# Median of mixture of two Gaussian distributions with equal weights

I am given a population $P$ that is equally divided into subsets $A$ and $B$. I know that a property $H$ of the population $P$ is normally distributed with mean $\mu_1$ and variance $\sigma_1^2$ for subset $A$ and $\mu_2$ and variance $\sigma_2^2$ for subset $B$. The task at hand is to find the exact median value of $H$ for the entire population $P$.

This is a Gaussian mixture with equal weights. I know that for a Gaussian distribution, the mean and the median are equal. I am trying to set up an equation based on the given data that is going to allow me to determine the median. I believe that the density of the Gaussian mixture is:

$$f(h) = 0.5f_A(h) + 0.5f_B(h)$$

where $f_A$ is the Gaussian for subset $A$ and $f_B$ the Gaussian of subset $B$.Now the standard way to handle this problem for distributions of continuous random variables is to solve for $m$ the equation:

$$\int_m^{\infty} f(h)dh = \frac{1}{2}$$

This does not look like a good strategy for this problem (for computational reasons -- although I may be totally off). If anyone could suggest a nicer, more elegant approach, I would deeply appreciate it!

• Is this a homework problem? If so, can you read and add the self-study tag? Oct 3, 2016 at 23:35
• @DilipSarwate I am stating in the beginning that the subpopulations are of the same size. Oct 3, 2016 at 23:45
• @AndrewM No, I encountered a version of the problem in machine learning context and was trying to see you approach it rigorously. Should I still add the tag, since it is, literally, self-study? Oct 3, 2016 at 23:47
• Well, you know that $F_A[x]+F_B[x]=1$, where $x$ is the mixture median and $F_A,F_B$ are the cumulative distribution functions (CDFs) of the mixture components. For a numerical solution, this is not too bad of a strategy, given that you know $F_k(x)=\Phi[(x-\mu_k)/\sigma_k]$ for $k=A,B$, where $\Phi$ is the CDF of the standard normal distribution. Oct 4, 2016 at 1:45
• @GeoMatt22 Could you please elaborate a bit on how we get $F_A[x] + F_B[x] = 1$? The rest of your argument is perfectly clear! Oct 4, 2016 at 2:12

Let $$m$$ denote the median of the mixture distribution whose CDF is $$\frac 12F_A(x) + \frac 12 F_B(x)$$. Then, \begin{align} \frac 12F_A(m) + \frac 12 F_B(m) &= \frac 12 \tag{1}\\ &\Downarrow\\ F_A(m) + F_B(m) &= 1\\ &\Downarrow\\ \Phi\left(\frac{m-\mu_A}{\sigma_A}\right) + \Phi\left(\frac{m-\mu_B}{\sigma_B}\right) &= 1\\ &\Downarrow\\ \frac{m-\mu_A}{\sigma_A} + \frac{m-\mu_B}{\sigma_B} &= 0 \tag{2} \end{align} where the last implication follows from the fact that $$\Phi(x)+\Phi(-x)=1$$. In short, the "computational reasons" that are deterring the OP are not really worrisome at all: solving $$(2)$$ for $$m$$ is trivial, and we get that $$m$$ is the linear combination $$\dfrac{\mu_A\sigma_B + \mu_B\sigma_A}{\sigma_A+\sigma_B}$$ of $$\mu_A$$ and $$\mu_B$$. Of course, if the mixture weights are $$p$$ and $$1-p$$ where $$p \neq \frac 12$$, that is, the mixture distribution is $$p\cdot F_A(x) + (1-p)\cdot F_B(x), \quad p \neq \frac 12,$$ then we need to solve $$p\cdot \Phi\left(\frac{m-\mu_A}{\sigma_A}\right) + (1-p)\cdot\Phi\left(\frac{m-\mu_B}{\sigma_B}\right) = \frac 12. \tag{3}$$ This will likely need numerical solution for $$m$$: at least, no straightforward analytical solution to (3) springs to my mind. I strongly suspect that it will turn out that, in general, $$m$$ is not a linear function of $$\mu_A$$ and $$\mu_B$$. My suspicion would be confirmed if I could produce just one specific instance of values of $$p, \mu_A, \sigma_A, \mu_B,$$ and $$\sigma_B$$ for which $$m$$ is not a linear function of $$\mu_A$$ and $$\mu_B$$ but don't have a specific instance to offer.