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I have a time series that has a power-law $A \cdot t^{b}$ behaviour for $t\xrightarrow{}\infty$, while for low $t$ it is far from that. Something like this: enter image description here How can I make a linear fit and find an estimation for $A$ and $b$? How many points shall be discarded?

Edit: as you can see from the picture, taken from 3 independent runs, for $t < t_0$ (let's say, 10,000) the behaviour is quite random, and unrelated, while for $t > t_0$ the 3 series are almost parallel, meaning that they have the same exponent. A simple linear fit will give the value of this exponent (about -0.0093). But the problem is: how to identify $t_0$, in order to make the regression only in the range $[t_0, \infty]$ where it follows the model.

Luca Petrone
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The best way should be to start from a physical model which allows to derive a convenient kind of function.

of course the function includes some parameters. With numerical data a non-linear regression would give approximate values of the parameters.

This is the best way because the parameters are likely to have physical significance, which is useful in practice.

If no physical model is available one have to guess some function susceptible to fit more or less correctly the data.

In the pesent case a linear piecewise function appears convenient (next figure).

enter image description here

The equation of the fitted curve drawn in red is : $$y(x)=(a_1x+b_1)H(\alpha-x)+(a_2x+b_2)\big(H(x-\alpha)-H(x-\beta)\big)+(a_3x+b_3)H(x-\beta))$$ $$\alpha=\frac{b_2-b_1}{a_1-a_2}\quad;\quad \beta=\frac{b_3-b_2}{a_2-a_3}$$ $H$ is the Heaviside function.

One can smooth the angular transition between the linear segments : One replace the Heaviside function by a approximate function of the logistic kind : $$H(\xi)\simeq \frac{1}{1+e^{-\gamma\:\xi}}\qquad \gamma>0$$ The more $\gamma$ is large the more this function is close to the Heaviside function.

Thus a smooth function to be fitted to the data can be : $$y(x)=\frac{a_1x+b_1}{1+e^{-\gamma_1\:(\alpha-x)}}+\frac{a_2x+b_2}{1+e^{-\gamma_1\:(x-\alpha)}}-\frac{a_2x+b_2}{1+e^{-\gamma_2\:(x-\beta)}}+\frac{a_3x+b_3}{1+e^{-\gamma_2\:(x-\beta)}}$$

The parameters to be found thanks to non-linear regression are : $a_1,b_1,a_2,b_2,a_3,b_3,\gamma_1,\gamma_2$. The two above equations relate $\alpha,\gamma$ to $a_1,b_1,a_2,b_2,a_3,b_3$.

From the graph provided in the question without numerical data one cannot proceed. Scannig the graph gives the coordinates of the pixels. But the numerical data obtained is not accurate enough and too scrambled.

On the figure below the red curve isn't fitted to data. This is only to show how the angular transitions are smoothed, for example with $\gamma_1=3$ and $\gamma_2=5$. These values must not be trusted.

enter image description here

JJacquelin
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  • the first two piecewise interpolations are meaningless, while the third is exactly what I am searching for (this is in fact Feder's Law of Random Sequential Adsorption) and the exponent -0.093 is what I was estimating myself, as well. But the real question is: how can I (automatically) estimate the value from which I can start the regression? – Luca Petrone Jan 31 '21 at 13:25
  • I edited the plot, hoping that now is clearer what I am asking : no smoothing, nor fitting for low values of time, only the exponent of decay and the time after which the points start obeying to the underlaying law – Luca Petrone Jan 31 '21 at 17:57
  • OK. I understand. That is why I wrote "Scannig the graph gives the coordinates of the pixels. But the numerical data obtained is not accurate enough and too scrambled." But that is how I compute the exponent -0.093 which is certainely not correct. Sorry that is all I can do to help you. – JJacquelin Jan 31 '21 at 19:20
  • anyhow, the exponent -0.093 is very precise, as I calculate the same value using the actual values. Nevertheless, it is the value of t0 what I am asking for. – Luca Petrone Jan 31 '21 at 21:06