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Using the well-known Chebyshev's inequality we can upper-bound the tail of the distribution of a random variable $X$ using its variance $\sigma^2_X$ moment as follows:

$$P(|X-\mu_X|\geq k\sigma_X)\leq 1/k^2$$

Obviously, only case when $k>1$ is useful.

I am wondering if there exists a moment-based lower bound which I can use to lower-bound

$$P(X\geq\mu_X+k\sigma_X)$$

where $0<k<1$. Unfortunately nothing except for the moments is known about $X$.

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A negative result, sorry about that...

Assume that $X$ is centered Bernoulli, that is, that $P[X=x]=1-x$ and $P[X=x-1]=x$ for some $x$ in $(\frac12,1)$, and let $A_k=[X\geqslant\mu_X+k\sigma_X]$. Then $\mu_X=0$ and $\sigma_X^2=x(1-x)$. Furthermore, $x\gt\frac12$ hence $x\gt\sqrt{x(1-x)}$ and one sees that, for every $k$ in $(0,1)$, $x\gt\mu_X+k\sigma_X$. Thus $A_k=[X=x]$, which implies $P[A_k]=1-x$. This can be as close to $0$ as one wants hence no general lower bound of $P[A_k]$ can hold.

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  • $\begingroup$ A negative result is still a result. Thank you for the enlightening explanation! $\endgroup$
    – M.B.M.
    Dec 7, 2013 at 9:00

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