# q-Means and the mode of a distribution

Let $$f:\mathbb{R} \rightarrow [0,\infty)$$ be a continuous probability density function on $$\mathbb{R}$$ such that $$$$\int_{\mathbb{R}} |x| f(x)\, dx < \infty,$$$$ and assume that $$f$$ has a strict global maximum $$x_0$$, that is $$f(x) < f(x_0)$$ for all $$x \neq x_0$$. For any fixed $$q \in (0,1]$$ consider the problem $$$$\min_{y \in \mathbb{R}} \int_{\mathbb{R}}|y-x|^{q} f(x) \,dx,$$$$

and let $$S_q$$ be the set of all minimizers for this problem. My question is: does $$S_q$$ "shrink" to $$x_0$$ for $$q \rightarrow 0$$, in the sense that for every $$\epsilon > 0$$ there is some $$\delta > 0$$ such that $$S_q \subset (x_0 - \epsilon, x_0 + \epsilon)$$ for every $$q \in (0, \delta)$$?

I found this last property stated in a monograph about applied statistics without any proof, but I am not so convinced of its validity.

NOTE (1). If we put for a fixed $$q \in (0,1]$$: $$$$F_q(y)=\int_{\mathbb{R}} |y-x|^q f(x) dx \quad (y \in \mathbb{R}),$$$$ then $$F_q$$ is a continuous function and $$\lim_{y \rightarrow - \infty} F_q(y) = \lim_{y \rightarrow \infty} F_q(y) = \infty$$, so the minimization problem

$$$$\min_{y \in \mathbb{R}} \int_{\mathbb{R}}|y-x|^{q} f(x) \,dx.$$$$

NOTE (2). For every $$q > 0$$ let $$g_q:\mathbb{R} \rightarrow \mathbb{R}$$ be the function: $$$$g_q(x) = \begin{cases} 1 &\quad &\text{if} &\quad |x| > q,\\ 0 &\quad &\text{if} &\quad |x| \leq q, \end{cases}$$$$ and let us put $$$$G_q(y)=\int_{\mathbb{R}} g_q(y-x)f(x) dx.$$$$ We have $$G_q \leq 1$$, $$G_q(y) < 1$$ for some $$y \in \mathbb{R}$$ and $$\lim_{y \rightarrow - \infty} G_q(y) = \lim_{y \rightarrow \infty} G_q(y) = 1$$, so that the problem $$$$\min_{y \in \mathbb{R}} G_q(y)$$$$ admits a non-empty compact set $$T_q$$ of minimizers. Moreover it is immediate to see arguing by contradiction that $$T_q$$ shrinks to $$x_0$$ as $$q \rightarrow 0$$ (in the precise sense explained above). This result might have suggested the conjecture about $$S_q$$ of our original problem.

• It looks like if it converges to anything at all, that anything should be a minimizer of $\int\log|y-x|f(x)\,dx$, which certainly does not need to be the mode even in the unimodal case. So I would bet on the claim being false. If nobody comes to a definitive conclusion in the next few days, I'll try to make an accurate computation. The suspicious case is when $f$ is essentially the characteristic function of $[-1,1]$ with a small upward bump near $1$. – fedja Jun 26 '18 at 2:38
• By the way, is it true that the mode is the limit of the maximizer as $q \to -\infty$? (maybe under some conditions on $f$) – usul Jun 26 '18 at 3:21
• @fedja Do you think you can find a counter-example in which the distribution is strongly unimodal, that is $f$ is non-decreasing on $(-\infty,x_0]$ and non-increasing on $[x_0, \infty)$? Henry said to suspect that in this case the conjecture could be true, but I am not convinced at all. – Maurizio Barbato Jun 27 '18 at 7:57
• @MaurizioBarbato The counterexample I suggested is unimodal. Check it. – fedja Jun 28 '18 at 9:31
• @fedja To compute $F_q(y)$ for the counterexample you have in mind is an easy (even though boring) task. But to study the behavior of $S_q$ while $q$ approaches 0 seems not trivial at all: of course we could approach this last task numerically, but I am looking for a rigorously examined counterexample, not a tentative one. Anyway, generally speaking, the more I think of this problem, the more it is clear to me that there is no compelling reason for which the conjecture should hold true. – Maurizio Barbato Jun 28 '18 at 17:59

I am not so convinced of the validity of the statement for the global maximum either. I suspect it could be true for the more restrictive case where the cumulative distribution function is convex below the mode and concave above it

Consider the following continuous density function (two triangular distributions stuck together, slightly higher on the negative side, but much wider on the positive side):

$$f(x)= \begin{cases} 0 & \text{when } x \le -2 \\ 0.1x + 0.2 &\text{when } -2 \le x \le -1\\ -0.1x &\text{when } -1 \le x \le 0\\ 0.009x &\text{when } 0 \le x \le 10\\ -0.009x+0.18 &\text{when } 10 \le x \le 20 \\ 0 & \text{when } 20 \le x \end{cases}$$

The global maximum of $f(x)$ would be $f(-1)=0.1$ compared with another local maximum of $f(10)=0.09$, but shrinkage would be towards $10$

Of course, when $q=2$ the minimising $y$ is the mean $8.9$, while when $q=1$ the minimising $y$ is the median $\sqrt{\frac{800}{9}}\approx 9.428$, so it is not particularly counter-intuitive that the minimising $y$ increases as $q$ decreases

Empirically it seems that in this example $F_q(y) \gt 1$ and $F_q(10)-1 \lt \frac12 \left(F_q(-1)-1\right)$ for all $q \in (0,1]$

• Thank you very much, Henry, for having kindly replied to my answer. Your counter-example seems to work, at least numerically. Do you have some idea to prove the conjecture under the restricted definition of unimodality you quoted? I tried to prove it without success, and now I am becoming to think that it is not true even under this strong assumption. – Maurizio Barbato Jun 27 '18 at 10:09

Finally, I discovered that this problem was deeply investigated by the great french mathematician Maurice Fréchet, who introduced in his remarkable essay Les éléments aléatoires de nature quelconque dans un espace distancié the general notion of q-mean, where $q$ is any positive real number, as a minimizer of $$F_q(y)=\int_{\mathbb{R}} |y-x|^q f(x) dx \quad (y \in \mathbb{R}).$$ Let us note that this objective function can be written in several interesting ways. First of all $F_q$ is simply the convolution of $f$ with the map $x \mapsto |x|^q$. Second, we have $F_q(y)= \mathbb{E} [|X-y|^q]$, where $X$ is a random variable with density function $f$. Finally, for $q \geq 1$ we also have $F_q(y)= || \mathbb{1} - y ||^{q}_q$ where $\mathbb{1}$ is the identity function on $\mathbb{R}$ and $||.||_q$ is the norm of the space $L^q(\mathbb{R},\mathcal{B}(\mathbb{R}), \mu)$, where $\mathcal{B}(\mathbb{R})$ is the Borel $\sigma$-algebra of $\mathbb{R}$ and $\mu$ the probability measure defined by $f$: $$\mu(A)=\int_{A} f(x)dx \quad (A \in \mathcal{B}(\mathbb{R})).$$ The notion of q-mean was extensively studied by Fréchet in his following essay Positions typiques d'un élément aléatoire de nature quelconque, while he dedicated an entire work Les valeurs typiques d'ordre nul ou infini d'un nombre aléatoire to study the limit cases $q=0$ and $q =\infty$.

In this last work , he introduces three different definitions for the case $q=0$, with the third one directly related to our conjecture. With the notation already introduced this last definition is the following.

$\textbf{Definition}$. A number $x \in \mathbb{R}$ is said to be a 0-mean (or mean of order zero) of the distribution defined by $f$ if there exists a sequence of positive numbers $(q_n)$ such that $\lim_{n \rightarrow 0} q_n=0$ and $\lim_{n \rightarrow 0} x_{q_n}= x$, where $x_{q_n} \in S_{q_n}$ for every $n$.

Then Fréchet finds a counterexample (discussed at pages 16-19 of the work) of a strongly unimodal $f$ (that is an $f$ which is non-decreasing on $(-\infty,x_0]$ and non-increasing on $[x_0,\infty)$) for which the 0-mean is unique and does not coincide with the mode. In particular, this implies that the conjecture of the post is false, as I suspected. Even though the $f$ in the counterexample found by Fréchet is very simple, the proof is not trivial at all and shows all the incredible analytical ability of this mathematical genius.