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".

onlycase you are interested? (Otherwise I would not know what to make of $n$.) Do you really meanfor any$C,R>0$? The order of quantifiers in a statement isvery important. If you expect meaningful answers, you need to ask unambiguous questions. $\endgroup$ – Liviu Nicolaescu Dec 21 '15 at 11:57