# How to find the point where two normal distributions intersect? [duplicate]

I have two normal distributions. Say for example, one with a mean of .76 and one with a mean of .62. Both standard deviations = .05. How do I find which point on the x-axis is where the two distributions cross? In other words, x-coordinate at the deepest part of the valley in between the two distributions?

P.S. I am using R, but non-R related answers would be welcome and appreciated too.

When the standard deviations are the same, the densities intersect at the midpoint of the means.

To answer the more general question in the title, presuming the distributions aren't identical, there may be either one or two intersection points (typically two, unless the means differ but the standard deviations don't, as discussed above). The two intersections are easiest to find on the log-density scale.

Keeping in mind that normal densities are everywhere positive,

\eqalign{ &f_1(x) = f_2(x)\\ &\implies \log f_1(x) = \log f_2(x)\\ &\implies \log f_1(x) - \log f_2(x)=0. }

If $f_i$ has mean $\mu_i$ and standard deviation $\sigma_i$ ($i=1,2$) then

$$\log(f_i(x))= -\frac12 \log(2\pi) - \frac12 \log(\sigma_i^2) -\frac12(x- \mu_i)^2/\sigma_i^2.$$

So \eqalign{ \log(f_1(x))-\log(f_2(x))&= \frac12 [\log(\sigma_2^2)-\log(\sigma_1^2) +(x- \mu_2)^2/\sigma_2^2-(x- \mu_1)^2/\sigma_1^2]\\ &=\frac12( Ax^2+Bx+C) } where

\eqalign{ A &= -1/\sigma_1^2+1/\sigma_2^2\\ B &= 2(-\mu_2/\sigma_2^2 + \mu_1/\sigma_1^2)\\ C &= \mu_2^2/\sigma_2^2 - \mu_1^2/\sigma_1^2+ \log(\sigma_2^2/\sigma_1^2). }

When $\sigma_1\ne\sigma_2$ we can simply apply the quadratic formula to find the (real) roots of the quadratic, which will give the x-values for the intersection points.

Since $\sigma_1^2-\sigma_2^2$ and $\log(\sigma_1^2/\sigma_2^2)$ have the same sign, the discriminant

$$\Delta = B^2 - 4 AC = \frac{4}{\sigma_1^2\sigma_2^2}\left((\mu_1-\mu_2)^2 + (\sigma_1^2-\sigma_2^2)\log\left(\frac{\sigma_1^2}{\sigma_2^2}\right)\right)$$

is nonnegative and equals zero only when $\mu_1=\mu_2$ and $\sigma_1=\sigma_2.$ Therefore, when $A\ne 0$ (that is, $\sigma_1\ne \sigma_2$) there are always two points of intersection (and, trivially, infinitely many when $\Delta=0$ which is the case of identical distributions).

• This is incorrect, since when variances are equal, it should be so that: $A$$C$ = $B^2$ And the factor of $m$ is $1$, and there is then only one intersection point, one solution to the quadratic equation. – Aku-Ville Lehtimäki Mar 4 '18 at 9:08
• Is it still incorrect? – sammosummo Mar 21 '18 at 20:36
• It looks like whuber fixed the algebraic error. – Glen_b Mar 21 '18 at 22:41