1
$\begingroup$

I have the following data points:

  data = {{1000, 9.58155251674433*10^(3848)}, {2000, 
    6.342086690693367*10^(8690)}, {3000, 
    3.363260189628007*10^(13920)}, {4000, 
    2.479454222921813*10^(19404)}, { 5000, 
    1.898605730736944*10^(25078)}, {6000, 
    1.529106621101898*10^(30904)}, {7000, 
    7.903765049528210*10^(36856)}, {8000, 
    2.733268673289474*10^(42918)}, {9000, 
    2.081989821506015*10^(49075)}, {10000, 
    1.131263617784419*10^(55317)}, {11000, 
    1.758447029798765*10^(61635)}, {12000, 
    1.042154296462331*10^(68032)}, {13000, 
    4.344886677434895*10^(74474)}, {14000, 
    1.775926151008434*10^(80985)}, {15000, 
    4.138917150233665*10^(87550)}};

Now I am trying to fit them, actually my fit model is pretty weird but is adjusted to the theoretically predicted outcome:

nlm = NonlinearModelFit[Log[data], 
   a*k*Log[k] - b*k*Log[Log[k]] - c*k + d*k*Log[Log[k]]/Log[k] + 
    f*k/Log[k] + l*k*(Log[Log[k]]/Log[k])^2, {a, b, c, d, f, l}, k];

What I get is:

nlm["BestFitParameters"]
{a -> 1.3939471292781074294*10^9, b -> 1.3333126023460171273*10^10, 
 c -> -2.4438759930779127494*10^10, d -> -1.2904358410322236152*10^10,
  f -> -2.5977630783453403124*10^10, l -> -4.3601191673382517145*10^9}

and upon plotting the fitted model I get (it is the lower Plot):

ListLogPlot[Exp[nlm["PredictedResponse"]]]

Okay, so my question: why is the x- Axis not starting at 1000 but instead starting at 1. Is mathematica really fitting the my data points. Also the variables $a,b,c,d,f,l$ are absolutely not what I expected. I went the other way round and inserted in the model what I expect for the parameters $(a=2,b=2,c=2,d=2,f=2,l=1)$ and got the following Plot (the upper one): [Fit of data points][1]Fit with already inserted expected parameters

enter image description here

$\endgroup$
6
  • $\begingroup$ The reason for x axis: the data Exp[nlm["PredictedResponse"]] provided to ListLogPlot only contain y values(a 1d list), so Range[number] is chosen for x. For the parameters, if use MapAt[Log, data, {All, 2}] instead of Log[data], with this setting I got the parameters around 10^5 $\endgroup$ – vapor Mar 9 '17 at 13:00
  • $\begingroup$ Hi. Thanks for the first explanation but I do not unterstand the part concerning the parameters. $\endgroup$ – user404302 Mar 9 '17 at 13:04
  • $\begingroup$ I read through the mathematica documentation of MapAt since I did not know that command however I do not really get what it does. could you explain that please? $\endgroup$ – user404302 Mar 9 '17 at 13:18
  • $\begingroup$ Take a look at N[nlm["CorrelationMatrix"] // MatrixForm, 4]. You'll see that all (not just some) of the correlations among the parameter estimators are nearly +1 or -1. The inference: your model is way overparameterized. You've got 15 data points and 7 parameters (a, b, c, d, f, l, and error variance). You'll get appropriate predictions but you won't get stable/expected values for the parameters. $\endgroup$ – JimB Mar 9 '17 at 17:07
  • $\begingroup$ Also see mathematica.stackexchange.com/questions/139038/…. $\endgroup$ – JimB Mar 9 '17 at 17:14
2
$\begingroup$

This is an extended comment as to why the model is overparameterized. In short each and every term in the model is a near perfect linear function of the others within the range of the predictor variable (from 1,000 to 15,000). Here is a figure that shows why that is true:

p[x_, y_] := ParametricPlot[{x, y}, {k, Log[1000], Log[15000]},
  AspectRatio -> 1, Frame -> True,
  FrameLabel -> (Style[ToString[#, InputForm], Bold, Larger] &) /@ {x, y},
  ImageSize -> 200]
GraphicsGrid[{
  {p[k Log[k], k Log[Log[k]]], p[k Log[k], k], 
   p[k Log[k], k Log[Log[k]]]},
  {p[k Log[k], k/Log[k]], p[k Log[k], k (Log[Log[k]]/Log[k])^2]}
  }]

Linear relationships among terms in the model

This means in a linear model (which is what you have because the coefficients enter linearly in the model) each of the terms bring the same predictor information and are nearly completely redundant of each other.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.