In a paper I'm trying to understand, the following time series is generated as "simulated data":

$$Y(i)=\sum_{j=1}^{1000+i}Z(j) \:\:\: ; \:\:\: (i=1,2,\ldots,N)$$

where $Z(j)$ is a Gaussian noise with mean $0$ and standard deviation $1$.

The paper is about estimating the fractal dimension of $Y$ (not to mention some other series). The author says that the fractal dimension of $Y$ is $1.5$, but doesn't explain why.

My first question is why should we expect the fractal dimension of $Y$ to be $1.5$?

My second question is: can anyone think of a reason why I keep getting a fractal dimension of about $2$ for $Y$, when using the technique described in the paper?

That technique, summarized, is:

1) Create a new set of time series based on the original $Y$ as follows:

$$X^m_k=Y(m),Y(m+k),Y(m+2k),\ldots,Y\left(m+\left \lfloor \frac{N-m}{k}\right \rfloor k\right) \:\:\:\: ; \:\:\:\: (m=1,2,\ldots,k)$$

where the notation $\left \lfloor \right \rfloor$ denotes the floor function, and $k=1,2,3,4,\ldots$. But for $k>4$, $k=\lfloor 2^{(j-1)/4} \rfloor$ where $j=11,12,13,\ldots$

2) Define and calculate the "length" of each "curve" $X^m_k$ as follows:

$$L_m(k)=\frac{1}{k} \frac{N-1}{Qk} \left( \sum_{i=1}^{Q} \left | X(m+ik)-X(m+(i-1)k) \right | \right)$$

where $Q=\left \lfloor\frac{N-m}{k} \right \rfloor$

3) For each $k$, define the average $L_m(k)$ (averaged over $m$) as $y(k)=\langle L_m(k) \rangle$, and then scatter plot $\ln(y(k))$ against $\ln(k)$ and fit a line via least squares. The line should be straight. And the slope of the line should be about $-1.5$, the negative of which is then interpreted as an estimation of the fractal dimension.

When I follow these steps, I get a straight line, but slope of $-2$.

The paper is Higuchi. 1988. "Approach to an irregular time series on basis of fractal dimension."

EDIT: Did has given an answer which I originally accepted, but then un-accepted. I un-accepted the answer because I do not completely follow the steps in his proof, especially the last step; and because it is well known within fractal dimension theory that any space filling curve should have a fractal dimension close to 2. Gaussian noise (a.k.a. white noise), is space filling, and my code gives me the slope of -2. I include my matlab code below. Appreciate if anyone can point out an error in the logic, but as far as I can tell it is exactly the algorithm described by Higuchi, and frankly it has been working quite well for me. For eg., it gives sensible results for so-called "Brown(ian)" noise (FD=1.5), which is halfway between the total randomness of white noise and the total determinism of a sine wave.

function [fractdim]=FractDim(Data,jstart,jend)

kvec=[1:4 floor(2.^([jstart:jend]./4))];
% Fractal Dimension
for indk=1:indkend
    for m=1:k
        Xsum=sum(abs(X(m+[1:Xend]*k)-[0; X(m+[1:Xend-1]*k)]));
  • $\begingroup$ Is there suddenly a problem with my answer, explaining that you (silently) unaccept it? $\endgroup$
    – Did
    Mar 5, 2014 at 22:07
  • $\begingroup$ @Did Yes. Thanks again for that answer, which is instructive in and of itself. However, recently I have had occasion to revisit this matter in light of new readings and must reconsider its acceptance. It is fractal Brownian noise (a.k.a. red noise) which should have a FD of 1.5. The term "Gaussian noise," which I used to think referred exclusively to white noise, can also refer to fBn. Even so, the noise formula $Y(i)$ Higuchi gives yields an FD of 2, not 1.5. This is confirmed by plotting the periodogram and seeing that it is flat (i.e. power law index a=0). $\endgroup$
    – ben
    Mar 6, 2014 at 4:01
  • $\begingroup$ Which part of the argument in this answer do you have trouble with? $\endgroup$
    – Did
    Mar 6, 2014 at 6:27
  • $\begingroup$ @Did Your answer asserts that the series $Y$ has FD=1.5 when in fact it has FD=2.0. $\endgroup$
    – ben
    Mar 6, 2014 at 12:14
  • $\begingroup$ No my answer does not "assert" that, it proves it. You, on the contrary, are invoking results which might or might not correspoond to the setting you described in your question. Please refer to what is written in my answer (did you actually read it?) to mention its specific faulty steps, if any. $\endgroup$
    – Did
    Mar 6, 2014 at 16:02

3 Answers 3


T. Higuchi, Approach to an irregular time series on basis of fractal dimension, Physica D: Nonlinear Phenomena, Volume 31, Issue 2, June 1988, Pages 277–283.

My second question is: can anyone think of a reason why I keep getting a fractal dimension of about 2 for Y, when using the technique described in the paper?

This question is difficult to answer since you do not show your computations but $D=\frac32$.

To show this, note that for every process with stationary increments such as $(Y(i))_i$, $$ \mathbb E(L(k))=\frac1k\frac{N-1}{Qk}Q\,\mathbb E(|Y(k)-Y(0)|), $$ and that in the present case $\mathbb E(|Y(k)-Y(0)|)=\sqrt{k}\,\mathbb E(|Z|)$ since $Y(k)-Y(0)$ is the sum of $k$ i.i.d. standard normal random variables hence equals $\sqrt{k}|Z|$ in distribution with $Z$ standard normal. Thus, $D=\frac32$ since $$ \mathbb E(L(k))=\frac{N-1}{k\sqrt{k}}\,\mathbb E(|Z|). $$ The same computation yields $D=2$ when $(Y(i))_i$ is i.i.d. and $D=1$ when $(Y(i))_i$ is regular, say $Y(i)=Y(0)+iT$ for some integrable random variable $T$.


I've repeated Higuchi calculations and got precisely the same answer as he promised D=-1.5143. Pay close attention to the number of times he divides the series length Lmk by k. That was my mistake for the first time, when I've lost the final averaging between the set of k series and we moved from Lkm to <Lkm>. Here the pointy brackets stands for additional averaging.

And if you want to know "why?" ... I think it does not matter here. As far as I understand you, you are frustrated that your code isn't working ))


Concerning your second question, in Higuchi's paper Y(i) is Brownian motion generated by integrating over a N = 2^17 + 1000 length Gaussian noise signal Z(j) such that

Y(1) = sum(Z(1:1001))


Y(end) = sum(Z(1:end)).

If you used a for loop to generate Y(i), and were (accidentally) recreating Z each loop, you would end up calculating a FD of ~2 for Y.

  • $\begingroup$ Using MathJax would certainly make your answer more readable $\endgroup$
    – Shailesh
    Nov 18, 2016 at 22:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .