# 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

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