# Fractional un-differencing of time-series

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.

• is d an irrational number – Aksakal Apr 23 '19 at 17:11
• 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. – Kiril E. Proykov Apr 23 '19 at 17:38

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 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... – Richard Hardy May 16 '19 at 19:36
• @RichardHardy, I'm using the same notation. For instance, $(1-B)^2x_t$ means $x_t+\phi_1x_{t-1}+\phi_2x_{t-2}$ – Aksakal May 16 '19 at 20:09
• 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). – Richard Hardy May 17 '19 at 6:32
• $\Theta(B)$ 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 $x_t+\phi_2x_{t-2}$ – Aksakal May 17 '19 at 13:19