# Overlapping Coefficient between for two Gaussian distributions where one of them has zero variance

Consider the case where we have two Gaussian distributions and we are trying to find the area of overlap between them. This is also called the Overlapping Coefficient and I think it is also related to the Weitzman Measure.

I am wondering about a specific limiting case when one (but not the other) of the Gaussians has a variance of zero, i.e. the PDF is a unit impulse function. In this case what would be the Overlapping Coefficient?

My intuition is that it will be zero, but I am not sure.

It must be something like this: $\int_{-\infty}^\infty f_{\mu_1,\sigma_1^2}(x)\delta(x-\mu_2)dx=f_{\mu_1,\sigma_1^2}(\mu_2)$, where $\mu$ and $\sigma^2$ are mean and variance of gaussians.

UPDATE. The above is wrong. The answer is ZERO. Here's why.

Two gaussians: fat and thin, different means. They'll intersect in exactly two points. Between the points the thin is above fat, beyond the points the fat is over thin. As we make thin thinner, these two points approach each other, and the area beyond the points keeps shrinking. In the limit it disappears completely, all the area of the thin gaussian will be above the fat gaussian. The max point of thin goes to infinity, so all the area will be concentrated above the fat gaussian.

I'm sure it's possible to put this in formulas.

UPDATE 2: @whuber showed how this can be easily proved. He computes the area of the thin gaussian, given that it is below the highest point of the fat gaussian; and shows that it shrinks to zero. This area covers the area of the overlap.

• I'm pretty sure the overlap coefficient is supposed to be restricted to the unit interval; this result is not. – Glen_b -Reinstate Monica Sep 12 '14 at 9:13
• @Glen_b - yes, overlap coefficient is supposed to be in [0 1]. Any suggestions for this then. – Tensored Plasma Sep 15 '14 at 12:19
• +1. To put this in formulas, let $y$ be the maximum of the PDF of one Gaussian and let the other have an SD of $\sigma$ which is so small that $u(\sigma)=-\log(y^2)-\log(\sigma^2)+\log(2\pi)$ is positive. By focusing on the part of the PDF of the other that rises above $y$ you can obtain an upper bound for the overlap equal to $2(1+\sigma y-\Phi(\sqrt{u(\sigma)}))$ (where $\Phi$ is the standard Normal CDF). As $\sigma\to 0$, $u(\sigma\to\infty)$, whence $\sigma y\to 0$ and $\Phi(\sqrt{u(\sigma)})\to 1$, showing that the upper bound decreases to $0$, QED. – whuber Sep 15 '14 at 18:08
• @whuber I am not sure I understand. Are you saying that Akskal solution above is bounded and hence correct? I feel I may be missing something. Could you pls elaborate. – Tensored Plasma Sep 25 '14 at 14:02
• He's showing that in the thin gaussian the part which is under the highest point of the fat gaussian is shrinking to zero. This part covers the overlapping part, and it is easier to evaluate its area. – Aksakal Sep 25 '14 at 14:07