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:


However, there doesn't seem to be any way to proceed further from here which will produce a contradiction. Any help is highly appreciated.

  • $\begingroup$ 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. $\endgroup$ May 9, 2021 at 12:27
  • $\begingroup$ @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. $\endgroup$
    – Peaceful
    May 9, 2021 at 12:41
  • 1
    $\begingroup$ 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. $\endgroup$ May 9, 2021 at 16:09

1 Answer 1


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.


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