caculate confidence interval from customized pdf

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

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

Thank you David for your valuable suggestion! However, as the pdf of gaussian mixture distribution is asymmetrical, [.05 .95] is not the shortest interval for the given confidence value(e.g. 90% in your example). I guess the function fmincon may help to find the shortest interval, but is there any function easy to efficienctly find the shortest interval by numerical methods?
See what you think of the modified answer above
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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