[Math] Gaussian noise have fractal dimension of 1.5

dimension-theory-analysisestimation-theoryfractalsnormal distribution

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))];
indkend=length(kvec);
%--------
% Fractal Dimension
for indk=1:indkend
    k=kvec(indk);
    for m=1:k
        Xend=floor((N-m)/k);
        Xsum=sum(abs(X(m+[1:Xend]*k)-[0; X(m+[1:Xend-1]*k)]));
        Lmk(m)=1/k*1/k*(N-1)/Xend*Xsum;
    end
    AvgLmk(indk)=mean(Lmk);
    Lmk=[];
end
%--------
x=log(kvec);y=log(AvgLmk);
p=polyfit(x,y,1);m=p(1);b=p(2);
fractdim=-m;

Best Answer

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 $T$.

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