DFA-1 on linear trend

$\smallskip$Let us suppose a linear time series $u(i)=A_{\rm L}i$. The integrated signal $y_{L}(i)$ is

$$y_{L}(i)=\sum_{j=1}^{i}A_{\rm L}j=A_{\rm L}\allowbreak \frac{i^{2}+i}{2}$$ (32)

Let as call $N_{max}$ the size of the series and $n$ the size of the box. The rms fluctuation $F_{\rm L}(n)$ as a function of $n$ and $N_{max}$ is

$$F_{\rm L}(n)=A_{\rm L}\sqrt{\frac{1}{N_{max}}\sum_{k=1}^{N_... ...1)n+1}^{kn}\left(\frac{i^{2}+i}{2}-(a_{k}+b_{k}i)\right)^{2}}$$ (33)

where $a_{k}$ and $b_{k}$ are the parameters of a least-squares fit of the $k$-th box of size $n$. $a_{k}$ and $b_{k}$ can be determined analytically, thus giving:

$$a_{k}=1-\frac{1}{12}n^{2}+\frac{1}{2}n^{2}k+\frac{1}{12}n-\frac{1}{2}%% k^{2}n^{2}$$ (34)

$$b_{k}=1-\frac{1}{2}n+kn+\frac{1}{2}$$ (35)

With these values, $F_{\rm L}(n)$ can be evaluated analytically:

$$F_{\rm L}(n)=A_{\rm L}\frac{1}{60}\sqrt{\left( 5n^{4}+25n^{3}+25n^{2}-25n-30\right) }$$ (36)

The dominating term inside the square root is $5n^{4}$ and then one obtains

$$F_{\rm L}(n)\approx \frac{\sqrt{5}}{60}A_{\rm L}n^{2}$$ (37)

leading directly to an exponent of 2 in the DFA. An important consequence is that, as $F(n)$ does not depend on $N_{max}$, for linear trends with the same slope, the DFA must give exactly the same results for series of different sizes. This is not true for other trends, where the exponent is 2, but the factor multiplying $n^{2}$ can depend on $N_{max}$.