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Let $f:\mathbb{R} \rightarrow [0,\infty)$ be a continuous probability density function on $\mathbb{R}$ such that \begin{equation} \int_{\mathbb{R}} |x| f(x)\, dx < \infty, \end{equation} 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 \begin{equation} \min_{y \in \mathbb{R}} \int_{\mathbb{R}}|y-x|^{q} f(x) \,dx. \end{equation}

(I) Can we find simple conditions on $f$ such that for some $\epsilon > 0$ and each given $q \in (0,\epsilon)$ there exists a unique solution to this problem?

Assume that for all $q$ small enough the problem has a unique solution, and put \begin{equation} x_q= \arg \min_{y \in \mathbb{R}} \int_{\mathbb{R}}|y-x|^{q} f(x) \,dx. \end{equation}

(II) Can we conclude that $x_q \rightarrow x_0$ for $q \rightarrow 0$?

I found this last property stated in a monograph about applied statistics without any proof.

Thank you very much in advance for your kind attention.

NOTE (1). For $q=1$ the solutions of our minimization problem are all the medians of the distributions defined by $f$: see Why does the median minimize $E[|X-c|]$. So a simple condition assuring that for $q=1$ our problem has a unique solution is that $f > 0$.

NOTE (2). Analogous questions can be asked about the midrange. Make the additional assumption that $f$ has compact support $S$ and put $a= \min S$ and $b = \max S$. If we define the probability measure \begin{equation} \mu(A)=\int_{A} f(x)dx \end{equation} for every set $A$ in the Borel $\sigma$-algebra $\mathcal{B}(\mathbb{R})$, and we consider for every measurable function $g:\mathbb{R} \rightarrow \mathbb{R}$ the norm $||g||_{\infty}= \operatorname{ess} \sup |g|$ with respect to measure space $(\mathbb{R}, \mathcal{B}(\mathbb{R}), \mu)$, then the problem \begin{equation} \min_{y \in \mathbb{R}} || \mathbb{1} -y||_{\infty}, \end{equation} where $\mathbb{1}(x)=x$ for all $x \in \mathbb{R}$, has as unique solution the midrange $x_{\infty}=(a+b)/2$.

Now we can ask:

(I') Are there simple conditions on $f$ such that for every $q$ greater then some fixed $M > 0$, there exists a unique solution to the problem \begin{equation} \min_{y \in \mathbb{R}} \int_{\mathbb{R}}|y-x|^{q} f(x) \,dx? \end{equation}

(II') Assume that these conditions are verified, and put \begin{equation} x_q= \arg \min_{y \in \mathbb{R}} \int_{\mathbb{R}}|y-x|^{q} f(x) \,dx. \end{equation}

Do we have $x_q \rightarrow x_{\infty}$ as $q \rightarrow \infty$?

Maurizio Barbato
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  • Similar to the median problem, you will need to solve $F(y)=\int_{-\infty}^y |x-y|^{q-1} f(x) dx - \int_y^\infty |x-y|^{q-1} f(x) dx=0$. Now suppose you have two solutions $y,z$ and compute $F(z)-F(y)$. – Ian Jun 22 '18 at 13:46
  • I think in fact even $f>0$ is not a sufficient condition when $0<q<1$. But the situation depends on what the space ${ g : \int g(x) f(x) dx = 0 }$ looks like, since solutions to the problem are precisely special members of this space (namely members of the form $g_y(x)=|x-y|^{q-1} \operatorname{sign}(x-y)$). – Ian Jun 22 '18 at 21:31
  • I see now that (I') has a trivial answer. Since for every fixed $x$ and every $q > 1$, $g(y)=|x-y|^q$ is strictly convex, $G(y)=\int_{\mathbb{R}} |x-y|^q f(x) dx$ turns out to be strictly convex. Moreover since $f$ has compact support, we have $G(y) \rightarrow \infty$ as $y \rightarrow \infty$ and as $y \rightarrow - \infty$. So $G$ has a unique global minimum. Anyway, I don't know for now if (II') has a positive answer. – Maurizio Barbato Jun 23 '18 at 07:56
  • Why is G convex for $q<1$? – Ian Jun 23 '18 at 08:51
  • @Ian Indeed I was referring to question (I') and not to question (I). – Maurizio Barbato Jun 23 '18 at 09:12
  • I see now that (II') has a positive answer. Assume not, then there exists $\delta \in (0, (b-a)/2)$ and a sequence $(q_n)$ such that $q_n \rightarrow \infty$ and $y_{q_n} > x_{\infty} + \delta$ for all $n$ or $y_{q_n} < x_{\infty} - \delta$ for all $n$. Consider e.g. the first case. Let $p_1 = \int_{x_{\infty} + \delta}^b f(x) dx$, $\bar{x} \in (0, 2 \delta)$, $p_2=\int_{a}^{a + \bar{x}} f(x) dx$, $m=(b-a)/2 - \delta$ and $M=(a+b)/2 + \delta - a - \bar{x}$. – Maurizio Barbato Jun 23 '18 at 09:18
  • Then we have $M > m$ and $\int_{y}^{\infty} (x-y)^{q-1}f(x) dx \leq m^{q-1} p_1$ and $\int_{-\infty}^y (y-x)^{q-1}f(x) dx \geq M^{q-1} p_2$, so that we cannot have $G'(y_{q_{n}})=0$ for all n. QED – Maurizio Barbato Jun 23 '18 at 09:19

1 Answers1

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As for question (I), there is essentially no simple condition on $f$ assuring the uniqueness of the solution of the problem \begin{equation} \min_{y \in \mathbb{R}} \int_{\mathbb{R}}|y-x|^{q} f(x) \,dx. \end{equation} Th fact is that the function $x \mapsto |x|^q$ is not convex for $q \in (0,1)$ nor $f$ can be a convex function, since it is a probability density function. So we cannot say very much about the convolution integral \begin{equation} F_q(y) = \int_{\mathbb{R}}|y-x|^{q} f(x) \,dx \end{equation} The only very special case I could envisage in which the uniqueness is garanteed is the following one. Suppose that $f \in C^1(\mathbb{R})$, that it has compact support and that $f$ is symmetric around $x_0$. Without loss of generality we can take $x_0=0$ to simplify the notation. Then we have \begin{equation} F'_q(y) = \int_{\mathbb{R}}|x|^{q} f'(y-x) \,dx = \int_{\mathbb{R}}|y-x|^{q} f'(x) \,dx = \int_{0}^{\infty} -f'(x) [|x+y|^q - |x-y|^q ] dx. \end{equation}

Now for any $x > 0$ we have $|x+y|^q - |x-y|^q > 0$ for $y >0$ and $|x+y|^q - |x-y|^q < 0$ for $y < 0$. So $F'_q(y) < 0$ for $y < 0$ and $F'_q(y) > 0$ for $y > 0$, and $x_0=0$ is the unique global minimum of $F_q$.

Anyway, essentially question (II) makes sense even if the uniqueness of the solution of our minimization problem is not assured. Indeed, if we call $S_q$ the set of all minimizers for the problem \begin{equation} \min_{y \in \mathbb{R}} \int_{\mathbb{R}}|y-x|^{q} f(x) \,dx, \end{equation} then we can ask whether $S_q$ "shrinks" to $x_0$ when $q \rightarrow 0$, in the sense that for every $\epsilon > 0$ there exists $\delta > 0$ such that $S_q \subset (x_0 - \epsilon, x_0 + \epsilon)$ for every $q \in (0,\delta)$.

This last question has been answered in the negative by Maurice Fréchet in his work Les valeurs typiques d'ordre nul ou infini d'un nombre aléatoire: see the counterexample at pages 16-19. For further discussion and references on this subject see my post q-Means and the Mode of a Distribution.

Maurizio Barbato
  • 3,638