$ (which can be computed but whose values are irrelevant for this discussion). Unless $X$ is constant, the left and right sides have different distributions, proving the $k^\text{th}$ difference is not stationary. Therefore no amount of differencing will make this time series stationary.

  • $\begingroup$ So given a time series(linear), how do you know if it can ever be differenced to form a stationary series? $\endgroup$
    – Victor
    Commented Nov 17, 2016 at 16:25
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    $\begingroup$ Please explain what you mean by a "linear" time series. In general, the process of fitting an AR model amounts to estimating the amount of differencing needed to make the series stationary. $\endgroup$
    – whuber
    Commented Nov 17, 2016 at 16:26
  • $\begingroup$ Thanks..let me think about that. I dont know how much I dont know $\endgroup$
    – Victor
    Commented Nov 17, 2016 at 16:29
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    $\begingroup$ This appears to be a consequence of the fact that the exponential function is its own derivative, and that immediately suggests to me that a time series can be made stationary by repeated differencing if and only if the "true" function it models is a polynomial (or, equivalently, its Taylor series expansion is finite). $\endgroup$
    – zwol
    Commented Nov 17, 2016 at 19:42
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    $\begingroup$ @zwol That's good insight--and it's why the exponential counterexample was the first to come to mind--but it's only part of the story. If the expectation is a polynomial function of time, then sufficient differencing will render the time series first order stationary: that is, the first moments of the distributions will be invariant over time. However, differencing will not necessarily make higher moments or multivariate moments stationary. $\endgroup$
    – whuber
    Commented Nov 17, 2016 at 20:04

The answer by whuber is correct; there are lots of time-series that cannot be made stationary by differencing. Notwithstanding that this answers your question in a strict sense, it might also be worth noting that within the broad class of ARIMA models with white noise, differencing can turn them into ARMA models, and the latter are (asymptotically) stationary when the remaining roots of the auto-regressive characteristic polynomial are inside the unit circle. If you specify an appropriate starting distribution for the observable series that is equal to the stationary distribution, you get a strictly stationary time-series process.

So as a general rule, no, not every time-series is convertible to a stationary series by differencing. However, if you restrict your scope to the broad class of time-series models in the ARIMA class with white noise and appropriately specified starting distribution (and other AR roots inside the unit circle) then yes, differencing can be used to get stationarity.

  • 3
    $\begingroup$ +1 Arguably, for some (many?) applications this is a more useful answer than the purely theoretical one I offered. $\endgroup$
    – whuber
    Commented Mar 13, 2018 at 15:02
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    $\begingroup$ Yes - I think sometimes its a matter of "Here's the answer to your question, and now here's the answer to a different question that you should also have asked." $\endgroup$
    – Ben
    Commented Mar 13, 2018 at 19:16

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