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I'm completely new to Mathematica and stuck on how to go about trying to fit a Gaussian function to my data. Pretty clueless and nothing I've tried has worked.

Can anyone help?

I've imported the data and plotted it, but not sure what to do next...

G' = Import["GP S3 F2 G' LX.txt", "Table"]
ListLinePlot[Derivative[1][G], PlotTheme -> "Scientific", PlotRange -> {{2620, 2730}, {0, 2750}}]

Plot of data

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    $\begingroup$ Check out FindDistributionParameters $\endgroup$ – Chris K Nov 27 '17 at 20:01
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    $\begingroup$ @ChrisK. I disagree. This is a regression problem where the fitted curve has a similar shape to a Gaussian probability function (a + b Exp[(x-c)^2/d]) rather than fitting a probability distribution from a random sample. $\endgroup$ – JimB Nov 27 '17 at 20:07
  • $\begingroup$ @JimB Good point -- I should avoid weighing in on statistical questions, which are outside my area of expertise! $\endgroup$ – Chris K Nov 27 '17 at 20:09
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    $\begingroup$ Try Normal@NonlinearModelFit[G', a+ b*Exp[ (x-c)^2/d], {a, b, c, d}, {x, y}]. You will get the fitting equation. A better reply could be provided if you give us access to the data G' $\endgroup$ – José Antonio Díaz Navas Nov 27 '17 at 20:22
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    $\begingroup$ If you have trouble with convergence following @JoséAntonioDíazNavas good suggestion, then you should include starting values (different from the default values of 1 for all parameters). You might try {{a, 100}, {b, 2500}, {c, 2675}, {d, -400}}. $\endgroup$ – JimB Nov 27 '17 at 20:49
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With NMinimize it's easy to solve the problem. Because there is no data provided I create some example data

data = Block[{mu = 1, sig = .73},Map[{#,Exp[-(1/2) ((# - mu)/sig)^2]/(sigSqrt[2 Pi])RandomReal[{0.9, 1.1}]} &, RandomReal[{-1, 3}, 100]]];
ListPlot[data]

of a perturbed gauss. The optimal approximation follows from

res = Map[(Exp[-(1/2) ((#[[1]] - mu)/sig)^2]/(sig Sqrt[2 Pi]) - #[[2]])^2 &, data];
J = res.res;
NMinimize[J, {mu, sig}]

{0.000044725, {mu -> 0.996546, sig -> 0.746079}}    

Alternativly the solution could be obtained with

NonlinearModelFit[data, Exp[-(1/2) ((x - mu)/sig)^2]/(sig Sqrt[2 Pi]), {mu, sig}, x]
Show[{bild, Plot[Normal[%], {x, -3, 3}]}]
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  • $\begingroup$ @UlrichNeumann thanks so much! Would it help more if I provided the data? I can't work out how to attach a file here $\endgroup$ – user53761 Nov 28 '17 at 12:10
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    $\begingroup$ I suggest using PDF[NormalDistribution[mu, sig], x] instead of typing out the PDF function. $\endgroup$ – Edmund Nov 28 '17 at 12:29
  • $\begingroup$ @user53761 You could reduce your data if necessary and provide it as an outputcell... $\endgroup$ – Ulrich Neumann Nov 28 '17 at 14:15
  • $\begingroup$ @UlrichNeumann I'm sorry, I realise this is a ridiculous way to share the data $\endgroup$ – user53761 Nov 28 '17 at 17:55
  • $\begingroup$ I agree ;-) ! What I meant is edit your question and append there(!) an outputcell with your data {...,{xi,yi},...}! $\endgroup$ – Ulrich Neumann Nov 29 '17 at 15:41
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as other users replied, you should provide the input data.

Anyway, note that the area underlying the graph is bigger than one- you should compute this total area, call it A, and divide every y by A, so that you have area=1. Then do compute the average and the standard deviation for the new distribution, put these two numbers (mu, sigma) into the formula for the normal distribution, then multiply back by the A factor. Try plotting this against your data.

Also.. You know that gaussian distributions tend to 0 for x->+-inf. In your case, not sure if the sample is shifted up by summing factor.. Do take that into consideration. (find the shifting factor a, subtract this number to all the points (to all the y's), before computing the total area).

Of course, @ChrisK and @JimB summed this up by telling to find the a,b,c,d parameters.

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  • $\begingroup$ While this addresses part of OP's concerns, a valid answer should probably contain a code sample to show how to apply mentioned notes. At the end it is about Mathematica, not math / stats. $\endgroup$ – Kuba Nov 28 '17 at 10:41

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