Show $mathbb{P}[X-m>alpha]leq frac{sigma^2}{sigma^2+alpha^2}$

Mathematics Asked on January 1, 2022

I found this problem in an old statistics book:

Suppose $X$ is a square integrable random variable with mean $m$ and variance $sigma^2$. For any $alpha>0$, show

mathbb{P}[X-m>alpha]leqfrac{sigma^2}{sigma^2 +alpha^2}

At first I thought that the inequality results from direct application of Markov-Chebyshev’s inequality, but when I actually tried it I realized it was not so. Does anybody know about this inequality and how to obtain it?

One Answer

That is is known as Cantelli's inequality. It can be obtained from Chebyshev's but with a twist.

For any $x>0$, $alpha+x>0$ and so, $$ begin{align} mathbf{P}[X-m>alpha]&=mathbf{P}[X-m+x>x+alpha]\ &leq frac{mathbf{E}[(X-m+x)^2]}{(alpha+x)^2}=frac{sigma^2+x^2}{(alpha +x)^2}=:g(x) end{align} $$

The game now is to find the best $x$. You can use differential Calculus to check that $x=frac{sigma^2}{alpha}$ does the trick.

Answered by Oliver Diaz on January 1, 2022

Add your own answers!

Related Questions

Radical of ideal: Radical of $4mathbb Z = 2mathbb Z$?

3  Asked on August 31, 2020 by dabofskateboarding


bounded Cantor pairing?

0  Asked on August 21, 2020 by some-guy


autocorrelation for truncated normal distribution

0  Asked on August 17, 2020 by bearcub


The width of a hypergraph

1  Asked on August 16, 2020 by erel-segal-halevi


False Position Method

1  Asked on August 15, 2020 by matheus-barreto-alves


Ask a Question

Get help from others!

© 2023 All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP