In Advances in Financial Machine Learning the author makes a case for fractionally differentiated price returns in chapter 5. The reason is to both maintain memory and to generate a stationary time series. The author argues that the first order derivative of prices, price returns, achieve stationarity but experience memory loss. He defines the backshift operator $B$ as

$B^{k}X_{n} = X_{n-k}$

and then applies the binomial expansion on this

$(1-B)^{d} = \sum_{n=1}^{\infty} {d \choose k} (-B)^{k}$

This then related to how memory can be preserved when d is a positive real

$\hat X_{t} = \sum_{k=0}^{\infty} w_{k} X_{t-k}$

Where the weights, $w$ are expanded using the same binomial method as $B$

FIGURE 5.1 𝜔k (y-axis) as k increases (x-axis). Each line is associated with a particular value of d ∈
[0,1], in 0.1 increments.

Other than the binomial expansion, what is the relationship between $B$ and $w$ and the relevance of the graph?

This is the caption of the graph

FIGURE 5.1 𝜔k (y-axis) as k increases (x-axis). Each line is associated with a particular value of d ∈ [0,1], in 0.1 increments.


1 Answer 1


Other than the binomial expansion, what is the relationship between $B$ and $w$?

As I understand correctly, there is no immediate relationship between the lag operator $B$ and the weights $w$. However, going from the fractional differencing operator, $(1-B)^d$ to $\hat{X}_t$ can be done as follows: \begin{align} (1-B)^d X_t &= \sum_{k=0}^{\infty} \begin{pmatrix} d \\ k\\ \end{pmatrix} (-B)^k \cdot X_t\\ &=X_t \sum_{k=0}^{\infty} (-B)^k \prod_{i=0}^{k-1}\frac{d-i}{k-i}\\ &= X_t \left(1-d\cdot B + \frac{d(d-1)}{2!}B^2 - \frac{d(d-1)(d-2)}{3!}B^3 + \cdots\right)\\ &= X_t - d\cdot X_{t-1} + \frac{d(d-1)}{2!} X_{t-2} - \frac{d(d-1)(d-2)}{3!} X_{t-3} + \cdots \\ &= \sum_{k=0}^{\infty} \omega_kX_{t-k}\\ &= \hat{X}_t \end{align} where $\omega=\left\{1,-d,\frac{d(d-1)}{2!},-\frac{d(d-1)(d-2)}{3!},\ldots\right\}$ is the binomial expansion, as noted. Furthermore, we have used (per definition) the lag-operator applied to $X_t$, namely $B^k X_t = X_{t-k}$, in the second last equality. I believe it was the authors intention to simplify the notation and reformulate it into something more recognizing for people within finance.

What is the relevance of the graph?

The graph shows you the memory-retainment (in terms of $k$ included lags of log-prices, $X_{t-1},X_{t-2},\ldots$ used to differencing the price-process) of fractional differenced processes, for different values of $d\in [0,1]$. As the graph depicts, there exist a memory cut-off for the first-differenced price process ($d=1$), in the sense that the history of the previous prices is disregarded when transforming the process into a stationary time-series. This is also one of the motivations for using fractional differencing (FD), since log-prices $X_t$, depends on many previous lags and not only the first lag, $X_{t-1}$. In order to preserve this dependence structure (memory), we resort to FD for $d<1$. If we depict the fractionally differentiated returns for two values of $d=\{0.5,1\}$ you get: \begin{align} &d=1: \qquad \; \: \: \hat{X}_t=X_t - X_{t-1}\\ &d=0.5: \qquad \hat{X}_t=X_t - 0.5X_{t-1} - 0.125 X_{t-2} - 0.0625 X_{t-3} -\cdots, \end{align} and conclusively for $d=1$, we have the standard first-differencing of the log-prices, and clearly see a cut-off at $k=2$, since $\omega_k=0$ for all $k\geq 2$. This does not happen for $d=0.5$, since $\omega = \{1,-0.5,-0.125,-0.0625,\ldots\}$.

Extra information: In one of his slidesets he argues that there exists a trade-off between stationarity and memory by choosing $d$ accordingly. However, as seen in his own example (example 2: He uses an Augmented Dickey-Fuller test to test the time-series under different values of $d$), the choice of $d$ is clearly sensitive to the choice of asset and asset-class. This makes it hard to hone in on one optimal choice of $d$, if you are a portfolio manager working with a large set of assets.

I hope this helps!

  • 1
    $\begingroup$ thanks! this is great, I'm not sure how to message on this website, but this was helpful and if you're on discord I was curious if you wanted to jump on a call and discuss. Alternatively, I'll think about it and post follow up questions here $\endgroup$
    – VVKK77
    Apr 13, 2021 at 1:20
  • 1
    $\begingroup$ Hi @VVKK77 and thank you. I do not have discord, but feel free to post more questions to quant SE :-). $\endgroup$
    – Pleb
    Apr 13, 2021 at 7:52
  • $\begingroup$ As a follow up question, what impact does the alternating sign have on the binomial expansion of the price differential? Why would we want to change the direction of the price deltas? $\endgroup$
    – VVKK77
    Apr 13, 2021 at 16:37
  • 1
    $\begingroup$ @VVKK77, The alternating sign for the algebraic formula in the first part, stems from the fact, that we need all of the binomial expansions to become negative and thus subtract $X_t$ with its previous lags. Therefore, we negate every second term in the binomial expansion. However, I made a mistake in my example (for $d=0.5$) and still alternated the lags, which is not true. I have corrected this and therefore no change in sign happens (change in direction). Sorry for the confusion. $\endgroup$
    – Pleb
    Apr 13, 2021 at 17:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.