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Comparison of Nakagami and Gamma distributions for a list of data

Alex Teymouri
Posted 1 year ago

Hi,

I tried to evaluate the Nakagami and Gamma distributions for monthly wind speed data.

Which distribution is better?

How to show two distributions in one plot for comparison.

I appreciate your support and time.

Alex

data = {9.9`, 9.`, 12.7`, 13.3`, 11.3`, 10.`, 9.8`, 8.8`, 8.`, 7.4`, 6.8`, 
   6.8`, 7.1`, 9.4`, 8.4`, 10.`, 15.2`, 11.2`, 9.3`, 9.7`, 8.8`, 8.4`, 6.2`, 
   7.9`, 7.4`, 8.9`, 10.5`, 9.5`, 9.3`, 11.5`, 7.1`, 8.9`, 6.5`, 5.4`, 7.1`, 
   7.6`, 8.8`, 10.2`, 14.7`, 11.9`, 9.9`, 9.8`, 10.8`, 10.2`, 8.6`, 8.9`, 
   8.5`, 8.7`, 8.3`, 10.`, 9.`, 10.5`, 10.2`, 11.5`, 9.6`, 9.9`, 7.8`, 6.8`, 
   7.3`, 5.7`, 7.6`, 5.9`, 7.4`, 8.1`, 9.2`, 10.7`, 9.9`, 9.`, 7.8`, 7.`, 5.`,
    5.7`, 6.4`, 7.1`, 9.7`, 9.9`, 7.3`, 10.7`, 10.7`, 8.2`, 8.2`, 8.8`, 4.7`, 
   5.9`, 7.4`, 6.4`, 8.`, 9.9`, 7.4`, 10.`, 9.3`, 10.8`, 9.6`, 8.6`, 8.4`, 
   8.4`, 7.`, 8.6`, 9.6`, 7.9`, 9.9`, 11.5`, 10.1`, 9.7`, 8.2`, 6.7`, 4.6`, 
   7.5`, 6.`, 7.8`, 9.9`, 10.3`, 8.4`, 10.1`, 11.6`, 12.`, 10.`, 7.9`, 6.5`, 
   5.9`, 10.6`, 9.`, 10.3`, 9.2`, 10.3`, 11.`, 11.2`, 10.4`, 10.3`, 9.5`, 
   8.3`, 10.1`, 8.1`, 9.`, 10.6`, 11.4`, 10.6`, 12.2`, 11.5`, 12.2`, 10.7`, 
   9.4`, 8.2`, 8.5`, 9.3`, 11.3`, 11.5`, 10.3`, 11.`, 12.5`, 12.2`, 12.9`, 
   10.6`, 9.2`, 7.8`, 8.8`, 9.5`, 9.9`, 11.`, 10.9`, 11.7`, 12.8`, 11.4`, 
   13.1`, 12.7`, 10.9`, 9.6`, 10.6`, 10.1`, 12.8`, 11.5`, 11.2`, 11.2`, 11.9`,
    12.2`, 11.3`, 11.2`, 10.`, 8.9`, 7.9`, 8.7`, 11.`, 11.1`, 13.4`, 13.8`, 
   12.5`, 12.1`, 12.`, 10.1`, 10.8`, 8.1`, 10.9`, 9.3`, 11.9`, 12.2`, 11.1`, 
   11.3`, 12.1`, 10.9`, 9.9`, 11.2`, 9.5`, 10.`, 9.9`, 8.6`, 9.6`, 11.3`, 
   11.2`, 11.5`, 11.3`, 10.9`, 10.9`, 10.1`, 10.8`, 8.9`, 6.9`, 9.9`, 10.3`, 
   12.1`, 11.9`, 10.7`, 11.3`, 12.5`, 11.1`, 10.3`, 9.8`, 10.`, 8.8`, 11.3`, 
   10.9`, 12.6`, 11.7`, 11.3`, 11.5`, 12.5`, 11.`, 11.4`, 9.7`, 7.2`, 8.7`, 
   9.8`, 12.6`, 13.2`, 11.8`, 12.2`, 12.4`, 11.5`, 10.2`, 10.3`, 8.1`, 9.7`, 
   7.7`, 8.4`, 10.1`, 13.`, 12.8`, 12.3`, 13.2`, 12.`, 11.6`, 11.8`, 9.`, 
   8.9`, 8.1`, 9.7`, 10.8`, 13.5`, 12.8`, 13.5`, 12.8`, 11.9`, 12.4`, 10.3`, 
   9.5`, 8.4`, 7.7`, 11.4`, 9.8`, 12.8`, 12.4`, 13.1`, 12.5`, 13.1`, 12.3`, 
   9.9`, 8.9`, 8.2`, 6.8`};

(*Gamma Distribution*)

gam = EstimatedDistribution[data, GammaDistribution[\[Alpha], \[Beta]]];

Show[
 Histogram[data, 15, "PDF"],
 Plot[PDF[GammaDistribution[23.81, 0.41], x], {x, Min[data], Max[data]}, 
  Filling -> Axis, Exclusions -> None]]

(*Nakagami Distribution*)

nak = EstimatedDistribution[data, NakagamiDistribution[\[Mu], \[Omega]]];

Show[
 Histogram[data, 15, "PDF"],
 Plot[PDF[NakagamiDistribution[6.39, 102.37], x], {x, Min[data], Max[data]}, 
  Filling -> Axis, Exclusions -> None]]
POSTED BY: Alex Teymouri
6 Replies
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Jim Baldwin
Jim Baldwin
Posted 1 year ago
POSTED BY: Jim Baldwin
Alex Teymouri
Alex Teymouri
Posted 1 year ago

Dear Rohit and Jim,

Thank you so much.

Is it possible to show that SmoothHistogram is better than the Nakagami and Gamma distributions by calculating the AIC values?

With best Regards
POSTED BY: Alex Teymouri
Jim Baldwin
Jim Baldwin
Posted 1 year ago
POSTED BY: Jim Baldwin

Hi Alex,

There are a couple of syntax errors. Try

aicgam = -2 LogLikelihood[GammaDistribution[\[Alpha], \[Beta]] /. mlegam, data] + 2*1

aicnak = -2 LogLikelihood[NakagamiDistribution[\[Mu], \[Omega]] /. mlenak, data] + 2*1
POSTED BY: Rohit Namjoshi
Alex Teymouri
Alex Teymouri
Posted 1 year ago

Thank you, Henrik.

As you know, we can evaluate these statistics distributions using the AIC.

Where I did I mistake?

(*Maximum Likelihood for Gamma*)

mlegam = 
 FindDistributionParameters[data, GammaDistribution[\[Alpha], \[Beta]]]

(*AIC for Gamma*)

aicgam = -2 LogLikelihood[
    GammaDistribution[{\[Alpha], \[Beta]} /. mlegam], data] + 2*1

(*Maximum Likelihood for Nakagami*)

mlenak = 
 FindDistributionParameters[data, 
  NakagamiDistribution[\[Mu], \[Omega]]]

(*AIC for Nakagami*)

aicnak = -2 LogLikelihood[
    NakagamiDistribution[{\[Mu], \[Omega]} /. mlenak], data] + 2*1
POSTED BY: Alex Teymouri

Hi Alex,

as it seems, Nakagami is slightly better. But if you let the system decide, the NormalDistribution winns:

gam = EstimatedDistribution[data, GammaDistribution[\[Alpha], \[Beta]]];
nak = EstimatedDistribution[data, NakagamiDistribution[\[Mu], \[Omega]]];
best = FindDistribution[data];

plot = Plot[{PDF[gam, x], PDF[nak, x], PDF[best, x]}, {x, Min[data], Max[data]}, PlotStyle -> {Red, Green, {Dashed, Blue}}, PlotLabels -> {"Gamma", "Nakagami", "best:Normal"}];

Show[Histogram[data, Automatic, "PDF"], plot, PlotRange -> {{4, 17.7}, Automatic}, ImageSize -> Large]

enter image description here

Does that help? Regards -- Henrik

POSTED BY: Henrik Schachner
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