# probability bound

Let $x_1,\ldots,x_n$ be i.i.d. random variables with continuous and concave distribution function F.

It is known that for $t\geq 0$ $$P\left(\sum_{i=1}^nx_i\leq t\right)\leq\\ \left\{ \begin{array}{rcl} \frac{1}{n!}\sum_{j=0}^k(-1)^j {n \choose j}\left(nF\left(\frac tn\right)-j\right)^n, &nF^{-1}\left(\frac kn\right)\leq t<F^{-1}\left(\frac{k+1}{n}\right), k=0,\ldots,n-1\\ 1,&t\geq nF^{-1}(1), \end{array}\right.$$ here $F^{-1}(t)=\inf\{x:F(x)\geq t\}, 0<t\leq 1$.

I am wondering if it is possible to bound the RHS of the inequality above (the sum). Is it possible by adding some assumptions on random variables $x_1,...,x_n$ to bound density function F?

Continuous distribution function $F$ with support $[0, \infty)$ is called concave, if $F(\lambda s+(1-\lambda t))\geq \lambda F(s)+(1-\lambda)F(t)$, for every $s,t\geq 0, 0\leq\lambda\leq 1$.

Any references and ideas would be very helpful.

Thank you.

-
Could you give a definition of a concave distribution function? Neither PDF nor CDF could be concave everywhere since there are no bounded concave functions. – Ilya Feb 2 '12 at 7:09
@Ilya There are bounded concave functions, but they are constant, and so not useful here. – Ben Derrett Feb 5 '12 at 14:13
@David Perhaps you only want concavity of the CDF on $[0\infty)$? – Ben Derrett Feb 5 '12 at 14:14
@Ben: thanks) I am working with non-constant bounded harmonic functions, so implicitly eliminated the trivial case – Ilya Feb 5 '12 at 21:22
@Ben Derrett: yes, thats what I wrot in the definition. I consider distribution function F with support $[0\infty)$. – David Feb 5 '12 at 21:28