I have this set of data from a paper:

data={{0.995, 0.142}, {3.003, 0.2}, {5.908, 0.25}, {10.525, 0.36}, {13.617, 
0.498}, {24.321, 0.616}, {33.917, 0.599}, {47.843, 0.7}, {64.172, 
0.835}, {91.353, 1.102}, {126.745, 1.083}, {174.118, 1.225}, 
{225.059, 1.133}, {292.998, 1.165}, {369.133, 1.298}, {640.295, 
1.365}, {828.169, 1.298}, {1255.39, 1.373}, {1496.61, 1.409}, 
{1942.79, 1.538}};

The proper way to plot it is with


and the result is somethins like this:

enter image description here

I'd like to know if is it possible to fit this experimental data with a CDF and, if so, if is it also possible to get a respective PDF of the CDF which fit the data Ps.: To contextualize, I'm tryin to reproduce the following result: enter image description here

I would be very happy if anyone could help me

Thanks in advance Jose

  • $\begingroup$ A CDF has values in [0, 1]. What does your data represent ? $\endgroup$ May 1, 2019 at 15:04
  • $\begingroup$ Hi, b.gatessucks. Thanks for your feedback. This data represent the change of the magnetizaion in rocks when these are exposed to a field which is not paralel to the magnetization. This non-in-equilibriun system will induce in the rocks a secondary magnetization component which is called viscous remanent magnetization. The data points represent the intensity of VRM varying with time. My suggestion is that the author of the paper had fitted the viscous increasing data with a CDF and generated a PDF which is related with the relaxation time disrtibution of the minerals within the rock $\endgroup$ May 1, 2019 at 15:14
  • 2
    $\begingroup$ You have a regression situation rather than a random sample from some particular probability distribution. So while you do have a cumulative function and can consider the rate of change of the cumulative function, you don't have a probability based CDF or PDF, respectively. Now you certainly can use a "Gaussian shape" for the cumulative function and it's associated rate of change. You should not only describe what your data represents but how it was collected. By "how it was collected" I mean was this a continuous set of measurements on a single set of rocks? Or something else? $\endgroup$
    – JimB
    May 1, 2019 at 15:15

1 Answer 1


While the cumulative function and its rate of change with respect to the $\log_{10}$ of time can be estimated, you'll have to ask the authors how they got way more than 20 bars in the inset graphic given only 20 data points.

First fit the cumulative function:

data = {{0.995, 0.142}, {3.003, 0.2}, {5.908, 0.25}, {10.525, 
    0.36}, {13.617, 0.498}, {24.321, 0.616}, {33.917, 0.599}, {47.843,
     0.7}, {64.172, 0.835}, {91.353, 1.102}, {126.745, 
    1.083}, {174.118, 1.225}, {225.059, 1.133}, {292.998, 
    1.165}, {369.133, 1.298}, {640.295, 1.365}, {828.169, 
    1.298}, {1255.39, 1.373}, {1496.61, 1.409}, {1942.79, 1.538}};

nlm = NonlinearModelFit[data, a + b CDF[NormalDistribution[c, d], Log10[t]], {a, b, c, d}, t];
Show[ListLogLinearPlot[data], LogLinearPlot[nlm[t], {t, 0.01, 20000}]]

Fit of cumulative function

For the rate of change of the cumulative function with respect to the $\log_{10}$ of time:

rateOfChange = D[nlm[10^log10t], log10t] /. 10^log10t -> t;
LogLinearPlot[rateOfChange, {t, 0.01, 20000}]

Rate of change of cumulative function

  • $\begingroup$ JimB, do you have any ideia of how to get an histogram from the rateOfChange curve? Other question: reading the paper again, I've notived that the author had fitted the data points with the following equation: Meq+(Mo-Meq) Exp[-t/tau], and the best fitting gave as an output the inset graphic with the distributions of tau, as well as the estimated value of Meq (which was calculated as 1.5). Do you have any ideia on how did they get it? Thanks in advance $\endgroup$ May 2, 2019 at 14:26
  • $\begingroup$ It's not that I don't know how to get a histogram from the rateOfChange curve, it's because you shouldn't. Are you suggesting that the authors converted the "cumulative dataset" (the dataset you give here) to a "non-cumulative dataset" and fit the function Meq+(Mo-Meq) Exp[-t/tau] ? It might be handy to give the reference to the publication. $\endgroup$
    – JimB
    May 2, 2019 at 15:49
  • $\begingroup$ By reading the paper, I think so. The data are from experimental acquisition, and in order to test the accuracy of the numerical approach, the authors suggested that the data are fitted by the given (theoretical) equation with Meq=1.5 and with the given Tau distribution. $\endgroup$ May 2, 2019 at 16:04
  • $\begingroup$ Here's the reference: sciencedirect.com/science/article/pii/S0031920106001610. $\endgroup$ May 2, 2019 at 16:05

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