LOG PEARSON TYPE III distribution

Could someone kindly tell me how to calculate the inverse of the CDF of a log pearson type III distribution?

In fact, I need to use Monte Carlo sampling to generate samples for certain log pearson type III distributed data. I should be able to take the inverse of the CDF of a log pearson type III distribution for this.

Answer 1

[David Goodmanson]

Hi Raj,

MODIFIED

Since the pearson type iii is a gamma distribution with a shifted and scaled variable, that means you can use the Matlab gamrnd function to produce random draws and proceed accordingly.

The pearson type iii distribution depends on three parameters. This answer assumes that the task is to create random draws from a distribution with three known parameters. For parameters a,b,p and for the log distribution, the pdf is 'pdfa' below. If the task is fitting the parameters from a known data set, then this answer does not apply.

Instead of gamrnd, you can use the inverse cdf method as you mentioned, which is shown in fig. 2 below. As is almost always the case with that method, there are some problems producing exteme outliers when the cdf is close to 1. But it works reasonably well.

% pearson type iii in log variable:
% pdf = (1/b)*logxs.^(p-1).*exp(-logxs)/gamma(p)
% where
% logxs = (log(x)-a)/b   is a shifted and scaled log(x),
% with log(x) >= a

n = 1e5;  % number of random draws
p = 3;
a = 4.7;
b = 2.4;
% constuct pdfa, the analytic pdf
logx = a:.01:50;
logxs = (logx-a)/b;         % shifted and scaled log(x)          [1]
pdfa = (1/b)*gampdf(logxs,p,1);

% random draws using gamrnd
randlogxs = gamrnd(p,1,1,n);          % shifted and scaled log(x)
randlogx = randlogxs*b +a;            % undo [1] for pearson type iii
figure(1)
histogram(randlogx,'Normalization','pdf'); grid on
hold on
plot(logx,pdfa)
hold off

% random draws using inverse of cdf
cdfa = cumtrapz(logx,pdfa);
r = rand(1,n);
r(r>max(cdfa)) = [];
randlogx = spline(cdfa,logx,r);  % or randlogx = interp1(cdfa,logx,r,'spline');
figure(2)
histogram(randlogx,'Normalization','pdf'); grid on
hold on
plot(logx,pdfa)
hold off

[Raj Arora]

Thankyou sir for your response. I have some questions to ask

1. Can you please let me know what are m, b0, b1, mu1 (primary constant)?
2. what is logx_min and lamda?

the plot looks like a log pearson type III distribution as it is left skewed but what are the randoms samples for the same

Example: I have attached some data for this I want to generate samples using MCS

number of sample i want = 1e7

Prob. Value

0.99 841

0.9 1200

0.5 2190

0.1 4960

0.05 6530

0.02 9110

0.01 11500

0.005 14500

0.002 19200

[David Goodmanson]

Hello Raj,

I assumed that the task was to create random draws from a pearson type iii distribution with four known parameters (called 'primary constants' berfore). I modified the answer to make that assumption clear. If the task is to find the parameters by fitting known data, then the answer does not apply.

Regardelss, lambda is a constant defined by the four paramters, and logx_min, which is defined by three of those parameters, is the value of logx at the left edge of the distribution.

[Raj Arora]

Thank you, sir. My actual objective is to generate random draws for pearson type III, but there are 3 variables to consider.

In reality, the gamma distribution has 2 parameters: shape and scale. In contrast, log pearson type III has 3 parameters: shape, scale, and location parameter.

On wiki, things aren't obvious about those 4 parameters, so I believe it's something different, but thanks for responding, sir.

[David Goodmanson]

Hello Raj,

I see what you mean. It looks like of the four parameters in Wikipedia, one of them is redundant. The much clearer Wolfram Mathworld definition is (adjusting for the variable being log)

pdf = (1/b)*logxs.^(p-1).*exp(-logxs)/gamma(p)
where logx denotes log(x) and
logxs = (logx-a)/b   is a shifted and scaled log(x)          [1]

The answer has been modified.

[Raj Arora]

Yes david your last comment is clear to me. I have some data with me with that data I already know all three paramter(given below a,b,p) and clearly I can take scaled value(log values) of all these. The question now revolves around the data provided as well as the three parameters that I have. I need to generate as many samples as I like using the same method, so let's say 1e7 (number of sample want to generate).

I simply needed those samples using the inverse CDF of the log pearson type III distribution, which will obviously be in the range of 841-19200.

Can you consider the below data and help me with this

Data : 841, 1200, 2190, 4960, 6530, 9110, 11500, 14500, 19200

b =-0.3161 ; p =12.4640 ; a = 12.4647