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

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Sim 2023-6-21

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评论： Sim 2023-6-22

采纳的回答： Aman

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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)

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Answer 1

Aman 2023-6-21

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https://ww2.mathworks.cn/matlabcentral/answers/1986134-usage-of-chi2gof-how-to-derive-the-expected-values-from-the-fitting-distribution-to-then-use-them#answer_1259859

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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!

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Sim 2023-6-22

编辑：Sim 2023-6-22

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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 2023-6-22