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!
Best Answer
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.