Consider a random variable $X$ whose variance is large. As a contrast to Markov's or Chebyshev's inequality, both of which measure the concentration of a probability distribution, is there a measure of how "spread out" a distribution is?
more specifically, I would like to have an inequality of the following sort:
$$
\Pr[|X-\mu|<r]\leq f(r)
$$
where $\mu=\mathbb{E}[X]$ and $f$ is some increasing function. This should be read: the probability of $X$ being close to its expectation is small. 


A case of particular interest to me is the following. Let $X=\sum\_{i=1}^n X\_i$ where $X\_1,...,X\_n$, are i.i.d. and non-constant. This time let $r$ be fixed, and I would like to have a bound that depends on the number of summands $n$:
$$
\Pr[|X-\mu|<r]\leq f(n)
$$
where $f$ goes to zero as $n$ goes to infinity. 


In fact, it is my intuition that for any fixed $C\in\mathbb{R}$ and $r>0$ it should be true that
$$
\Pr[|X-c|<r]\leq f(n)
$$
where $f$ goes to zero as $n$ goes to infinity, since as we have more summands, $X$ is more "smoothened out". 



A measure of how "spread out" a probability measure is
Here is one crude calculation. The claim is that if the distribution of $X$ has bounded density $f$ then $\sup\_c \mathsf{P}\{|X - c| \le 1\} = O(1/\sqrt n)$. The assumption is probably far too strong, but the $1/\sqrt{n}$ asymptotics is clearly optimal, by comparing to the Gaussian.


Let $\gamma\_{\sigma^2}$ be the Gaussian of variance $\sigma^2$, and denote by $\Vert \cdot \Vert\_p$ the $L^p$ norm for $1 \le p \le \infty$. I will use Young's convolution inequality in the form



> 
> $$\Vert f\_1 \ast \dots \ast f\_n \Vert\_{q} \le \frac{C\_{p\_1} \dots C\_{p\_n}}{C\_q} \Vert f\_1 \Vert\_{p\_1} \dots \Vert f\_n \Vert\_{p\_n},$$
>  $$1/q + n - 1 = 1/p\_1 + \dots + 1/p\_n$$
>  where $C\_p^2 = \frac{|p|^{1/p}}{|p^\prime|^{1/p^\prime}}$, $1/p + 1/p^\prime = 1$, and in the limiting case $C\_\infty = 1$
> 
> 
> 


(see e.g. Gardner "The Brunn-Minkowski inequality")


By this inequality for $p\_1 = \dots = p\_n = \frac{n}{n-1}, q = \infty$,
$$\Vert f^{\ast n} \ast \gamma\_1 \Vert\_\infty = \Vert (f \ast \gamma\_{1/n})^{\ast n} \Vert\_\infty \le \frac{C\_{\frac{n}{n-1}}^n}{C\_\infty} \Vert f \ast \gamma\_{1/n} \Vert\_{\frac{n}{n-1}}^n$$


By an explicit calculation,
$$C\_{\frac{n}{n-1}}^n = \left( \frac{\left(1 - \frac{1}{n}\right)^{-(n-1)}}{n} \right)^{1/2} \sim \sqrt{e / n}$$


On the other hand, by Holder, $\Vert f \ast \gamma\_{1/n} \Vert\_{\frac{n}{n-1}} \le \Vert f \ast \gamma\_{1/n} \Vert\_\infty^{1/n} \le \Vert f \Vert\_\infty^{1/n}$, so if $f$ is bounded then
$$\Vert f^{\ast n} \ast \gamma\_1 \Vert\_\infty = O(1 / \sqrt n) $$


In particular, on bounded intervals the mass of $f^{\ast n}$ is at most $O(1/\sqrt n)$.


