Superposition law for DFA

For two uncorrelated signals $f(i)$ and $g(i)$, their root mean square fluctuation functions are $F_f(n)$ and $F_g(n)$ respectively. We want to prove that for the signal $f(i)+g(i)$, its rms fluctuation

$$F_{f+g}(n) = \sqrt{F_f(n)^2+F_g(n)^2}$$ (27)

Consider three signals in the same box first. The integrated signals for $f$, $g$ and $f+g$ are $y_f(i)$, $y_g(i)$ and $y_{f+g}(i)$ and their corresponding trends are $y^{fit}_{f}$, $y^{fit}_{g}$,$y^{fit}_{f+g}$ ($i=1,2,...,n$, $n$ is the box size). Since $y_{f+g}(i)=y_{f}(i)+y_g(i)$ and combine the definition of detrended fluctuation function Eq.3, we have that for all boxes

$$Y_{f+g}(i)=Y_f(i)+Y_g(i),$$ (28)

where $Y_{f+g}$ is the detrended fluctuation function for the signal $f+g$, $Y_f(i)$ is for the signal $f$ and $Y_g(i)$ for $g$. Furthermore, according to the definition of rms fluctuation, we can obtain

$$F_{f+g}(n) = \sqrt{\frac{1}{N_{max}} \sum\limits_{i=1}^{N_{... ...m\limits_{i=1}^{N_{max}}\left[Y_{f}(i)+Y_{g}(i)\right ]^2},$$ (29)

where $\ell$ is the number of boxes and $k$ means the $k$th box. If $f$ and $g$ are not correlated, neither are $Y_{f}(i)$ and $Y_{g}(i)$ and, thus,

$$\sum\limits_{i=1}^{N_{max}}Y_{f}(i)Y_{g}(i)=0.$$ (30)

From Eq.30 and Eq.29, we have

$$F_{f+g}(n)=\sqrt{\frac{1}{N_{max}} \sum\limits_{i=1}^{N_{ma... ...} =\sqrt{\left[F_{f}(n)\right ]^2+\left[F_{g}(n)\right ]^2}.$$ (31)