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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.

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  • $\begingroup$ is d an irrational number $\endgroup$ – Aksakal Apr 23 at 17:11
  • $\begingroup$ 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. $\endgroup$ – Kiril E. Proykov Apr 23 at 17:38
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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.

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  • $\begingroup$ 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. $\endgroup$ – Kiril E. Proykov Apr 23 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... $\endgroup$ – Richard Hardy May 16 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 May 16 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). $\endgroup$ – Richard Hardy May 17 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 May 17 at 13:19

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