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I'm having trouble writing a Nonlinear Model fit where the model is a numerical integral evaluated with NIntegrate. I have read other questions and answers about more or less the same problem, but for some reason I cannot make sense of this. Here goes:

I define a function containing the integral:

    LumDist[zmax_?NumericQ, Om_?NumericQ, Ol_?NumericQ, Ot_?NumericQ, 
  n_?NumericQ] := NIntegrate[1/\[Sqrt](Om*(1 + x)^3 + Ol + Ot*(1 + x)^n), {x, 0, zmax}]

and from there define the model:

DistMod[zmax_?NumericQ, Om_?NumericQ, Ol_?NumericQ, Ot_?NumericQ, 
  n_?NumericQ, mu0_?NumericQ] := 
 5*Log10[LumDist[zmax, Om, Ol, Ot, n]] + mu0

From here I write a NonlinearModelFit for this model (data) is an (x,y) List:

lmvar = NonlinearModelFit[
  data, {DistMod[z, 0.28, 0.71, Ot, 10^(-3), mu0], 
   1 < Ot >= 0}, {{Ot, 0.08}, {mu0, 23.89}}, x, Weights -> 1/errors^2,
   VarianceEstimatorFunction -> (1 &), Method -> "NMinimize"]

But this does not work, and I get lots of errors such as:

NIntegrate::nlim: x = z is not a valid limit of integration. >>

and

NIntegrate::itraw: Raw object 0.01012` cannot be used as an iterator. >>

Here is a sample of the data and errors:

data = {{0.015, 34.1114},{0.0277, 35.705},{0.048948, 36.7316},{0.0651, 37.3067},{0.100915, 38.4567},{0.159, 39.4164},{0.248508, 40.2722},{0.455, 42.3239},{0.655, 42.3151},{0.75, 43.243},{0.84, 43.5143},{0.961, 44.2642},{1.188, 44.6076},{1.34, 45.0675},{1.414, 44.8038}}

errors={0.215239,0.118114,0.175773,0.242628,0.121087,0.21481,0.152213,0.306006,0.188809,0.198402,0.471047,0.1534, 0.421,0.213,0.1345}

Can anyone tell me what I am doing wrong here?

Moreover, I can even write another simpler model (it's completely wrong, but I want to try to isolate the problem):

testfunc[a_?NumericQ, x_?NumericQ] := NIntegrate[a*t, {t, 1, x}];

Which I can try to fit using:

NonlinearModelFit[data, {testfunc[b], b > 0.}, b, x, 
 Method -> "NMinimize"]

but this also throws lots of errors, such as:

NonlinearModelFit::nrnum: The function value 1/2 ((-26.877+testfunc[0.0914636])^2+(-26.2011+testfunc[0.0914636])^2+(-26.1609+testfunc[0.0914636])^2+(-26.0592+testfunc[0.0914636])^2+(-26.0071+testfunc[0.0914636])^2+(-25.9072+testfunc[0.0914636])^2+(-25.7394+testfunc[0.0914636])^2+(-25.6034+testfunc[0.0914636])^2+<<35>>+(-24.5574+testfunc[0.0914636])^2+(-24.548+testfunc[0.0914636])^2+(-24.5443+testfunc[0.0914636])^2+(-24.5128+testfunc[0.0914636])^2+(-24.4717+testfunc[0.0914636])^2+(-24.4442+testfunc[0.0914636])^2+(-24.4217+testfunc[0.0914636])^2+<<996>>) is not a real number at {b} = {0.0914636}. >>
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  • $\begingroup$ In the first case, don't you need to give z a numeric value? $\endgroup$ – Chris K Jul 30 '18 at 15:29
  • $\begingroup$ Since we do not have data, we can not help anything. Or do we have to come up with data? $\endgroup$ – Alex Trounev Jul 30 '18 at 17:12
  • $\begingroup$ I provided a data set with errors in my post. $\endgroup$ – koldrakan Jul 30 '18 at 17:25
  • $\begingroup$ There are 15 data points and 11 errors. Adding in the additional errors is necessary as we can't know which data points have missing errors. $\endgroup$ – JimB Jul 30 '18 at 18:05
  • $\begingroup$ That's my mistake, I added some errors. It doesn't really matter what they are. $\endgroup$ – koldrakan Jul 30 '18 at 18:26
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The fundamental problem with your simplified example is that you don't explicitly incorporate x in NonlinearModelFit. A solution, albeit a completely incorrect one, can be found with a slight modification to NonlinearModelFit:

testfunc[a_?NumericQ, x_?NumericQ] := NIntegrate[a*t, {t, 1, x}];
NonlinearModelFit[data, {testfunc[b,x], b > 0.}, b, x, Method -> "NMinimize"]

To really test what is going on here, a slightly more flexible model could be introduced. Changing testfunc just a bit:

testfunc[a_?NumericQ, b_?NumericQ, x_?NumericQ] := NIntegrate[a*t^b, {t, 0, x}];

Now, NonlinearModelFit can be used to find a reasonable fit:

nlm = NonlinearModelFit[data, {testfunc[a, b, x], a > 0, b < 0}, {a, b}, x]
(*FittedModel[testfunc[2.65151,-0.939799,x]]*)

nlm["ParameterTable"]
(*  
Estimate    Standard Error  t-Statistic P-Value
a   2.65151 0.0537729   49.3094 3.59017*10^-16
b   -0.939799   0.00114251  -822.574    4.78556*10^-32
*)

Showing the solution:

enter image description here

There is a similar error in the definition of the actual function of interest. You should be able to find a fit with:

LumDist[zmax_?NumericQ,Om_?NumericQ,Ol_?NumericQ,Ot_?NumericQ,n_?NumericQ]:=NIntegrate[1/\[Sqrt](Om*(1+x)^3+Ol+Ot*(1+x)^n),{x,0,zmax}]
DistMod[zmax_?NumericQ,Om_?NumericQ,Ol_?NumericQ,Ot_?NumericQ,n_?NumericQ,mu0_?NumericQ]:=5*Log10[LumDist[zmax,Om,Ol,Ot,n]]+mu0

lmvar=NonlinearModelFit[data,{DistMod[z,0.28,0.71,Ot,10^(-3),mu0],0<= Ot<1},{{Ot,0.08},{mu0,23.89}},z,Weights->1/errors^2,VarianceEstimatorFunction->(1&)]

(*FittedModel[DistMod[z,0.28,0.71,0.999999,1/1000,44.5454]]*)

lmvar["ParameterTable"]
(*
    Estimate    Standard Error  t-Statistic P-Value
Ot  0.999999    0.727728    1.37414 0.192627
mu0 44.5454 0.3562  125.057 2.05428*10^-21
*)

Show[Plot[lmvar[x], {x, 0, 2}], ListPlot[data]]

enter image description here

Note, the problem with the original call to NonlinearModelFit did not include the dependent variable (in this case z). You might be able to find a better fit playing around with the options of NonlinearModelFit.

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  • $\begingroup$ testfunc will no longer need NIntegrate as Integrate can be used. Further a reasonable fit would be with taking logs of both the the response and predictor variable and fitting a straight line. The errors vector might not even be needed or appropriate: looking at the residuals would help with that decision. $\endgroup$ – JimB Jul 30 '18 at 19:54
  • $\begingroup$ How long did that take you to run? My real dataset has just over 1000 points and it is taking a long time to run. $\endgroup$ – koldrakan Jul 31 '18 at 8:30
  • $\begingroup$ Using the 15 points from the original data, RepeatedTiming is showing somewhere in the neighborhood of 12 seconds on my laptop. $\endgroup$ – Marchi Aug 2 '18 at 18:26
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Update

One of the problems is that the model is over-parameterized by 2 or more parameters. Below are some algebraic steps that shows at least two of the parameters can be replaced by a single parameter. (That at least two additional parameters can be replaced by a single parameter is not shown but the evidence is in examining the parameter CorrelationMatrix from NonlinearModelFit.) This suggests that either just taking the logs of the response and predictor variable and fitting a straight line is what should be done or rethinking of the original model is necessary.

Here is the definition of LumDist where we extract Om:

Extracting Om

The final model (DistMod) can therefore be written as follows:

Description of DistMod

We see that we can only at best estimate $-10 \log_{10} O_m+\mu_0$ and not $O_m$ and $\mu_0$ separately.

End of update

There are (still) a few typos in the original question. If x is changed to z in the NonlinearModelFit statement, the restriction 1 < Ot >= 0 is changed to 0 <= Ot < 1 (or as written it might even be equivalent to Ot > 1 - can't really tell), and VarianceFunctionEstimation is removed, the model converges:

lmvar = NonlinearModelFit[data, {DistMod[z, 0.28, 0.71, Ot, 10^(-3), mu0], 0 <= Ot < 1},
   {{Ot, 0.08}, {mu0, 23.89}}, z, Weights -> 1/errors^2];
lmvar["ParameterTable"]

$$\begin{array}{l|llll} \text{} & \text{Estimate} & \text{Standard Error} & \text{t-Statistic} & \text{P-Value} \\ \hline \text{Ot} & 0.999999 & 2.42865 & 0.411751 & 0.68723 \\ \text{mu0} & 44.5454 & 1.18875 & 37.4725 & \text{1.2436775876447106$\grave{ }$*${}^{\wedge}$-14} \\ \end{array}$$

It appears that the use of the parameter Ot is not very useful as the fit is poor and the estimate is at the boundary.

A much better fit can be found by noting that a log-log plot appears to be a simple straight line

ListLogLogPlot[data]

Log log plot of data

Fitting such a model we have

lmvar2 = NonlinearModelFit[Log[data], a + b logx, {a, b}, logx];
lmvar2["ParameterTable"]

$$\begin{array}{l|llll} \text{} & \text{Estimate} & \text{Standard Error} & \text{t-Statistic} & \text{P-Value} \\ \hline a & 3.7851 & 0.00207508 & 1824.07 & \text{1.526517371336953$\grave{ }$*${}^{\wedge}$-36} \\ b & 0.0601371 & 0.00104813 & 57.3755 & \text{5.051558118117479$\grave{ }$*${}^{\wedge}$-17} \\ \end{array}$$

Show[ListPlot[data], 
 Plot[{lmvar[z], Exp[lmvar2[Log[z]]]}, {z, Min[data[[All, 1]]], 
   Max[data[[All, 1]]]},
  PlotLegends -> {"Original fit", "Better fit"}]]

Comparison of fits

Assuming that the full dataset has a similar look and feel, this new approach not only provides a better fit but is much faster.

(I note that despite Wolfram producing an example as to how to fit curves when there is measurement error, that example assumes (generally inappropriately) that there is only measurement error and assumes the more typical lack-of-fit error is zero. So given that the measurement errors in this example (which I assume are standard deviations) are pretty similar among themselves, I'd even drop the use of Weights.)

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  • $\begingroup$ This is indeed a much better fit, and I have thought about this, but then I need a way to translate this fit to parameter constraints on my original model, and I haven't figured out how to do that. $\endgroup$ – koldrakan Aug 2 '18 at 7:13
  • $\begingroup$ I think the problem is that the model is overparameterized by at least 2 parameters. (Very loosely the data only supports the estimation of 2 parameters.) One can definitely see that at least one of the parameters needs to be removed. Consider factoring out Om from LumDist so that the integral is f[Om]*integral. Taking the logs as DistMod does means that the model has the form 5*Log10[f(Om)] + 5*Log10[integral] + mu0. This means that you can only estimate 5*Log10[f(Om)]+mu0 but not mu0 and Om separately. That an additional parameter needs to be removed is more subtle. $\endgroup$ – JimB Aug 2 '18 at 17:56

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