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caculate confidence interval from customized pdf

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Hi
I'm wondering How can I caculate the confidence interval of customized pdf e.g. Gaussian mixture distribution?
pdf=@(x) w1*normpdf(x,mu1,sigma1)+w2*normpdf(x,mu2,sigma2);
cdf=@(x) integral(pdf,-Inf,x);
As icdf function only support specified distribution, I'm wondering how to caculate the shortest confidence interval?

Accepted Answer

David Goodmanson
David Goodmanson on 19 Mar 2024
Edited: David Goodmanson on 19 Mar 2024
Hello SX,
Ordinarily to find the cdfs you would have to use numerical integration. In this case for the normal distributions, the cdf function is available. Then you can interpolate using the cdf as the independent variable. Here is an example. In the plot you get a wide minimum which you might expect.
mu1 = 1;
mu2 = 2;
sig1 = 1;
sig2 = 3;
w1 = .3;
w2 = .7;
c = .9; % confidence span, there is probably a better name for this
x = -20:.00001:20;
cdf = w1*normcdf(x,mu1,sig1) +w2*normcdf(x,mu2,sig2);
cdn = linspace(min(cdf),max(cdf)-c,1e4);
xdn = interp1(cdf,x,cdn);
cup = linspace(min(cdf)+c,max(cdf),1e4);
xup = interp1(cdf,x,cup);
figure(1); grid on
plot(xup-xdn)
[x0 ind] = min(xup-xdn);
xdn(ind) % lower end of confidence interval
xup(ind) % upper end of confidence interval
cdn(ind) % lower cdf value
cup(ind) % upper cdf value
% ans = -2.4087
% ans = 6.3858
% ans = 0.0497
% ans = 0.9497
D = xup(ind)-xdn(ind) % the result
cup(ind)-cdn(ind) % check, should be c = confidence span
% D = 8.7945
% ans = 0.9000
% try a different case, get a larger confidence interval
xtest = interp1(cdf,x,[.07 .97]);
Dtest = diff(xtest)
% xtest = -1.8602 7.1554
% Dtest = 9.0156
  3 Comments
David Goodmanson
David Goodmanson on 19 Mar 2024
See what you think of the modified answer above
苏越 徐
苏越 徐 on 19 Mar 2024
Thank you David! It' good idea to narrow the range for ends of confidence span and then search for the shortest.

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