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.

  • 2
    $\begingroup$ (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. $\endgroup$
    – Danica
    Oct 14, 2018 at 16:50
  • $\begingroup$ 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. $\endgroup$
    – Danica
    Oct 14, 2018 at 16:51
  • $\begingroup$ What you are stating is markov's inequlality, chebyshev's inequality uses the variance. en.wikipedia.org/wiki/Chebyshev%27s_inequality. $\endgroup$
    – seanv507
    Oct 14, 2018 at 16:58
  • $\begingroup$ @seanv507, I have a precautionary statement about this point. Chebyshev's inequality follows directly from Markov's inequality, so... $\endgroup$
    – Yes
    Oct 14, 2018 at 17:00
  • $\begingroup$ .... 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) $\endgroup$
    – Glen_b
    Oct 15, 2018 at 9:10

1 Answer 1


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.


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