7

I have the following code that I wish to estimate the parameters of a custom distribution. For more details on the distribution. Then using the estimated parameters I want to see if the estimated PDF resembles the distribution of the given the data (it is supposed to match the distribution of the given data).

[EDIT]: 'x' now holds a sample of data and not a PDF

The main code is:

x           = [0.0320000000000000   0.0280000000000000  0.0280000000000000  0.0270000000000000  0.0320000000000000  0.0320000000000000  0.0480000000000000  0.0890000000000000  0.0500000000000000  0.0620000000000000  0.0480000000000000  0.0300000000000000  0.0520000000000000  0.0460000000000000  0.0540000000000000  0.0520000000000000  0.0510000000000000  0.0310000000000000  0.0330000000000000  0.0330000000000000  0.0380000000000000  0.0850000000000000  0.102000000000000   0.0290000000000000  0.0530000000000000  0.0590000000000000  0.0320000000000000  0.0800000000000000  0.0410000000000000  0.0280000000000000  0.0670000000000000  0.0350000000000000  0.0420000000000000  0.0280000000000000  0.0370000000000000  0.0480000000000000  0.0330000000000000  0.101000000000000   0.0420000000000000  0.0840000000000000  0.0340000000000000  0.0900000000000000  0.0900000000000000  0.0460000000000000  0.0290000000000000  0.0330000000000000  0.0350000000000000  0.0330000000000000  0.0320000000000000  0.0420000000000000  0.0600000000000000  0.0500000000000000  0.0390000000000000  0.0480000000000000  0.0680000000000000  0.0330000000000000  0.0510000000000000  0.0430000000000000  0.0270000000000000  0.0330000000000000  0.0590000000000000  0.0380000000000000  0.0270000000000000  0.0600000000000000  0.0310000000000000  0.0520000000000000  0.0350000000000000  0.0640000000000000  0.0570000000000000  0.0520000000000000  0.0330000000000000  0.0480000000000000  0.0530000000000000  0.0380000000000000  0.0320000000000000  0.0340000000000000  0.0380000000000000  0.0470000000000000  0.0950000000000000  0.0510000000000000  0.0280000000000000  0.124000000000000   0.0360000000000000  0.0670000000000000  0.0380000000000000  0.0760000000000000  0.0440000000000000  0.0390000000000000  0.0500000000000000  0.0500000000000000  0.0370000000000000  0.0350000000000000  0.0490000000000000  0.0570000000000000  0.0560000000000000  0.0500000000000000  0.0350000000000000  0.0390000000000000  0.0390000000000000  0.0310000000000000  0.0260000000000000  0.0350000000000000  0.0610000000000000  0.0280000000000000  0.0480000000000000  0.0560000000000000  0.0650000000000000  0.0400000000000000  0.131000000000000   0.0600000000000000  0.0310000000000000  0.0620000000000000  0.0320000000000000  0.0510000000000000  0.0510000000000000  0.0480000000000000  0.0420000000000000  0.0470000000000000  0.0690000000000000  0.0590000000000000  0.0550000000000000  0.0580000000000000  0.0410000000000000  0.0320000000000000  0.0440000000000000  0.0370000000000000  0.0390000000000000  0.0620000000000000  0.0290000000000000  0.0340000000000000  0.0490000000000000  0.0300000000000000  0.0540000000000000  0.0420000000000000  0.0610000000000000  0.159000000000000   0.0520000000000000  0.0450000000000000  0.0670000000000000  0.0590000000000000  0.111000000000000   0.0250000000000000  0.0660000000000000  0.0210000000000000  0.0370000000000000  0.0330000000000000  0.0510000000000000  0.0330000000000000  0.0450000000000000  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0.0440000000000000  0.0290000000000000  0.0280000000000000  0.0350000000000000  0.0840000000000000  0.0660000000000000  0.0410000000000000  0.0300000000000000  0.0440000000000000  0.0450000000000000  0.0470000000000000  0.0620000000000000  0.0420000000000000  0.0300000000000000  0.0330000000000000  0.0320000000000000  0.0440000000000000  0.0700000000000000  0.0340000000000000  0.0420000000000000  0.0480000000000000  0.0360000000000000  0.0590000000000000  0.106000000000000   0.0280000000000000  0.0540000000000000  0.0870000000000000  0.0300000000000000  0.0300000000000000  0.0370000000000000  0.0210000000000000  0.0360000000000000  0.0910000000000000  0.126000000000000   0.0780000000000000  0.0510000000000000  0.0500000000000000  0.0370000000000000  0.0540000000000000  0.0380000000000000  0.0350000000000000  0.0480000000000000  0.0300000000000000  0.0340000000000000  0.133000000000000   0.0330000000000000  0.0340000000000000  0.0480000000000000  0.0590000000000000  0.0460000000000000  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0.0530000000000000];

Censored    =  ones(1,size(x,2));% 
custpdf     =  @eval_custpdf;
custcdf     =  @eval_custcdf;
options     =  statset('Display','iter','MaxFunEvals',1000,'MaxIter',1000,...
                'FunValCheck','off','TolX',1.0e-10,'TolFun',1.0e-10);
phat        =  mle(x,'pdf', custpdf,'cdf', custcdf,...
               'start',[0.6,0.02,1.01,2,4,-10],...
               'lowerbound',[0 0 0 0 0 -inf],...
               'upperbound',[inf inf inf inf inf inf],...
               'Censoring',Censored,...
               'Options',options);;

% Checking how close the estimated PDF and CDF match with those from the data x
figure();
h        =  histogram(x,'Normalization','probability');hold on
x_times  =  h.BinEdges(1:end-1) + h.BinWidth/2 ;
y_vals   =  custpdf(x_times,  phat(1), phat(2), phat(3), phat(4), phat(5), phat(6))./...
           sum(custpdf(x_times,  phat(1), phat(2), phat(3), phat(4), phat(5), phat(6)),'omitnan');
plot(x_times,y_vals,'linewidth',2)
legend('Data','Estimated PDF')

The functions are:

function out = eval_custpdf(x,myalpha,mytheta,mybeta,a,b,c)
  first_integral      =  integral(@(x) eval_K(x,a,b,c),0,1).^-1;
  theta_t_ratio       =  (mytheta./x);
  incomplete_gamma    =  igamma(myalpha,theta_t_ratio.^mybeta);
  n_gamma             =  gamma(myalpha);
  exponent_term       =  exp(-theta_t_ratio.^mybeta-(c.*(incomplete_gamma./n_gamma)));


  numerator   =  first_integral.* mybeta.*incomplete_gamma.^(a-1).*...
               theta_t_ratio.^(myalpha*mybeta+1).*exponent_term;
  denominator =  mytheta.* n_gamma.^(a+b-1).* (n_gamma-incomplete_gamma.^mybeta).^(1-b);

  out         =  numerator./denominator;
end

function out = eval_custcdf(x,myalpha,mytheta,mybeta,a,b,c)
 out = zeros(size(x));
   for i = 1: length(x)
     first_integral      =  integral(@(x) eval_K(x,a,b,c),0,1).^-1;
     theta_t_ratio       =  mytheta./x(i);
     incomplete_gamma    =  igamma(myalpha,theta_t_ratio.^mybeta);
     n_gamma             =  gamma(myalpha);
     second_integral     =  integral(@(x) eval_K(x,a,b,c),0,...
                            incomplete_gamma.^mybeta./n_gamma);
    %  second_integral     =  integral(@(x) eval_K(x,a,b,c),0,2);
     out(i)     =  first_integral*second_integral;
   end
end 

function out = eval_K(x,a,b,c)

  out =  x.^(a-1).*(1-x).^(b-1).*exp(-c.*x);

end

However, I have not been successfull to obtain the desired PDF. As you can see in the figure, the estimated PDF (orange line) does not trace the histogram of 'x' (blue bars).

[UPDATE] Normalized histogram

enter image description here Note that I have varied the initial values of the parameters. But this is very time consuming. I also increased the number of iterations and minimized the tolerance, but no luck yet. Is there a better way to estimate the paratemters other than mle?

Any help would be appreciated.

Thanks in advance.

nashynash
  • 375
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    What output do you get and how is it different than what you desire? MLE is expecting sample data. It looks like you're feeding it a PDF? If you want to fit the PDF and not a sample then MLE is the wrong tool. What you're looking for is a curve fit rather than a distribution fit. – John Jun 17 '19 at 20:05
  • @John You are right. I am feeding a PDF and not a sample. I will change that and see if feeding a sample works. Thanks a lot for pointing that out. – nashynash Jun 18 '19 at 00:41
  • @John I updated 'x' with a sample data. I ran the code but still, I could not get the desired results. – nashynash Jun 18 '19 at 07:27
  • What results are you getting now? Are you assessing success by comparing the PDF and histogram? If so, are they normalized in the same way? – John Jun 24 '19 at 16:44
  • @John The figure shows the result i get now. Yes, I am comparing the PDF to the histogram to asses the success. I believe they are normalised in the same way. – nashynash Jun 24 '19 at 20:35
  • The histogram isn't normalized. The PDF is probably normalized to sum to 1, but the sum of the histogram is the total number of data points. Divide the histogram counts by their sum and see how the graph looks then. – nhowe Jun 25 '19 at 00:12
  • @nhowe I have included an additional figure that shows a normalized histogram and normalized PDF as well. No change can be seen in the outcome – nashynash Jun 25 '19 at 07:04
  • I believe the mistake is in your custom cdf/pdf, which is more or less impossible for us to debug as we do not know which function you intended to write. So to proceed, I suggest that you write down the PDF and CDF very carefully and then check equally carefully that you have entered them correctly, if this does not work, you can upload a picture of the function you intended to write. Also, where is the rest of the mass? From the histogram, I can see less than 3% of the probability mass (Could this be the mistake?) – Nicky Mattsson Jul 02 '19 at 14:19
  • @NickyMattsson Thank you for taking interest. A link to the function is already provided in “For more details on the distribution”. I am not sure what you mean by ‘less than 3% of the probability mass’. Could you elaborate more? I plotted the histogram of ‘x’ and the estimated PDF. The estimated PDF is generated at each values on the x-axis. – nashynash Jul 02 '19 at 14:28
  • Sorry I missed the link. By "less than 3% of the probability mass" I meant that if you plot the pdf, its integral should be 1, yes? A very generous bound for your plot is `0.16*0.15=0.025`. However, this is not the problem, but simply fixed by using `histogram(x,'Normalization','pdf')` instead. – Nicky Mattsson Jul 02 '19 at 14:34
  • @NickyMattsson It’s funny you point that out because I had plotted the histogram with ‘pdf’ as normalisation. However, user nhowe suggested that I normalise it with the sum and so I used ‘probability’ insisted. And that does not seem to fix the problem. – nashynash Jul 02 '19 at 14:40
  • A little late and not relating to the ultimate solution, but for normalizing the histogram the integral needs to be 1 so the normalization factor is the sum of the counts multiplied by the bin width. – John Jul 08 '19 at 17:25

1 Answers1

5

Noting this from the help

If the 'censoring' name/value pair is not present, you may omit the 'cdf' name/value pair.

gives us a first suggestion for debugging. So removing the censoring part and the CDF from the input list and running

phat =  mle(x,'pdf', @eval_custpdf,'start',[0.6,0.02,1.01,2,4,-10]);
phat =  mle(x,'pdf', @eval_custpdf,'start',phat); %Restart for better result

yields the figure

plot1

Telling us that the problem might be in the CDF function. Comparing with the link given in the question, we see that this line

second_integral     =  integral(@(x) eval_K(x,a,b,c),0,incomplete_gamma.^mybeta./n_gamma);

should have been

second_integral =  integral(@(x) eval_K(x,a,b,c),0,incomplete_gamma./n_gamma);
Nicky Mattsson
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  • That is awesome Nicky and I love the way you broke down the explanation. It works now. Thank you very much. I also found that in the PDF function `denominator = mytheta.* n_gamma.^(a+b-1).* (n_gamma-incomplete_gamma.^mybeta).^(1-b);` should have been `denominator = mytheta.* n_gamma.^(a+b-1).* (n_gamma-incomplete_gamma).^(1-b);`. – nashynash Jul 03 '19 at 05:57