Usage of chi2gof: how to derive the "expected" values from the fitting distribution to then use them as argument of chi2gof?

How to derive the "expected" values from the fitting distribution to then use them as argument of chi2gof?

observed_data = exprnd(2, 100, 1);
xgrid = linspace(0,100,1000)';
pd = fitdist(observed_data,'Exponential'); % <-- fitting distribution
hold on
line(xgrid,pdf(pd,xgrid),'Linewidth',2,'color','b')
histogram(observed_data,100,'Normalization','pdf','facecolor','blue')
hold off

% Desired output
% [h, p] = chi2gof(observed_data, 'Expected', expected_counts)

Answer 1

[Aman]

Hi Sim,

To generate the expected counts from a fitted distribution, you can use the probability density function (PDF) of the fitted distribution to generate the expected values for each bin.

Here's an example for the same,

observed_data = exprnd(2, 100, 1);

pd = fitdist(observed_data,'Exponential');

num_bins = 10;
bin_edges = linspace(min(observed_data), max(observed_data), num_bins+1);

expected_values = numel(observed_data) * diff(cdf(pd,bin_edges));
% since the area of the pdf must be 1, it is better to obtain the expected_counts by multiplying with the total number of observations accumulated on every single bin
expected_counts = expected_values * numel(observed_data);

[h, p] = chi2gof(observed_data, 'Expected', expected_counts);

Hope this helps!

[Sim]

Hi, I did a test, and it looks like that by using "expected_counts"

expected_counts = expected_values * numel(observed_data);

within chi2gof looks like to give a wrong answer (if possible, I would be grateful if you could check it as well). I therefore, removed that line from the code and I replaced "expected_counts" with "expected_values" inside chi2gof:

[h,p,st] = chi2gof(observed_data, 'Expected', expected_values)

Test 1, with corrections

observed_data = exprnd(2, 1000, 1);        % <-- Exponential random numbers
pd = fitdist(observed_data,'Exponential'); % <-- fitting exponential distribution

xgrid = linspace(0,10,1000)';
num_bins = 20;

figure
hold on
line(xgrid,pdf(pd,xgrid),'Linewidth',2,'color','k')
histogram(observed_data,num_bins,'Normalization','pdf','facecolor','b')
hold off

bin_edges = linspace(min(observed_data), max(observed_data), num_bins+1);
expected_values = numel(observed_data) * diff(cdf(pd,bin_edges));
[h,p,st] = chi2gof(observed_data, 'Expected', expected_values)

h = 0

p = 0.0934

st = struct with fields:
    chi2stat: 20.0719
          df: 13
       edges: [1.1578e-04 0.7060 1.4119 2.1178 2.8237 3.5296 4.2355 4.9414 5.6473 6.3531 7.0590 7.7649 8.4708 9.1767 14.1180]
           O: [304 178 157 107 62 47 36 36 21 14 17 10 2 9]
           E: [285.6580 204.0530 145.7604 104.1205 74.3760 53.1287 37.9513 27.1096 19.3651 13.8330 9.8813 7.0584 5.0420 11.4111]

Test 2, with corrections

observed_data = normrnd(5, 1, 1000,1);     % <-- Normal random numbers
pd = fitdist(observed_data,'Normal');      % <-- fitting normal distribution

xgrid = linspace(0,10,1000)';
num_bins = 20;

figure
hold on
line(xgrid,pdf(pd,xgrid),'Linewidth',2,'color','k')
histogram(observed_data,num_bins,'Normalization','pdf','facecolor','b')
hold off

bin_edges = linspace(min(observed_data), max(observed_data), num_bins+1);
expected_values = numel(observed_data) * diff(cdf(pd,bin_edges));
[h,p,st] = chi2gof(observed_data, 'Expected', expected_values)

h = 0

p = 0.8183

st = struct with fields:
    chi2stat: 10.8560
          df: 16
       edges: [2.0865 2.7114 3.0239 3.3363 3.6488 3.9612 4.2737 4.5861 4.8986 5.2110 5.5235 5.8359 6.1484 6.4608 6.7733 7.0857 7.3982 8.3355]
           O: [6 18 20 38 61 79 115 120 117 131 96 83 55 30 17 11 3]
           E: [7.3848 11.2279 21.7932 38.0906 59.9498 84.9640 108.4324 124.6127 128.9570 120.1730 100.8435 76.2022 51.8518 31.7714 17.5300 8.7096 6.0358]

As you can see, for both Tests, I now get

h = 0

which means that both tests FAIL to rejects the null hypothesis at the 5% significance level, which means that both fitting distributions follow their corresponding datasets, as we can also notice by a visual inspection of both plots.

Do you agree?

[Aman]

Hi Sim, Yes, sorry for confusion, that was for PDF, but we needed CDF, so I think expected_values should be used.

[Sim]

thanks a lot @Aman :-)