# An inequality giving a sharper bound than that given by the Chebyshev's?

Let $$X > 0$$ be a random variable; let $$P$$ be the underlying probability measure; let $$\delta > 0$$. I wonder if there is already in probability literature a known result giving a sharper bound for $$P ( X > \delta )$$ than that given by the Chebyshev's. (By Chebyshev's inequality I simply mean the relation $$P(X > \delta) \leq \delta^{-1}EX$$; one may call it Markov's inequality, which does not affect the discussion here.)

A major motive for this question is that, besides application convenience, there is a result in probability theory (in Chung's probability text, for example) stating that $$\sum_{n \in \mathbb{N}}P(X > n) \leq EX \leq 1 + \sum_{n \in \mathbb{N}}P(X > n)$$. This is stunning, as if $$X$$ is integrable-$$P$$ then an upper bound for $$P(X > n)$$ in terms of $$E X$$ becomes "infinitely much" sharper than that given by Chebyshev's.

This triggers my interest as to if there is already a result that may be already well-known for probability theorists but somehow less known for applied probability people. If possible, I would love to be guided to the original and/or related literature.

• (Deleted my previous answer because I realized it wasn't really an answer.) One point of note is that the inequality is at least tight for some cases; see e.g. this example which refers to it as Markov's inequality. So of course there is no bound which can always be tighter. Oct 14, 2018 at 16:50
• You also might be aware of the continuous version of the stated inequality, that if $P(X \ge 0) = 1$, then $$E X = \int_0^\infty P(X > x) \mathrm{d}x .$$ Of course in either the discrete or the continuous form, Markov's inequality gives a vacuous (infinite) bound here. I don't know of a general bound for $P(X > x)$ making the sum/integral converge. Oct 14, 2018 at 16:51
• What you are stating is markov's inequlality, chebyshev's inequality uses the variance. en.wikipedia.org/wiki/Chebyshev%27s_inequality. Oct 14, 2018 at 16:58
• @seanv507, I have a precautionary statement about this point. Chebyshev's inequality follows directly from Markov's inequality, so...
– Yes
Oct 14, 2018 at 17:00
• .... so just call it by what it is commonly called, which is Markov's, not Chebyshev's (unless you have a particular reason not to, which you'd need to explain) Oct 15, 2018 at 9:10

Cantelli's inequality gives a better bound in many cases. Simply stated, for $$k>0,$$
$$P \left[X \geq \mu + k \sigma \right] \leq \frac{1}{k^2+1}$$
For a thorough treatment, see B.K. Ghosh's "Probability Inequalities Related to Markov's Theorem," $$\it{The \ American \ Statistician},$$ August 2002, Vol. 56, No. 3, pp. 186-190.