I need to lower bound the tail probability of a non-negative random variable. I have a lower bound on its expected value. I am aware of a reverse markov's inequality that does the job when the random variable is bounded above. Unfortunately that is not my case.
Is there any other inequality that may be useful to me in this regard?
thanks
NR
Best Answer
You want a lower bound of the probability of $[X\gt x]$ hence an upper bound of the probability of the event $A=[X\leqslant x]$. As explained by others there is little hope to achieve such a bound depending on $\mathrm E(X)$ only, which would be valid for every nonnegative random variable $X$.
However, for every decreasing bounded function $u$, $A=[u(X)\geqslant u(x)]$ hence Markov's inequality yields $$ \mathrm P(A)\leqslant u(x)^{-1}\mathrm E(u(X)). $$ Two frequently used cases are $$u(x)=\mathrm e^{-tx}$$ and $$u(x)=\frac1{1+tx}$$ for some positive $t$, related to Laplace and Stieltjes transforms, respectively. In both cases, one can choose the value of the parameter $t$ which yields an optimal, or nearly optimal, upper bound.
This yields $$ \mathrm P(X\gt x)\geqslant 1-u(x)^{-1}\mathrm E(u(X)). $$ A simple consequence is the fact that, for every positive $s$ (and for $s=0$ as well, provided $1/X$ is integrable), $$ \mathrm P(X\gt x)\geqslant \mathrm E\left(\frac{X-x}{s+X}\right). $$