TITLE: A measure of how "spread out" a probability measure is
QUESTION [5 upvotes]: 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".

REPLY [11 votes]: What you need is a standard inequality for the concentration function for sums of independent random variables; see e.g. [Petrov],
Ch. III, Section 2, Theorem 3 (due to Esseen), which implies the following. 
For a random variable (r.v.) $X$ and real $c>0$, let 
$$Q(X;c):=\sup_{x\in\mathbb R} P(x\le X\le x+c). 
$$ 
Let $S_n:=X_1+\dots+X_n$, where $X,X_1,\dots,X_n$ are any independent identically distributed r.v.'s. Then for any real $c>0$ 
$$Q(S_n;c)\le \frac A{\sqrt{n D(\tilde X;c)}}, 
$$ 
where $A$ is a universal constant, $\tilde X:=X-X_1$,
$$D(\tilde X;c):=\frac1{c^2}\,E\,\tilde X^2\,I\{|\tilde X|<c\}+P(|\tilde X|\ge c)=E\Big(1\wedge\frac{\tilde X^2}{c^2}\Big), 
$$
and $I\{\cdot\}$ is the indicator function. Note that, for any real $c>0$, one has $D(\tilde X;c)=0$ iff $P(\tilde X=0)=1$ iff the r.v. $X$ is degenerate (i.e., $P(X=a)=1$ for some real $a$). Hence, if $X$ is non-degenerate, then $D(\tilde X;c)\in(0,\infty)$.  
A simpler but a bit less precise bound is due to [Rogozin]: 
$$Q(S_n;c)\le \frac A{\sqrt{n(1-Q(X;c))}}.  
$$
Addendum: If you only care about the concentration of $S_n$ around its expectation, you may want to use a Berry--Esseen type of bound (see e.g. Theorem 7 in [3]
), which implies 
$$P\Big(\Big|\frac{S_n-n\mu}{\sigma}\Big|\le c\Big)\le P\Big(|Z|\le\frac c{\sqrt n}\Big)+A\frac{\beta}{\sigma^3\sqrt n}
\le\frac C{\sqrt n}, 
$$
where $\mu:=EX_1$, $\sigma:=\sqrt{E(X_1-\mu)^2}$, $\beta:=E|X_1-\mu|^3$, $c\in[0,\infty)$, $Z$ is a standard normal r.v.,
$C:=\frac{2c}{\sqrt{2\pi}}+A\frac{\beta}{\sigma^3}$, and $A\in(0,96/100)$ is a universal constant. 
Similar but more direct and a bit more general bounds on the concentration are given e.g. in Proposition 6.1 in [4]
and Proposition 2.1 in [5].  
All these bounds are of the optimal order $O(1/\sqrt n)$ in $n$. It cannot be improved in general even for the concentration of $S_n$ around its expectation. E.g., let $P(X_i=\pm1)=1/2$. Then, by Stirling's formula, $P(S_{2n}=0)>A/\sqrt n$ for some universal constant $A>0$.