0

When I was looking at time series I noticed that a common approach to time series modelling is the ARIMA model which basically does differencing until a stationary series is found and then fits the ARMA parameters.

Is it possible to construct time series that are infinitely often differentiable but never stationary when differenced?

I did not see how to construct one and was thinking of a series similar to the Weierstrass never differentiable function but this did not seem fruitful.

Ideas would be appreciated!

Jan
  • 255
  • 1
  • 7
  • 1
    How about a time series that monotonically increases through time? For exp. one generated from the exponential function? – user2974951 Oct 17 '18 at 11:51
  • Yes I was also thinking about the exponential function as it is infintely often differentiable. – Jan Oct 17 '18 at 11:56
  • I was hoping to also know if such a series could exist if the series is bounded – Jan Oct 17 '18 at 11:58
  • Relevant/related: https://stats.stackexchange.com/questions/180270/nice-example-where-a-series-without-a-unit-root-is-non-stationary – Glen_b Oct 17 '18 at 13:09
  • Thanks for the link @Glen_b. It is an interesting post and somewhat relevant, however that post is about a time series example that has no unit root but is non-stationary. In my question we can have as many unit roots as we want, I am looking for a series that never becomes sationary when differencing. This is not tackled in the post you linked and thus deserves a new question! – Jan Oct 17 '18 at 13:17
  • Since "time series" is such a broad class of processes, perhaps you could be more specific about what kinds of examples you might be seeking? For instance--to make previous comments explicit--let $K$ be *any* random variable and construct the time series $(e^{tK}),\ t\in\mathbb{R}.$ You could build on that by adding any stationary or bounded or slowly growing series, according to taste. – whuber Oct 17 '18 at 13:48
  • @Jan It is a different question but I believe that two of the answers would be answers to the question here. – Glen_b Oct 17 '18 at 14:29

0 Answers0