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?

probability probability theory statistics

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

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

One Answer

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