# Generation of large data sets (10^4) for log Pearson type III distribution and finding its paramters

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Raj Arora on 6 Sep 2022
Commented: Jeff Miller on 7 Sep 2022
Dear All, I have a data set which is log pearson type III distributed fitted. Using that data how can I find the parameters associated with that distribution. further using that paramters I have to generate more number of samples. That samples which I have generated should overlap twith the data given below. Can anyone please help me with this I am working on this from a long time.
Q PDF(LP3)
14.162562 0.000011
28.325123 0.000723
31.403941 0.001391
33.866995 0.002169
36.945813 0.002904
39.408867 0.003638
41.871921 0.004439
45.566502 0.005262
49.261084 0.006063
53.571429 0.006875
59.729064 0.007632
70.197044 0.008288
89.901478 0.007654
100.369458 0.006875
113.300493 0.005996
125.000000 0.005173
137.315271 0.004350
150.862069 0.003571
167.487685 0.002815
192.118227 0.002025
225.985222 0.001246
275.246305 0.000679
326.970443 0.000345
370.073892 0.000211
418.103448 0.000133
459.359606 0.000089
500 5.56242E-05
Torsten on 6 Sep 2022
Too much to read and understand - sorry.

Jeff Miller on 7 Sep 2022
If the distribution you have in mind is the same one described here then this is a good problem for Cupid (see especially the file DemoLogPearsonIII.m).
% Initialize distribution information
Q = [14.162562, 28.325123, 31.403941, 33.866995, 36.945813, 39.408867, 41.871921, 45.566502, 49.261084, 53.571429, 59.729064, 70.197044, 89.901478, 100.369458, 113.300493, 125.000000, 137.315271, 150.862069, 167.487685, 192.118227, 225.985222, 275.246305, 326.970443, 370.073892, 418.103448, 459.359606, 500];
PDF = [0.000011, 0.000723, 0.001391, 0.002169, 0.002904, 0.003638, 0.004439, 0.005262, 0.00606, 0.006875, 0.007632, 0.008288, 0.007654, 0.006875, 0.005996, 0.005173, 0.004350, 0.003571, 0.002815, 0.002025, 0.001246, 0.000679, 0.000345, 0.000211, 0.000133, 0.000089, 5.56242E-05];
% Create a starting LogPearsonIII distribution with initial guesses
% for the parameters (which will be estimated later):
GammaShape = 5;
GammaRate = 5;
ShiftConst = 2;
% RNGamma, AddTrans, and ExpTrans are all parts of Cupid
startGamma = RNGamma(GammaShape,GammaRate);
% Shift the gamma by adding in a constant. The resulting distribution
% is a Pearson III distribution:
% The Log-Pearson III is a distribution whose log is Pearson III.
% That means that the Log-Pearson III distribution is an
% exponential transformation of the original PearsonIII.
% So, we can make the desired Log-Pearson III by using the Exp transform:
LogPearsonIII = ExpTrans(PearsonIII);
% LogPearsonIII is now a distribution object with the starting guesses for the parameter estimates.
% Next estimate the distribution's parameters by trying to match the PDF values at the Q points:
LogPearsonIII.EstDensity(Q,PDF);
% Print the estimated parameter values:
LogPearsonIII.ParmValues
% ans =
% 26.803 8.8801 1.6729
% Display the PDF and CDF of the estimated distribution
LogPearsonIII.PlotDens;
% Generate 1000 random numbers from the fitted distribution:
randlp = LogPearsonIII.Random(1000,1);
% Compare the histogram of the random numbers with the PDF originally specified.
figure;
histogram(randlp,'normalization','pdf');
hold on
plot(Q,PDF);
% You can also generate the random numbers like this:
randlp2 = exp( gamrnd(26.803,1/8.8801,1000,1)+1.6729 );
Jeff Miller on 7 Sep 2022
You are welcome, Raj. I am glad that Cupid was useful to you.

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