Minimal conditions on random vector $X \in R^n$ to ensure that $\lim_{t\to 0^+}\sup_{\|w\|_p = 1}\sup_{u \in \mathbb R}\mathbb P(|X'w-u| \le t)=0$

Question

Let $X$ be a random variable on $\mathbb R^n$ and let $S_p^n := \{w \in \mathbb R^n \mid \|w\|_p = 1\}$ be the unit-sphere w.r.t to the $\ell_p$-norm in $\mathbb R^n$. We will be particularly interested in the case $p \in \{1,2\}$. For any $w \in B_p^n$ and $t \ge 0$, let $$ L(X'w,t):= \sup_{u \in \mathbb R}\mathbb P(|X' w - u| \le t). $$ be the Lévy concentration function of the random variable $X' w$. Finally, define $$ L(S_p^n,t) := \sup_{w \in S_p^n} L(X'w,t). $$

Question. Under what minimal assumptions on $X$ is it true that $\lim_{t \to 0^+} L(S_p^n, t) \to 0 $ ?

For example, does it suffice to assume that

• $X$ has density which is sufficiently smooth (e.g continuous) ?
• What if we simply assume that the distribution of $X$ is atomless ?

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Some solved cases

• (1) Bounded Radon transform of density of $X$. Suppose there exists a positive constant $b$ such that for every $w \in B_p^n$, the random variable $X'w$ has density bounded by $b$. Then, $$ L(S_p^n,t) = \sup_{w \in S_p^n}\sup_{u \in \mathbb R}\mathbb P(|X'w-u| \le t) \le b\cdot 2t = 2bt \to 0. $$ For example, this is the case when the distribution of $X$ is a mixture of multivariate Gaussians $\sum_k\pi_kN(\mu_k,\Sigma_k)$; in this case each $X'w$ has distribution $\sum_k \pi_k N(\mu_k'w,w'\Sigma_k w)$, and the condition clearly holds with $b=1$. Finally, note that if $f$ is the density of $X$, then the density $\rho_w$ of $X'w$ evaluated at a point $c$ is nothing but the Radon transform $R[f]$ of $f$ w.r.t the hyperplane $H_{w,c} := \{x \in \mathbb R^n \mid x'w = c\}$. Thus, the assumption that $\sup_{w \in S_p^n}\|\rho_w\|_\infty \le b$, is really reminiscent of demanding that $R[f] \in L^\infty$. So the question now is, under what minimal conditions if the Radon transform of a density function bounded ?

• (2) $X$ is an transformation of log-concave random vector. C

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onsider the scenario where $X \overset{D}{=}AZ + \mu$, where $A$ is a deterministic $m \times n$ matrix, $\mu$ is a determistic vector in $\mathbb R^m$, and $Z$ is random variable which is isotropic, log-concave, whose coordinates are $b$-subGaussian, then thanks to Grigoris Paouris' Small Ball Probability Estimates for Log-Concave Measures, we know that $$ L(X'w,t) = L((Aw)'X,t) \le (t/\|Aw\|_2)^{(c/b)^2},\text{ for sufficiently small }t $$ where $c$ is an absolute positive constant. Thus, if $A$ is non-degenerate in the sense that $\kappa_p(A) := \inf_{w \in S_p^n} \|Aw\|_2 \ge 1/C > 0$, then

$$ L(S_p^n,t) \le (t/\kappa_{p}(A))^{(c/b)^2} = (Ct)^{(c/b)^2} \to 0. $$

Answer 1

In Hengartner, W. and Theodorescu, R. (1973). Concentration Functions. Academic Press, New York. MR033144

they study the continuity properties and when the concentration function is zero even for the multivariate case in THEOREM 1.7.4..

The main idea goes through showing that the maximum is attained for each $t>0$ and so as $t\to 0$, the concentration function goes to zero.

The article "Maximal Inequalities and Some Applications" has many other references too.

Even with the extra supremum over the sphere, the same argument should work i.e. for each $t>0$ attaining attaining the supremum and using continuity of $X'$.

Some details. Let $f(x,t):=P(|X-x|\leq t)$. Since the set $S_{t}:=\{f(x,t):x\in \mathbb{R}^{n}\}$ is bounded, the supremum $L_{t}=\sup_{x}f(x,t)$ is finite i.e. there are sequences $\epsilon_{n}\to 0, y_{n}=f(x_{n},t)\in S_{t}$ such that

$$L_{t}-\epsilon_{n}\leq f(x_{n},t)\leq L_{t}.$$

So by boundedness of $y_{n}:=f(x_{n},t)\in [0,1]$, we can apply Bolzano-Weirstrass and continuity to get subsequence $x_{n_k}\to x^{*}_{t}$ and

$$L_{t}\leq f(x^{*}_{t},t)\leq L_{t}\Rightarrow L_{t}=f(x^{*}_{t},t).$$

From here we simply use continuity in $t$-variable to get limit zero.

Answer 2

Claim. If $X$ has density, then $L(S_p^n,t) \longrightarrow 0$ in the limit $t \to 0^+$.

Indeed, if $X$ has density, then so does $F(X)$, for any continuous function $F:\mathbb R^n \to \mathbb R^m$. In particular, for every $w \in \mathbb R^n$, the random variable $X' w$ has density, and hence a continuous CDF. Let $R$ be a large positive number. By compacity of $S_p^n \times [-R,R]$ and the preceding argument, the function $t \mapsto \underset{w \in S_p^n,\,|u| \le R}{\sup}\mathbb P(|X'w-u| \le t)$ is continuous, and so $$ \tag{1} \lim_{t \to 0^+}\sup_{w \in S_p^n}\sup_{|u| \le R}\mathbb P(|X'w-u| \le t) = \sup_{w \in S_p^n,\,|u| \le R}\lim_{t \to 0^+}\mathbb P(|X'w-u| \le t) = 0. $$

On the other hand, if $|u| \gt R$, then $|X'w-u| \ge ||X'w| - |u|| \ge |u| - |X'w| \gt R-|X'w|$, and so for any $t \ge 0$, one computes $$ \begin{split} \sup_{w \in S_p^n}\sup_{|u| \gt R} \mathbb P(|X'w-u| \le t) &\le \mathbb P(R - |X'w| \le t) \le \sup_{w \in S_p^n} \mathbb P(|X'w| \ge R - t)\\ & \le \mathbb P(\sup_{w \in S_p^n} |X'w| \ge R - t)\\ &= \mathbb P(\|X\|_q \ge R - t) \longrightarrow 0 \text{ in the limit }R \to \infty. \end{split} \tag{2} $$

In the last step, we have used the fact that the CDF of $\|X\|_q$ is continuous (because $X$ has density and so $\|X\|_q$ does too, by continuity of the $\ell_q$-norm on $\mathbb R^n$). Combining (1) and (2) completes the claim.