Superposition rule

Here we show how the DFA results for any two signals $f$ and $g$ [denoted as $F_f(n)$ and $F_g(n)$] relate with the DFA result for the sum of these two signals $f+g$ [denoted as $F_{f+g}(n)$, where $n$ is the box length (scale of analysis)]. In the general cases, we find $\vert F_f-F_g\vert\leq F_{f+g} \leq F_f+F_g$. When the two signals are not correlated, we find that the following superposition rule is valid: $F^2_{f+g}=F^2_{f}+F^2_{g}$. Here we derive these relations.
First we summarize again the procedure of the DFA method[3]. It includes the following steps: starting with an original signal $u(i)$ of length $N_{max}$, we integrate and obtain $y(k)=\sum\limits_{j=1}^{k}(u(j)-\langle u\rangle)$, where $\langle u \rangle$ is the mean of $u(i)$. Next, we divide $y(k)$ into non-overlapping boxes of equal length $n$. In each box we fit the signal $y(k)$ using a polynomial function $y_n(k)=a_0+a_1x(k)+a_2x^2(k)+...+a_sx^s(k)$, where $x(k)$ is the $x$ coordinate corresponding to the $k$th signal point. We calculate the r.m.s. fluctuation function $F(n)=\sqrt{\frac{1}{N_{max}}\sum\limits^{N_{max}}_{k=1} [y(k)-y_{n}(k)]^2}$.
To prove the superposition rule, we first focus on one particular box along the signal. In order to find the analytic expression of best fit in this box, we write

$$I(a_0,...,a_s) =\sum\limits_{k=1}^{n}[y(k)-(a_0+...+a_sx^s(k))]^2,$$ (19)

where $a_m, m=0,...,s$ are the same for all points in this box. ``Best fit'' requires that $a_m, m=0,...,s$ satisfy

$$\frac{\partial I}{\partial a_m}=0,m=0,...s$$ (20)

Combining Eq. (19) with (20) we obtain $s+1$ equations

$$y_{m}=a_0t_{m0}+a_1t_{m1}...+a_st_{ms}, m=0,...,s$$ (21)

where

$$y_{m}=\sum\limits_{k=1}^{n} y(k) x^m(k), t_{mj}=\sum\limits_ {k=1}^{n} x^{m+j}(k), j=0,...,s$$ (22)

From Eqs. (21) we determine $a_0,a_1,..,a_s$.
For the signals $f$, $g$ and $f+g$ after the integration, in each box we have

$$f_{m}=a_0t_{m0}+a_1t_{m1}...+a_st_{ms},m=0,...,s$$

$$g_{m}=a_0^\prime t_{m0}+a_1^\prime t_{m1}...+a_s^\prime t_{ms},m=0,...,s$$

$$(f+g)_{m}=a_0^{\prime \prime}t_{m0}+a_1^{\prime \prime} t_{m1}...+a_s^{\prime \prime} t_{ms},m=0,...,s$$ (23)

where $f_{m}$, $g_{m}$ and $(f+g)_{m}$ correspond to $y_{m}$ in Eqs. (21).
Comparing the three groups of equations in Eqs. (23), we find that, when we add the first two groups together, the left side becomes $f_{m}+g_{m}=(f+g)_{m}$ , which is precisely the left side of the third group of equations. Thus we find

$$a_m^{\prime \prime}=a_m+a_m^\prime,m=0,...,s$$ (24)

and for each point $k$ in every box, the polynomial fits for the signals $f$, $g$ and $f+g$ satisfy

$$(f+g)_{n}(k)=f_{n}(k)+g_{n}(k).$$ (25)

This result can be extended to all boxes in the signals. For the signal $f+g$ we obtain

$$F^2_{f+g}=\frac{1}{N_{max}}\sum\limits_{k=1}^{N_{max}}[f(k)- f_{n}(k)]^2+[g(k)-g_{n}(k)]^2+2[f(k)-f_{n}(k)][g(k)-g_{n}(k)].$$ (26)

After the substitutions $f(k)-f_{n}(k)=Y_f(k)$ and $g(k)-g_{n}(k)=Y_g(k)$, we rewrite the above equation as

$$F^2_{f+g}=\frac{1}{N_{max}}\Bigl [\sum\limits_{k=1}^{N_{max... ...+\frac{2}{N_{max}}\sum\limits_{k=1}^{N_{max}}Y_{f}(k)Y_{g}(k).$$ (27)

In the general case, we can utilize the Cauchy inequality

$$\left\vert\sum\limits_{k=1}^{N_{max}}Y_f(k)Y_g(k)\right\vert... ...1/2}\left( \sum\limits_{k=1}^{N_{max}} (Y_g(k))^2\right)^{1/2}$$ (28)

and we find

$$(F_f-F_g)^2 \leq F^2_{f+g} \leq (F_f+F_g)^2 \Longrightarrow \vert F_f-F_g\vert\leq F_{f+g} \leq F_f+F_g.$$ (29)

From Eqs. (21) for $m=0$, in every box we have $\sum\limits_{k=1}^{n} y(k)=\sum\limits_{k=1}^{n} y_{n}(k)$. Thus we obtain $\sum\limits_{k=1}^{N_{max}}Y_f(k)=\sum\limits_{k=1}^{N_{max}}Y_g(k)=0$ where $Y_f(k)$ and $Y_g(k)$ fluctuate around zero. When $Y_f(k)$ and $Y_g(k)$ are not correlated, the value of the third term in Eq. (27) is close to zero and we obtain the following superposition rule

$$F^2_{f+g}=F^2_{f}+F^2_{g}.$$ (30)