I have time series y that has long-memory 0 < d < 1.

I have differenced y with order d, resulting into y'.

I would like to now restore y' back into y.

Do you guys know how to do that?

P.S. Functions in R are also welcomed if you know any.

  • $\begingroup$ is d an irrational number $\endgroup$
    – Aksakal
    Commented Apr 23, 2019 at 17:11
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    $\begingroup$ Yes it can be irrational. I have noticed that to integrate y' back to y I can do the following: y'' = fracdiff(y',-d) + y'[1]. Where y'' is the reintegrated series, fracdiff is a fractional differencing function that takes two arguments: series and differencing parameter and y'[1] is the first element of the y' variable. So I just reverse the sign to minus and add the first element. However, I have no mathematical proof that this is the proper way of doing + it gets some noise when I do that. $\endgroup$ Commented Apr 23, 2019 at 17:38
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    $\begingroup$ @KirilE.Proykov, any chance you found a solution? I have the same problem, only worse - when I do fracdiff(y', -d) + y'[1] what I get back is not even close to the original series. I'm using R's fracdiff package. What are you using? $\endgroup$
    – Parzival
    Commented May 21, 2021 at 19:29
  • $\begingroup$ @KirilE.Proykov Hey did you manage to find an answer? I just tried the method you suggested and the results are a little of. Also I find that if I do it this way then I loose a lot of entries.. $\endgroup$ Commented Mar 17, 2022 at 23:53
  • $\begingroup$ Not sure if still of interest, but I applied the same methodology as in this post: quant.stackexchange.com/questions/17661/unsmoothing-of-returns , and it seems to be working. Comparing the un-differenced with original data is basically the same (MSE close to 0). The only caveat is that this only works if you (i) use a fixed sized window when differencing, and (ii) keep track of the first window original data points. $\endgroup$
    – Tingiskhan
    Commented Nov 17, 2022 at 20:24

1 Answer 1


This paper offers one way to integrate the fractional integrated process where $d\in(-1/2,1/2)$: Reisen, V. A. and Lopes, S. (1999) Some simulations and applications of forecasting long-memory time series models; Journal of Statistical Planning and Inference, 80, 269–287 You can find it here.

The idea's that when you have ARFIMA(p,d,q) process like: $$\Phi(B)(1-B)^dX_t=\Theta(B)\varepsilon_t$$ with $B$ - backshift operator, you can represent it as infinite AR process: $$X_{t+k}=-\sum_{j=1}^\infty\pi_jX_{t+k-j}+\varepsilon_{t+l}$$ They have an equation for coefficients $\pi_j$ as a function of integration order. For instance, for a ARFIMA(0,d,0) you'd get: $$\pi_1=-\frac 1 {d+1}$$ then $$X_{t+1}=\frac 1 {d+1}X_{t} -\dots +\varepsilon_{t+l}$$

I wound't do it manually, and instead get a stat package that does it for me.

  • $\begingroup$ Thanks! Nice answer. I noticed that when I reverse the sign of the differencing parameter it integrates the series back to the original with some noise and shift equal to the first element of the series. However, I wanted to see a formal explanation of how the process works. $\endgroup$ Commented Apr 23, 2019 at 17:42
  • $\begingroup$ I suppose you define $B$ differently in this post than in your answer to this one. Coming back to the question on novel and potentially misleading notation there... $\endgroup$ Commented May 16, 2019 at 19:36
  • $\begingroup$ @RichardHardy, I'm using the same notation. For instance, $(1-B)^2x_t$ means $x_t+\phi_1x_{t-1}+\phi_2x_{t-2}$ $\endgroup$
    – Aksakal
    Commented May 16, 2019 at 20:09
  • $\begingroup$ So what does $\Theta(B)$ mean then? The same as $(1-B)^k$ for some $k$? I am still curious to see any other source (textbook, paper, whatever) that interprets $(1-B)^k$ the way you do. I mean, why not use the commonly accepted time series notation of backshift operators since we are on a statistics site and everyone else around here uses it (as far as I have seen). $\endgroup$ Commented May 17, 2019 at 6:32
  • $\begingroup$ $\Theta(B)$ is a polynomial on differencing, a more general form form than $(1-B)^k$, in the latter all orders of differencing are present, while in former some orders can be dropped, like $x_t+\phi_2x_{t-2}$ $\endgroup$
    – Aksakal
    Commented May 17, 2019 at 13:19

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