Inequality of probablity.

Question

Let $(S,F,P)$ be probability space and $X :S \rightarrow \mathbb{R}$ be random variable with mean 0 and variance 1.

Let $c \geq 0$.

I want to show that $P(X \geq c) \leq \frac{1}{1+c^2}.$

I already know Chebychev inequality $P(X\geq) \leq 1/c^2$ when $c > 0$.

If this inequality holds, $1/1+c^2$ will be better estimates.

Could you help me?

score 1 · Accepted Answer · answered Sep 12 '16 at 03:15

1

I think Cantelli's inequality is what you need.

answered Sep 12 '16 at 03:15

BGM

1 Answers1