Given a distribution obeying the power-law (fractal) relation, such as the cumulative distribution function $L_{cf}(> X) = CR^{-D}$, if $X$ is given, how does one find the constant $C$ from a given data set?

For example, given this table:

Source: C. Li et al. / Geomorphology 130 (2011)

[Source: C. Li et al. / Geomorphology 130 (2011)]

how does one determine how the values for $L_{cf}$ were obtained? It appears that they used the method of least squares fitting to obtain the values for $D$, but I don't seem to understand where the values for $C$ are coming from. Any help would be appreciated.

  • $\begingroup$ Are you looking for a mathematical method (and you assume that the necessary data is already available)? $\endgroup$ Apr 16 '16 at 11:27
  • $\begingroup$ That's exactly right. $\endgroup$
    – sequence
    Apr 16 '16 at 17:04

There are two solutions using the least square method for calculating $C$ and $D$. Both methods yield different results for your constants. There is no correct method.

Method of least Squares

We define the least square error as follows: $$\text{lse} = \sum_{i}{\left(y_i - f(x_i)\right)^2}$$ The $y_i$ and $x_i$ are our data through that we want to fit a function $f(x)$. The aim is to reduce our error $\text{lse}$.

The solution (minimal $\text{lse}$) for a linear function $f(x) = a \cdot x + b$ is described here. The basic idea for calculating the minimum of $\text{lse}$ is to set $\partial \text{lse}/\partial a$ and $\partial \text{lse}/\partial b$ equal to $0$ and solve the resulting equation system to $a$ and $b$.

Having said that we come to two solutions for calculating $C$ and $D$ in $f(x) = C\cdot x^{-D}$

Solution 1: Logarithmize the Function

We rewrite $$y_i = C \cdot x_i^{-D}$$ to $$\ln{y_i} = \ln{\left(C \cdot x_i^{-D}\right)} = \ln{C} - D\cdot \ln{x_i} $$

Now we have a function of the form $\tilde{y_i} = a\cdot \tilde{x_i} + b$ with $\tilde{y_i} = \ln{y_i}$, $\tilde{x_i} = \ln{x_i}$, $a=-D$ and $b=\ln{C}$. Thus, we logarithmize our measured $x_i$ and $y_i$ values and put them into the formulas of the linear least square error method. From the resulting $a$ and $b$ we than calculate $D$ and $C$.

Solution 2: Insert the Function as it is

We set $f(x) = C\cdot x^{-D}$, insert it into our $\text{lse}$ formula: $$ \text{lse} = \sum_i{\left(y_i - C\cdot x^{-D}\right)^2}$$ and minimize the resulting formula with respect to $C$ and $D$ for the given set of $x_i$ and $y_i$. This is a tricky and not as straight-forward as it is in the linear case. You could make $\partial \text{lse}/\partial C$ and $\partial \text{lse}/\partial D$ and look how far you come.

I personally, prefer the answer of Gordon Stanger :-) .


C is purely empirical for any given situation. BEWARE of such equations! Using coefficients to six significant figures gives the illusion of high precision when, in fact, the whole approach is extremely 'rubbery', and highly dependent upon the local geology / soil type / local hydrology. The expression is only valid for the location in which it was calibrated. A small change in drainage can make a huge change in the equation so, personally, I think such stochastic simplifications of complex processes are not worth the paper they are written on! Intelligent eye-balling the potential landslide conditions, with a view to geology and drainage, may be just as good a guide as any (generally poorly calibrated) equation.

  • $\begingroup$ I think that there is no issue with dependency of $C$ on location. $\endgroup$
    – sequence
    Apr 17 '16 at 0:50

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