Classification of complex systems by their sample-space scaling exponents

Imagine the ordinary random walk with two possibilities at any timestep—a step to the left, or to the right. The number of possible configurations (i.e. possible paths) after N steps is

$$\begin{eqnarray}&&{W}_{}(N+1)=2{W}_{}(N),\end{eqnarray} \tag{ 23 }$$

which means exponential phasespace growth, ${W}_{}(N)={2}^{N}$. We obtain l = 1, ${c}_{0}^{(1)}\,=\,1$ and ${c}_{j}^{(1)}=0$, for j ≥ 1, and for the exponents of the entropy d₀ = 0, ${d}_{1}\equiv {d}_{0}^{(1)}\,=\,1$ and d_j=0, for j ≥ 2. This set of exponents belongs to the class of (c, d)-entropies described in [7] for c = 1 − d₀ = 1, and d = d₁ = 1. They correspond to the scaling exponents of the Shannon entropy: from (18) we obtain that $g(x)\sim x\mathrm{log}x$ and from (19) we get $S(W)\sim \mathrm{log}W$, which is Boltzmann entropy. It is not immediately apparent what the entropy of a random walk should be. However, the random walk is equivalent to spin system of N independent spins, the 2^N different paths correspond one-to-one to the 2^N configurations in the spin model, where the role entropy of it is clear. Obviously, for the random walk, (SK1–3) are applicable.

In this variation of the random walk we impose correlations on the walk. After the first random choice (left or right) the walker goes one step in that direction. The second random choice is followed by two steps in the same direction, the next step is followed by three steps in the same direction, etc. For k independent choices, one has to make $N={\sum }_{i=1}^{k-1}i=1/2k(k-1)$ steps. For this walk, we get that the number of possible paths is

$$\begin{eqnarray}&&{W}_{}(N+k)=2{W}_{}(N),\end{eqnarray} \tag{ 24 }$$

which leads to $W(N)={2}^{N/k}\sim {2}^{k/2}$. For N ≫ 1, we have $k\approx \sqrt{N}$, and we obtain a stretched exponential (sub-exponential) asymptotic behavior, ${W}_{}(N)\sim {2}^{\sqrt{N}}$. The order is again l = 1 and the exponents are ${c}_{0}^{(1)}\,=\,1/2$ and ${c}_{j}^{(1)}=0$, for j ≥ 1. In terms of the d exponents we have d₀ = 0 and ${d}_{1}\equiv {d}_{0}^{(1)}\,=\,2$. Therefore, the three SK axioms are applicable and the resulting extensive entropy belongs to the class of entropies characterized by the Anteneodo–Plastino entropy, since we have $g(x)\sim x{(\mathrm{log}x)}^{2}$ and $S(W)\sim {(\mathrm{log}W)}^{2}$. This entropy is the special case of the (c, d)-entropy for c = 1 and d = 2, see [7].

Consider N coins with two states (up or down). These coins are magnetic, so that any two can stick to each other to create a pair which is a third state obtained by interactions of elements (one possible configuration). As mentioned before, in [11] it is shown that the phasespace volume can be obtained recursively

$$\begin{eqnarray}&&{W}_{}(N+1)=2{W}_{}(N)+{{NW}}_{}(N-1).\end{eqnarray} \tag{ 25 }$$

For $N\gg 1$, we get ${W}_{}(N)\sim {N}^{N/2}{{\rm{e}}}^{2\sqrt{N}}$, which yields l = 1, and the scaling exponents ${c}_{0}^{(1)}\,=\,1$, ${c}_{1}^{(1)}\,=\,1$ and ${c}_{j}^{(1)}=0$, for $j\geqslant 2$. The scaling exponents of the entropy are d₀ = 0, ${d}_{1}\equiv {d}_{0}^{(1)}\,=\,1$, and ${d}_{2}\equiv {d}_{1}^{(1)}=-1$. For the entropy this means, that $g(x)\sim x{\rm{log}}x/{\rm{log}}({\rm{log}}x)$ and $S(W)\sim {\rm{log}}W/{\rm{log}}({\rm{log}}W)$. This case is not contained in the class of (c, d)-entropies, because the third exponent, corresponding to the doubly-logarithmic correction, is not zero. Actually we obtain c = 1 and d = 1, which would naively indicate Shannon entropy. However, the correction makes the system clearly super-exponential. The SK axioms are still applicable, the class of accessible entropy formulas is restricted by (SK2). For example, for the representative entropy equation (20) we find that a₀ ≥ 0 and a₁ ≥ 0, see appendix.

Imagine a random network with N nodes. When a new node is added, there emerge N new possible links, which gives us 2^N new possible configurations for each configuration of the network with N links. We obtain the recursive growth equation

$$\begin{eqnarray}&&{W}_{}(N+1)={2}^{N}{W}_{}(N),\end{eqnarray} \tag{ 26 }$$

which leads to ${W}_{}(N)={2}^{\displaystyle \left(\genfrac{}{}{0em}{}{N}{2}\right)}$, as expected. For this phasespace growth, we obtain l = 1, ${c}_{0}^{(1)}=2$ and ${c}_{j}^{(1)}=0$ for $j\geqslant 1$, and d₀ = 0 and ${d}_{1}\equiv {d}_{0}^{(1)}=\tfrac{1}{2}$. The corresponding entropy can be expressed by $g(x)\sim x{(\mathrm{log}x)}^{1/2}$, and $S(W)\sim {(\mathrm{log}W)}^{1/2}$. The entropy corresponds to the class of compressed exponentials, which are super-exponential, however, the entropy belongs to the class of (c, d)-entropies for c = 1 and d = 1/2. Because all exponents are positive the entropy observes the SK axioms.

Consider a generalization of the random walk, where a walker can take a left or right step, but it can also split into two walkers, one of which then goes left, the other to the right. Each walker can then go left, right, or split again (multiple walkers can occupy the same position). The number of possible paths after N steps is

$$\begin{eqnarray}&&{W}_{}(N+1)=2{W}_{}(N)+{W}_{}{(N)}^{2},\end{eqnarray} \tag{ 27 }$$

where the first term reflects the left/right decisions, the second the splittings. We have ${W}_{}(N)={2}^{({2}^{N-1})}-1$, and find that l = 2, ${c}_{0}^{(2)}=1$ and ${c}_{j}^{(2)}=0$, for $j\geqslant 1$, and ${d}_{0}=0$, ${d}_{1}=0$ and ${d}_{2}\equiv {d}_{0}^{(2)}\,=\,1$. The corresponding extensive entropy is $g(x)\sim x\,\mathrm{loglog}(x)$ and scales as $S(W)\sim \mathrm{loglog}W$. Because the coefficients are not negative, SK axioms are applicable. However, even though all correction scaling exponents are zero, the system cannot be described in terms of (c, d)-entropies, because l = 2. We would naively obtain that c = 1 and d = 0, which would wrongly correspond to Tsallis entropy. Alternatively, we can think of an example of a spin system with the same scaling exponents. In this case, N would not describe the size of a system, but its dimension. For N = 1, we would have two particles on the line, for N = 2 we have 4 particles forming a square, for N = 3 we have a cube with 8 particles in its vertices, etc. In general, we can think of a spin system of particles sitting on the vertices of a N-dimensional hypercube. The number of particles is naturally 2^N and for two possible spins we obtain $W(N)={2}^{({2}^{N})}$.

The SK axioms read:

• (SK1) Entropy is a continuous function of the probabilities p_i only, and should not explicitly depend on any other parameters.
• (SK2) Entropy is maximal for the equi-distribution ${p}_{i}=1/W$.
• (SK3) Adding a state $W+1$ to a system with ${p}_{W+1}=0$ does not change the entropy of the system.
• (SK4) Entropy of a system composed of 2 sub-systems A and B, is $S(A+B)=S(A)+S(B| A)$.

They state requirements that must be fulfilled by any entropy. For ergodic systems all four axioms hold. For non-ergodic ones the composition axiom (SK4) is explicitly violated, and only the first three (SK1–SK3) hold. If all four axioms hold the entropy is uniquely determined to be Shannon's; if only the first three axioms hold, the entropy is given by the (c, d)-entropy [7, 10]. The SK axioms were formulated in the context of information theory but are also sensible for many physical and complex systems.

Given a trace form of the entropy as in equation (9), the SK axioms imply the restrictions on g(x): (SK1) implies that g is a continuous function, (SK2) means that g(x) is concave, and (SK3) that $g(0)=0$. For details, see [7].

We first prove a theorem which determines the general form of rescaling relations in the thermodynamic limit for any general function.

Theorem. Let g(x) be a positive, continuous function on ${{\mathbb{R}}}^{+}$. Let us define the function $z(\lambda ):{{\mathbb{R}}}^{+}\to {{\mathbb{R}}}^{+}$

$$\begin{eqnarray}&&z(\lambda ):= \mathop{\mathrm{lim}}\limits_{x\to \infty }\displaystyle \frac{g({r}_{\lambda }^{(n)}(x))}{g(x)}.\end{eqnarray} \tag{ A.1 }$$

Then, $z(\lambda )={\lambda }^{c}$ for some $c\in {\mathbb{R}}$.

Proof. From the definition of $z(\lambda )$, it is straightforward to show that $z(\lambda \lambda ^{\prime} )=z(\lambda )z(\lambda ^{\prime} )$, because

$$\begin{eqnarray*}z(\lambda \lambda ^{\prime} ) & = & \mathop{\mathrm{lim}}\limits_{x\to \infty }\displaystyle \frac{g({r}_{\lambda \lambda ^{\prime} }^{(n)}(x))}{g(x)}=\mathop{\mathrm{lim}}\limits_{x\to \infty }\displaystyle \frac{g({r}_{\lambda \lambda ^{\prime} }^{(n)}(x))}{g({r}_{\lambda }^{(n)}(x))}\displaystyle \frac{g({r}_{\lambda }^{(n)}(x))}{g(x)}\\ & = & \mathop{\mathrm{lim}}\limits_{{r}_{\lambda }^{(n)}(x)\to \infty }\displaystyle \frac{g({r}_{\lambda ^{\prime} }^{(n)}[{r}_{\lambda }^{(n)}(x)])}{g({r}_{\lambda }^{(n)}(x))}\mathop{\mathrm{lim}}\limits_{x\to \infty }\displaystyle \frac{g({r}_{\lambda }^{(n)}(x))}{g(x)}=z(\lambda ^{\prime} )z(\lambda ).\end{eqnarray*}$$

For the computation we used the group property of rescaling in equation (3) and the continuity of g. The only class of functions satisfying the functional equation above are power functions, $z(\lambda )={\lambda }^{c}$. □

Let us take the first scaling relation of the sample space $W({r}_{\lambda }^{(0)}(N))=W(\lambda N)$. From the previous theorem we obtain

$$\begin{eqnarray}&&\displaystyle \frac{W(\lambda N)}{W(N)}\sim {\lambda }^{{c}_{0}}\ \Rightarrow \ W({r}_{\lambda }^{(0)}(N))\sim {r}_{{\lambda }^{{c}_{0}}}^{(0)}(W(N)).\end{eqnarray} \tag{ A.2 }$$

It may happen that c₀ is infinite. Thus, we may need to use higher-order scaling for the sample space, i.e., ${r}_{{\lambda }^{{c}_{0}}}^{(l)}(W(N))$, as shown in the main text. l is determined by the condition that the scaling exponent should be finite. The first correction term is given by the scaling $W({r}_{\lambda }^{(1)}(N))=W({N}^{\lambda })$. To obtain the sub-leading correction, we have to factor out the leading growth term. This means that the scaling relation for the first sub-leading correction looks like

$$\begin{eqnarray}&&\displaystyle \frac{({\mathrm{log}}^{(l)}W({N}^{\lambda }))/{N}^{{c}_{0}^{(l)}\lambda }}{({\mathrm{log}}^{(l)}W(N))/{N}^{{c}_{0}^{(l)}}}\sim {\lambda }^{{c}_{1}^{(l)}},\end{eqnarray} \tag{ A.3 }$$

which is again a consequence of the above theorem. To obtain the corresponding scaling relations for higher-order scaling exponents for the sample space (A.4), we need to factor out all previous terms corresponding to lower-order scalings, so the scaling relation looks like

$$\begin{eqnarray}&&\displaystyle \frac{{\mathrm{log}}^{(l)}W({r}_{\lambda }^{(k)}(N))}{{\mathrm{log}}^{(l)}W(N)}\displaystyle \prod _{j=0}^{k-1}{\left(\displaystyle \frac{{\mathrm{log}}^{(j)}({r}_{\lambda }^{(k)}(N))}{{\mathrm{log}}^{(j)}(N)}\right)}^{-{c}_{j}^{(l)}}\sim {\lambda }^{{c}_{k}^{(l)}}.\end{eqnarray} \tag{ A.4 }$$

Because the left-hand side of this relation has the form of the function z appearing in the theorem, the validity of the relation is satisfied for $N\to \infty$. Similarly, we can deduce the relations for scaling exponents that are associated with the extensive entropy.

The asymptotic representation of W(N) is obtained by the rescaling that corresponds to the Poincaré asymptotic expansion [13] of ${\mathrm{log}}^{(l+1)}(W)$ in terms of ${\phi }_{n}(N)={\mathrm{log}}^{(n+1)}(N)$ for $N\to \infty$. Let us consider a function f(x) with a singular point at x₀. It is possible to express its asymptotic properties in the neighborhood of x₀ in terms of the asymptotic series of functions ${\phi }_{n}(x)$, if $f(x)={ \mathcal O }({\phi }_{0}(x))$ and ${\phi }_{n+1}(x)={ \mathcal O }({\phi }_{n}(x))$. The series is given as

$$\begin{eqnarray}&&f(x)=\displaystyle \sum _{j=0}^{k}{c}_{j}{\phi }_{j}(x)+{ \mathcal O }({\phi }_{j}(x)).\end{eqnarray} \tag{ A.5 }$$

The coefficients can be calculated from the formulas in [13]

$$\begin{eqnarray}&&{c}_{k}=\mathop{\mathrm{lim}}\limits_{x\to {x}_{0}}\displaystyle \frac{f(x)-{\displaystyle \sum }_{j=0}^{k-1}{c}_{j}{\phi }_{j}(x)}{{\phi }_{k}(x)}.\end{eqnarray} \tag{ A.6 }$$

In our case, i.e., for $N\to \infty$ and ${\phi }_{n}(N)={\mathrm{log}}^{(n+1)}(N)$ the function ${\mathrm{log}}^{(l+1)}(W)$ can be expressed (for appropriate l) in terms of this series, and the coefficients ${c}_{k}^{(l)}$ are given by

$$\begin{eqnarray*}{c}_{k}^{(l)} & = & \mathop{\mathrm{lim}}\limits_{N\to \infty }\displaystyle \frac{{\mathrm{log}}^{(l+1)}(W)-{\displaystyle \sum }_{j=0}^{k-1}{c}_{j}^{(l)}{\mathrm{log}}^{(j+1)}(N)}{{\mathrm{log}}^{(k+1)}(N)}\\ & = & \mathop{\mathrm{lim}}\limits_{N\to \infty }\displaystyle \frac{\mathrm{log}({\mathrm{log}}^{(l)}(W)/{\displaystyle \prod }_{j=0}^{k-1}{\mathrm{log}}^{(j)}{(N)}^{{c}_{j}^{(l)}})}{{\mathrm{log}}^{(k+1)}(N)}.\end{eqnarray*}$$

Using L'Hospital's rule and the derivative of the nested logarithm

$$\begin{eqnarray}&&\displaystyle \frac{{\rm{d}}\,{\mathrm{log}}^{(n)}(x)}{{\rm{d}}x}=\displaystyle \frac{1}{{\displaystyle \prod }_{j=0}^{n-1}{\mathrm{log}}^{(j)}(x)},\end{eqnarray} \tag{ A.7 }$$

a straightforward calculation yields equation (7).

Which entropy functional that fulfills axioms (SK1–3)? The choice is not unique, but a concrete entropy functional serves as a representative of the class in the thermodynamic limit. The requirements imposed by the first three SK axioms are: g(x) is continuous, g(x) is concave, and $g(0)=0$. From equation (18) we have, $g(x)\sim x{\prod }_{j=0}^{n}{\left[{\mathrm{log}}^{(j+l)}\left(\tfrac{1}{x}\right)\right]}^{{d}_{j}^{(l)}}$ for $x\to 0$, which gives us the scaling for the values around zero. Unfortunately, the presented form cannot be extended to the full interval $[0,1]$, because the domain of ${\mathrm{log}}^{(n)}(1/x)$ is $(0,1/{\exp }^{(n-2)}(1))$. This can be fixed by replacing ${\mathrm{log}}^{(n)}$ by ${[1+\mathrm{log}]}^{(n)}=1+\mathrm{log}(1+\mathrm{log}(...))$, which is defined on the whole domain $(0,1]$, where ${\mathrm{lim}}_{x\to 0}{[1+\mathrm{log}]}^{(n)}(1/x)=+\infty$ and ${[1+\mathrm{log}]}^{(n)}(1)=1$. The scaling remains unchanged for $x\to 0$.

The second problem is that in general the function is not concave. For this we introduce the transformation

$$\begin{eqnarray}&&{f}^{\star }(x)={\int }_{0}^{x}\displaystyle \frac{f(y)}{y}{\rm{d}}y.\end{eqnarray} \tag{ A.8 }$$

The original function can be obtained by

$$\begin{eqnarray}&&f(x)=x\displaystyle \frac{{\rm{d}}{f}^{\star }(x)}{{\rm{d}}x}.\end{eqnarray} \tag{ A.9 }$$

This transform turns an increasing/decreasing function to a convex/concave function, while the scaling for $x\to 0$ remains unchanged. Let us write the function g in the form of the transform

$$\begin{eqnarray}&&g(x)\sim {\int }_{0}^{x}\displaystyle \prod _{j=0}^{n}{\left[[1+{\mathrm{log}}^{(n)}]\left(\displaystyle \frac{1}{y}\right)\right]}^{{d}_{j}}\,{\rm{d}}y.\end{eqnarray} \tag{ A.10 }$$

Axiom (SK3) means $g(0)=0$. This requires that the integrand should not diverge faster than $1/x$ for $x\to 0$. This can be fulfilled for ${d}_{0}\equiv {d}_{0}^{(0)}\lt 1$.

Because ${[1+\mathrm{log}]}^{(n)}(1/x)$ is a decreasing function, g(x) is automatically concave if ${d}_{n}\geqslant 0$, since a product of positive, decreasing functions is also decreasing. However, for ${d}_{n}\lt 0$, ${[1+\mathrm{log}]}^{(n)}{(1/x)}^{{d}_{n}}$ is an increasing function from zero to one and the whole product may not be decreasing. In order to solve this issue, we introduce a set of constants a_i and write g(x) in the form

$$\begin{eqnarray}&&g(x)={\int }_{0}^{x}\displaystyle \prod _{j=0}^{n}{\left[{a}_{j}+{[1+\mathrm{log}]}^{(n)}\left(\displaystyle \frac{1}{y}\right)\right]}^{{d}_{j}}\,{\rm{d}}y.\end{eqnarray} \tag{ A.11 }$$

The constants a_i can be chosen to ensure that the integrand is a decreasing function. We assume ${a}_{i}\geqslant -1$ to avoid problems with powers of negative numbers. The second derivative of g(x), i.e., the first derivative of the integrand is an increasing function and $\tfrac{{{\rm{d}}}^{2}g(x)}{{\rm{d}}{x}^{2}}{| }_{x\to {0}^{+}}=-\infty$ for ${d}_{l}\gt 0$. For ${d}_{l}\lt 0$, the entropy cannot be concave, so ${d}_{l}\gt 0$ is the restriction given by (SK2). To obtain a negative second derivative on the whole domain $[0,1]$, it is therefore enough to investigate $\tfrac{{{\rm{d}}}^{2}g(x)}{{\rm{d}}{x}^{2}}{| }_{x=1}$, which leads to the condition

$$\begin{eqnarray}&&\left(\displaystyle \prod _{j=l}^{n}{(1+{a}_{j})}^{{d}_{j}-1}\right)\left(-\displaystyle \sum _{j=l}^{n}\displaystyle \frac{{d}_{j}}{1+{a}_{j}}\right)\leqslant 0.\end{eqnarray} \tag{ A.12 }$$

Because ${d}_{0}^{(l)}\equiv {d}_{l}\gt 0$, we can choose a_l = 0. In the following terms, i.e., for $i\gt l$, d_i can be both positive and negative. Positive d_i pose no problem, because the term corresponding to d_i, i.e. $-{d}_{i}/(1+{a}_{i})$ is negative, so we can choose a_i = 0. When all d_i are negative we can compensate the positive contribution of the negative terms by diminishing them through choice of appropriate a_i. If we choose

$$\begin{eqnarray}&&1+{a}_{i}=-\displaystyle \frac{{d}_{i}^{(l)}}{{{nd}}_{0}^{(l)}},\end{eqnarray} \tag{ A.13 }$$

then equation (A.12) becomes zero. If this is given together with previous results and we summarize it as

$$\begin{eqnarray}&&{a}_{i}=\max \left\{-1-\displaystyle \frac{{d}_{i}^{(l)}}{{{nd}}_{0}^{l}},0\right\},\end{eqnarray} \tag{ A.14 }$$

which has been presented in equation (21) in the main text. Clearly, this is not the only possible choice. Note that for all ${d}_{i}^{(l)}\gt 0$, one may even choose ${a}_{i}=-1$. On the other hand, for the case of the magnetic coin model, one obtains that for ${a}_{0}=0$, ${a}_{1}=0$ as well.

Finally, let us show the connection to (c, d)-entropy derived in [7]. In this case, we assume only d₀ and d₁ can be non-zero, which leads to

$$\begin{eqnarray}&&{g}_{({d}_{0},{d}_{1})}^{(0,1)}(x)={\int }_{0}^{x}{(1/y)}^{{d}_{0}}{(1+{a}_{1}+\mathrm{log}(1/x))}^{{d}_{1}}\,{\rm{d}}y.\end{eqnarray} \tag{ A.15 }$$

By the choice ${a}_{1}=-1+\tfrac{1}{1-{d}_{0}}$, we get

$$\begin{eqnarray}&&{g}_{(c,d)}(x)=\displaystyle \frac{e}{{c}^{d+1}}{\rm{\Gamma }}(1+d,1-c\mathrm{log}x),\end{eqnarray} \tag{ A.16 }$$

for $c=1-{d}_{0}$ and $d={d}_{1}$, which is nothing else than the gamma entropy of [7].

The set of scaling exponents form natural classes of equivalence with natural ordering. Consider two discrete random processes X(N) and Y(N) with sample spaces W_X(N) and W_Y(N), respectively. The corresponding sets of scaling exponents are denoted by ${{ \mathcal C }}_{X}=\{{c}_{0}^{(l)},{c}_{1}^{(l)},...\}$, and ${{ \mathcal C }}_{Y}=\{{\tilde{c}}_{0}^{(\tilde{l})},{\tilde{c}}_{1}^{(\tilde{l})},...\}$. One can introduce an ordering based on the scaling exponents. We write

$$\begin{eqnarray}&&X\prec Y\ (\ {{ \mathcal C }}_{X}\prec {{ \mathcal C }}_{Y})\ \mathrm{if}\ \left\{\begin{array}{l}l\lt l^{\prime} \\ l=l^{\prime} ,{c}_{0}^{(l)}\lt {\tilde{c}}_{0}^{(\tilde{l})}\\ l=l^{\prime} ,{c}_{0}^{(l)}={\tilde{c}}_{0}^{(\tilde{l})},{c}_{1}^{(l)}\lt {\tilde{c}}_{1}^{(\tilde{l})}\\ \mathrm{etc}.\end{array}\right..\end{eqnarray} \tag{ A.17 }$$

This is equivalent to lexicographic ordering. One can also introduce an ordering, which takes into account only certain a number of correcting terms. So, for example

$$\begin{eqnarray}&&X{\prec }_{0}Y\ (\ {{ \mathcal C }}_{X}{\prec }_{0}{{ \mathcal C }}_{Y})\ \mathrm{if}\ \left\{\begin{array}{l}l\lt l^{\prime} \\ l=l^{\prime} ,{c}_{0}^{(l)}\lt {\tilde{c}}_{0}^{(\tilde{l})}\end{array}\right..\end{eqnarray} \tag{ A.18 }$$

Similarly, one can define ≺_k, which takes into account only k correction terms. Additionally, it is possible to introduce an equivalence relation

$$\begin{eqnarray}&&X\sim Y\ \mathrm{if}\ {{ \mathcal C }}_{X}\equiv {{ \mathcal C }}_{Y}\ \Rightarrow \ l=l^{\prime} ;{c}_{i}^{(l)}={\tilde{c}}_{i}^{(\tilde{l})}\ \forall i\end{eqnarray} \tag{ A.19 }$$

and also equivalence up to certain correction

$$\begin{eqnarray}&&X{\sim }_{k}Y\ \mathrm{if}\ {{ \mathcal C }}_{X}\equiv {{ \mathcal C }}_{Y}\ \Rightarrow \ l=l^{\prime} ;{c}_{i}^{(l)}={\tilde{c}}_{i}^{(\tilde{l})}\ \forall i\leqslant k.\end{eqnarray} \tag{ A.20 }$$

As an example, for magnetic coin model and random walk we have that ${X}_{\mathrm{MC}}{\sim }_{0}{X}_{\mathrm{RW}}$, but ${X}_{\mathrm{MC}}{/}\!\!\!\!\!\!\!{\sim }{X}_{\mathrm{RW}}$.

To understand the mechanism of how the scaling exponents correspond to the structure of a random process, let us discuss a simple procedure to generally obtain processes with given scaling exponents ${c}_{k}^{(l)}$. We start with a random variable X₀ with N possible outcomes, so that ${W}_{{X}_{0}}(N)=\{1,\ldots ,N\}$. The scaling exponents of this process are naturally ${c}_{0}^{(0)}\,=\,1$ and ${c}_{k}^{(0)}=0$ for $k\geqslant 1$. Let us construct a new variable by choosing subsets of ${W}_{{X}_{0}}(N)$.

First we can create all possible subsets of ${W}_{{X}_{0}}(N)$. This defines a new variable X₁ with ${W}_{{X}_{1}}(N)={2}^{{W}_{{X}_{0}}(N)}$, and we get ${c}_{0}^{(1)}\,=\,1$. Generally, the transform

$$\begin{eqnarray}&&{\mathfrak{2}}:X\to {{\mathfrak{2}}}^{X},\end{eqnarray} \tag{ A.21 }$$

where ${{\mathfrak{2}}}^{X}$ denotes a variable on all subsets of X. One can easily show that this results in a shift of scaling exponents ${c}_{k}^{(l)}\to {c}_{k}^{(l+1)}$, and ${d}_{k}^{(l)}\to {d}_{k}^{(l+1)}$, because ${W}_{{{\mathfrak{2}}}^{X}}(N)={2}^{{W}_{X}(N)}$. The interpretation of this transformation is the following: consider an ordinary random walk with two possible steps. If ${X}_{0}(N)$ denotes a number of steps of a random walker, then ${X}_{1}(N)={{\mathfrak{2}}}^{{X}_{0}(N)}$ denotes the number of possible paths. When we apply the transform again, we obtain ${X}_{2}(N)={{\mathfrak{2}}}^{{X}_{1}(N)}$. This denotes the number of possible configurations of a random walk cascade, etc. As a result, by more applications of ${\mathfrak{2}}$, we obtain processes with more complicated structure of the respective phasespace.

To construct processes with arbitrary exponents, let us think about a procedure, where we create only partial subsets, which number p(N) can be between N (no partitioning) and ${2}^{N}$ (full partitioning). We denote this procedure by ${\mathfrak{P}}$. This can be understood as a process corresponding to a correlated random walk. This means that not every step of the walk is independent, but some steps can be determined by the previous steps, which diminishes the number of possible configurations when compared to the uncorrelated random walk. The resulting random process is obtained as the composition of l uncorrelated random walks (full partitioning) and a correlated random walk

$$\begin{eqnarray}&&X={{\mathfrak{2}}}^{(l)}[{\mathfrak{P}}({X}_{0})].\end{eqnarray} \tag{ A.22 }$$

Let us now focus on the construction of correlated random walk with a pre-determined number of states given by p(N).

First we consider the full set of subsets of N elements with natural ordering,

$$\begin{eqnarray}&&{W}_{{{\mathfrak{2}}}^{X}}=\{\{\},\{1\},\{2\},\ldots ,\{1,\ldots ,n\}\}.\end{eqnarray} \tag{ A.23 }$$

The correlations can be represented by merging subsets to p(N) sequences of length $\{s(1),\ldots ,s(p(N))\}$, i.e.,

$$\begin{eqnarray}&&{W}_{{\mathfrak{P}}(X)}=\{\mathop{\underbrace{\{\{\},\{1\},\{2\},...\}}}\limits_{s(1)},\ldots ,\mathop{\underbrace{\{...,\{1,\ldots ,n\}\}}}\limits_{s(p(N))}\}.\end{eqnarray} \tag{ A.24 }$$

This means that after one independent step, there are $s(1)-1$ dependent steps, after the second independent step, there are $s(2)-1$ dependent steps, etc. Let us determine the form of function s for given p(N). The function s can be obtained from

$$\begin{eqnarray}&&\displaystyle \sum _{i=1}^{p(N)}s(i)={2}^{N}.\end{eqnarray} \tag{ A.25 }$$

In the limit of large N we can assume that the function s does not depend on N, i.e., is a priori given by the scaling exponents of the system. Let us also assume, without loss of generality, that s is an increasing function (we can neglect the last cell, because its size is determined by the size of previous cells). For $N\gg 1$, we approximate the sum by the integral and obtain

$$\begin{eqnarray}&&{\int }_{0}^{p(N)}s(i){\rm{d}}i\sim {2}^{N}.\end{eqnarray} \tag{ A.26 }$$

Denoting $S(m)={\int }_{0}^{m}s(y){\rm{d}}y$, and substituting $x=p(N)$, we recast the previous equation as, $S(x)={2}^{{p}^{-1}(x)}$, where ${p}^{-1}$ denotes the inverse function of p. The function s(x) can be therefore determined as

$$\begin{eqnarray}&&s(x)=\displaystyle \frac{{\rm{d}}\,{2}^{{p}^{-1}(x)}}{{\rm{d}}x}=\displaystyle \frac{{2}^{{p}^{-1}(x)}}{p^{\prime} ({p}^{-1}(x))}.\end{eqnarray} \tag{ A.27 }$$

Some examples for s(x) for a corresponding p(N) are

• $p(N)={2}^{N}$, i.e., full partitioning corresponding to uncorrelated random walk. In this case, we obtain that $s(x)={\rm{const}}.$, as expected.
• $p(N)=N$, i.e., no partitioning to maximally correlated random walk. We obtain that $s(x)\sim {2}^{x}$, which can be seen from the relation ${\sum }_{i}^{N}{2}^{i}\sim {2}^{N}$.
• $p(N)=N\mathrm{log}N$, which corresponds to the correction in the magnetic coin model. In this case, $s(x)\sim {2}^{W(x)}/\mathrm{log}(W(x))$, where W(x) is the Lambert W-function.