How to fit a Distribution to data that already looks like a distribution?

Question

I have a set of data that is already ordered as a distribution. Here is a picture worth 999 words:

I tried FindDistributionParameters, and EstimatedDistribution and I keep getting the message "One or more data points are not in support of the process...".

Any advise? Thank you.

P.S.: Here is the data. Note, the max range of the data is not known, so it could be from {0,inf} - but the range of the data shown is [0,10].

0.0258899, 0.047329, 0.0683456, 0.0861312, 0.103529, 0.118554, \
0.124337, 0.127011, 0.131217, 0.134408, 0.132016, 0.130485, 0.127444, \
0.122495, 0.117607, 0.112261, 0.108703, 0.10322, 0.100889, 0.0955182, \
0.0902072, 0.085152, 0.079376, 0.0752159, 0.0722947, 0.0686795, \
0.0653517, 0.062192, 0.0598642, 0.0569383, 0.0548917, 0.0513092, \
0.0485989, 0.0472422, 0.0458829, 0.0434763, 0.0406302, 0.03872, \
0.0371639, 0.0370529, 0.0353547, 0.0332693, 0.0320803, 0.0302454, \
0.0300313, 0.0294249, 0.0281058, 0.0271762, 0.0256051, 0.0257485, \
0.0250776, 0.0237297, 0.0227674, 0.0225286, 0.0221572, 0.0213558, \
0.0206458, 0.0199427, 0.0191173, 0.0172471, 0.017145, 0.0170844, \
0.016805, 0.0162183, 0.0150205, 0.0149596, 0.0148139, 0.0142276, \
0.0141969, 0.0139716, 0.0132321, 0.0128165, 0.0129189, 0.0126378, \
0.0123681, 0.0122255, 0.0121722, 0.0117113, 0.010789, 0.0105248, \
0.0100393, 0.0103667, 0.0102826, 0.0102321, 0.0104435, 0.0100861, \
0.00972111, 0.00957839, 0.00966471, 0.00963415, 0.00880081, \
0.00873474, 0.00859673, 0.00816511, 0.00792854, 0.0075657, \
0.00747507, 0.00734244, 0.00729432}

Answer 1

You don't seem to have a random sample from any probability distribution but rather have measurements/observations that might have a similar functional form as a probability density function.

Your curve has an area of about 0.43 going from 0 to 10. While the tail of the distribution you envision might be heavy enough to hold the remaining 0.57 of a probability distribution, I don't believe it.

So as @Young suggested, you can certainly fit a curve that happens to have a similar form as a probability distribution (but you can't pretend you're fitting a probability distribution).

To fit such a curve I'm assuming that the data presented are the "y-values" and that the "x-values" are uniformly spaced from 0.1 to 9.9. So the first step is to make a data set that reflects that:

data = {0.0258899, 0.047329, 0.0683456, 0.0861312, 0.103529, 0.118554,
    0.124337, 0.127011, 0.131217, 0.134408, 0.132016, 0.130485, 
   0.127444, 0.122495, 0.117607, 0.112261, 0.108703, 0.10322, 
   0.100889, 0.0955182, 0.0902072, 0.085152, 0.079376, 0.0752159, 
   0.0722947, 0.0686795, 0.0653517, 0.062192, 0.0598642, 0.0569383, 
   0.0548917, 0.0513092, 0.0485989, 0.0472422, 0.0458829, 0.0434763, 
   0.0406302, 0.03872, 0.0371639, 0.0370529, 0.0353547, 0.0332693, 
   0.0320803, 0.0302454, 0.0300313, 0.0294249, 0.0281058, 0.0271762, 
   0.0256051, 0.0257485, 0.0250776, 0.0237297, 0.0227674, 0.0225286, 
   0.0221572, 0.0213558, 0.0206458, 0.0199427, 0.0191173, 0.0172471, 
   0.017145, 0.0170844, 0.016805, 0.0162183, 0.0150205, 0.0149596, 
   0.0148139, 0.0142276, 0.0141969, 0.0139716, 0.0132321, 0.0128165, 
   0.0129189, 0.0126378, 0.0123681, 0.0122255, 0.0121722, 0.0117113, 
   0.010789, 0.0105248, 0.0100393, 0.0103667, 0.0102826, 0.0102321, 
   0.0104435, 0.0100861, 0.00972111, 0.00957839, 0.00966471, 
   0.00963415, 0.00880081, 0.00873474, 0.00859673, 0.00816511, 
   0.00792854, 0.0075657, 0.00747507, 0.00734244, 0.00729432};
(* Include the "x" value *)
data = Table[{i/10, data[[i]]}, {i, Length[data]}];

Now try a curve that has the same shape as a log-normal density function:

nlm = NonlinearModelFit[data, 
   k Exp[-(Log[x] - m)^2/(2 s^2)]/x, {m, s, k}, x];
nlm["BestFitParameters"]
(* {m -> 0.795688, s -> 0.951072, k -> 0.186508} *)
Show[ListPlot[data], Plot[nlm[x], {x, 0, 10}]]

The fit is not great (in my mind) but you need to determine if it meets your needs.

Answer 2

I agree with Jim Baldwin, but here is the code I put together based on this data originating from a random sample (Just in case that helps you for some reason). Your original problem was the 0 data point causing the functions not to evaluate.

dist = {0.0258899, 0.047329, 0.0683456, 0.0861312, 0.103529, 0.118554,
    0.124337, 0.127011, 0.131217, 0.134408, 0.132016, 0.130485, 
   0.127444, 0.122495, 0.117607, 0.112261, 0.108703, 0.10322, 
   0.100889, 0.0955182, 0.0902072, 0.085152, 0.079376, 0.0752159, 
   0.0722947, 0.0686795, 0.0653517, 0.062192, 0.0598642, 0.0569383, 
   0.0548917, 0.0513092, 0.0485989, 0.0472422, 0.0458829, 0.0434763, 
   0.0406302, 0.03872, 0.0371639, 0.0370529, 0.0353547, 0.0332693, 
   0.0320803, 0.0302454, 0.0300313, 0.0294249, 0.0281058, 0.0271762, 
   0.0256051, 0.0257485, 0.0250776, 0.0237297, 0.0227674, 0.0225286, 
   0.0221572, 0.0213558, 0.0206458, 0.0199427, 0.0191173, 0.0172471, 
   0.017145, 0.0170844, 0.016805, 0.0162183, 0.0150205, 0.0149596, 
   0.0148139, 0.0142276, 0.0141969, 0.0139716, 0.0132321, 0.0128165, 
   0.0129189, 0.0126378, 0.0123681, 0.0122255, 0.0121722, 0.0117113, 
   0.010789, 0.0105248, 0.0100393, 0.0103667, 0.0102826, 0.0102321, 
   0.0104435, 0.0100861, 0.00972111, 0.00957839, 0.00966471, 
   0.00963415, 0.00880081, 0.00873474, 0.00859673, 0.00816511, 
   0.00792854, 0.0075657, 0.00747507, 0.00734244, 0.00729432};

fdist = FindDistribution[dist]
paramsA = FindDistributionParameters[dist, LogNormalDistribution[μ, σ]]
paramsB = FindDistributionParameters[dist, UniformDistribution[{min, max}]]

Plot[PDF[LogNormalDistribution[μ, σ] /. paramsA, {x}], {x, 0, 0.1}, 
     Filling -> Axis, PlotRange -> All]
Plot[PDF[UniformDistribution[{min, max}] /. paramsB, {x}], {x, 0, 0.1},
     Filling -> Axis, PlotRange -> All]
Plot[PDF[fdist, {x}], {x, 0, 0.1}, Filling -> Axis, PlotRange -> All]

MixtureDistribution[{0.747238, 0.252762}, {LogNormalDistribution[-3.99373, 0.633476], UniformDistribution[{0.00729432, 0.134408}]}]

Answer 3

data = {0.0258899, 0.047329, 0.0683456, 0.0861312, 0.103529, 0.118554, 
   0.124337, 0.127011, 0.131217, 0.134408, 0.132016, 0.130485, 0.127444, 
   0.122495, 0.117607, 0.112261, 0.108703, 0.10322, 0.100889, 0.0955182, 
   0.0902072, 0.085152, 0.079376, 0.0752159, 0.0722947, 0.0686795, 0.0653517, 
   0.062192, 0.0598642, 0.0569383, 0.0548917, 0.0513092, 0.0485989, 0.0472422,
    0.0458829, 0.0434763, 0.0406302, 0.03872, 0.0371639, 0.0370529, 0.0353547,
    0.0332693, 0.0320803, 0.0302454, 0.0300313, 0.0294249, 0.0281058, 
   0.0271762, 0.0256051, 0.0257485, 0.0250776, 0.0237297, 0.0227674, 
   0.0225286, 0.0221572, 0.0213558, 0.0206458, 0.0199427, 0.0191173, 
   0.0172471, 0.017145, 0.0170844, 0.016805, 0.0162183, 0.0150205, 0.0149596, 
   0.0148139, 0.0142276, 0.0141969, 0.0139716, 0.0132321, 0.0128165, 
   0.0129189, 0.0126378, 0.0123681, 0.0122255, 0.0121722, 0.0117113, 0.010789,
    0.0105248, 0.0100393, 0.0103667, 0.0102826, 0.0102321, 0.0104435, 
   0.0100861, 0.00972111, 0.00957839, 0.00966471, 0.00963415, 0.00880081, 
   0.00873474, 0.00859673, 0.00816511, 0.00792854, 0.0075657, 0.00747507, 
   0.00734244, 0.00729432};

Try a mix of distributions

distModel = MixtureDistribution[{w, 1 - w},
   {LogNormalDistribution[m, s1],
    RayleighDistribution[s2]}];

dist = distModel /.
   FindDistributionParameters[data,
    distModel];

Integrate[PDF[dist, x], {x, 0, Infinity}]

(*  1.  *)

Show[
 Histogram[data, 11, "PDF"],
 Plot[PDF[dist, x], {x, 0, Max[data]}],
 PlotRange -> All]