Finding the intersection of Gaussian distributions - matlab

Question

How can I find all the points where Gaussian mixture distributions intersect in MATLAB?

Answer 1

The generic category of your question is finding the intersection of two curves, which is a manageable but non-trivial task (the hardest part is making sure you catch all the intersections).
But your problem is very specific: you're looking for the intersection of two Gaussians. This is very good: we both have an analytic formula for your function, and it's guaranteed that there are exactly two intersections unless some of your parameters are the same.*
Let's assume that your distributions are characterized by means mu1, mu2 and scale sigma1, sigma2. Then your Gaussians at position x are defined by the function
1/sqrt(2*pi*sigma^2) * exp(-(x-mu)^2/2/sigma^2)
Turns out that we can fully solve this equation for x on paper:
1/sqrt(2*pi*sigma1^2)*exp(-(x-mu1)^2/2/sigma1^2) == 1/sqrt(2*pi*sigma2^2)*exp(-(x-mu2)^2/2/sigma2^2)
sigma2/sigma1 == exp((x-mu1)^2/2/sigma1^2) * exp(-(x-mu2)^2/2/sigma2^2)
log(sigma2/sigma1) == (x-mu1)^2/2/sigma1^2) - (x-mu2)^2/2/sigma2^2
which results in the parabolic equation ax^2 + bx + c == 0 where
a = 1/(2*sigma1^2) - 1/(2*sigma2^2);
b = mu2/(sigma2^2) - mu1/(sigma1^2);
c = mu1^2/(2*sigma1^2) - mu2^2/(2*sigma2^2) - log(sigma2/sigma1);
It can be easily proved that the D = b^2 - 4 a c discriminant is non-negative, so indeed the equation has two real roots when the parameters aren't degenerate. So the two intersection points are, with the above definitions,
D = b^2 - 4 * a * c;
x1 = (-b + sqrt(D))/(2*a);
x2 = (-b - sqrt(D))/(2*a);
Using two gaussians with the pseudo-random parameters:
% define parameters and Gaussian
mu1=1; sigma1=3; mu2=2; sigma2=4;
% intersections
a = 1/(2*sigma1^2) - 1/(2*sigma2^2);
b = mu2/(sigma2^2) - mu1/(sigma1^2);
c = mu1^2/(2*sigma1^2) - mu2^2/(2*sigma2^2) - log(sigma2/sigma1)
D = b^2 - 4 * a * c;
x1 = (-b + sqrt(D))/(2*a);
x2 = (-b - sqrt(D))/(2*a);
Just to prove that the above intersection points are correct:
>> f = #(x,mu,sigma) 1/sqrt(2*pi*sigma^2) * exp(-(x-mu).^2/2/sigma^2);
>> f(x1,mu1,sigma1) - f(x1,mu2,sigma2)
ans =
2.7756e-17
>> f(x2,mu1,sigma1) - f(x2,mu2,sigma2)
ans =
1.0408e-17
The above means that the values of the two Gaussians at points x1 and x2 are equal to one another within machine precision, which is as good as any numerical answer will get.
*I originally claimed that we always have two intersections as long as the Gaussians aren't exactly the same. Clearly if both the means and the variances are the same we have two degenerate curves and intersections become meaningless. But as Cris Luengo pointed out in a comment, it might happen that there's only one intersection: when the variances are the same and the means are different (i.e. we have two curves of the exact same shape shifted along x). In this case a=0, consequently the corresponding equation is b*x + c == 0, giving us x0 = -c/b for the intersection. So a more exact (but a bit pseudo-codey) answer (given a, b and c) is
if a == 0 % or allow some tolerance... <=> sigma1 == sigma2
if b == 0 % or allow some tolerance... <=> mu1 == mu2
% degenerate curves: a == b == c == 0, f1(x)==f2(x) for all x
disp('curves are degenerate...')
else
% single intersection: mu1 ~= mu2
x1 = -c/b;
end
else
% two intersections; both parameters are different
D = b^2 - 4 * a * c;
x1 = (-b + sqrt(D))/(2*a);
x2 = (-b - sqrt(D))/(2*a);