A proof that the median is a nonlinear statistical functional This question is with reference to the top answer (by @StephanKolassa) to this question. Let $F$ and $G$ be CDFs and define $$H(x)=aF(x)+(1-a)G(x)$$ with $a\in [0, 1]$. Now suppose $F$ and $G$ are CDFs corresponding to Gaussian distributions with means $\mu_1$ and $\mu_2$. Also, assume that $T(F)$ is a statistical function that gives median for the CDF $F$. I want to prove that $T$ is not linear. That is: $$T(H) \neq aT(F)+(1-a)T(G)=a\mu_1+(1-a)\mu_2$$.
The answer mentioned above claims that any quantile of a Gaussian mixture is a nonlinear functional. I tried proving this by contradiction as follows. If $m$ is the median of $H$, and if the equality holds, we have: $$m=a\mu_1+(1-a)\mu_2$$. In that case, we will have:
$$aF[a\mu_1+(1-a)\mu_2]+(1-a)G[a\mu_1+(1-a)\mu_2]=\frac{1}{2}$$
However, there doesn't seem to be any way to proceed further from here which will produce a contradiction. Any help is highly appreciated.
 A: Here is a very simple and almost-rigorous argument.
The medians of the two components are $m_1=\mu_1$ and $m_2=\mu_2$. In particular, they are independent of the components' standard deviations $\sigma_1$ and $\sigma_2$. So any weighted linear combination of the components' medians will also be independent of $\sigma_1$ and $\sigma_2$.
Now, consider our mixture. We can assume $\mu_1<\mu_2$. The median $m$ of the mixture will lie between $m_1=\mu_1$ and $m_2=\mu_2$.
We keep all parameters fixed and reduce $\sigma_1$. This shifts probability mass for our mixture to the left, so $m$ will get smaller. (*) Thus, $m$ is a (non-trivial) function of $\sigma_1$. But above, we have seen that any weighted linear combination of $m_1$ and $m_2$ will be independent of both $\sigma_1$ and $\sigma_2$, a contradiction. Thus, $m$ cannot be a weighted linear combination of $m_1$ and $m_2$.
Of course, it's statement (*) that is intuitively obvious but not yet rigorous. We still need to prove that $m$ is an increasing function of $\sigma_1$ if all other parameters are kept fixed. It may be possible to show that the defining function of $m$ per equation (3) in this earlier answer defines an $m$ that is a differentiable function of $\sigma_1$, then do some implicit differentiation (where we still probably need to argue that we can differentiate under the integral for the improper integral $\int_{-\infty}^m$) and finally find that $\frac{dm}{d\sigma_1}>0$.
Alternatively, it would be easy to take specific normals, say $N(0,\sigma_1^2)$ and $N(2,1)$ with equal weights. For $\sigma_1=1$, we have a symmetric situation, so $m=1$. We note that the $N(2,1)$ has probability mass of $0.067$ to the left of $0.5$ (R: pnorm(0.5,2,1)), so if we reduce $\sigma_1$ far enough that the mass of the $N(0,\sigma_1^2)$ to the left of $0.5$ is large enough, we can estimate that the new median $m'$ of the mixture with $\sigma_1\ll1$ definitely satisfies $m'<0.5$. For this, we just need to trust quantiles, or tables.
