# 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.

• Take a look at this answer, in particular equation (3), which shows the equation the median $m$ would need to satisfy. We need to show that there is no linear function $m=p\mu_A+(1-p)\mu_B$ that works in general. At some point, we will need to make use of the specific form of the normal CDF, which is where things probably get painful. May 9, 2021 at 12:27
• @StephanKolassa : I already had a look at it. However, as you said, probably one needs to use a particular form of CDFs. But then, I am wondering how you made the general statement about nonlinearity of quantiles. May 9, 2021 at 12:41
• This is a nice question, thank you! I have to admit that I like my answer. I edited my answer to the motivating question to link here. May 9, 2021 at 16:09

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.