Mathepsilon's Blog

Let $t>0$ and $X$ be a real-numbered random variable. Then

$\mathbb{P}(X-\mathbb{E}(X)\geq t) \leq \frac{Var(X)}{Var(X)+t^2}$

Proof: Without loss of generality, we assume $\mathbb{E}(X)=0$ . Then we know for all $t$ ,

$t=\mathbb{E}[t-X]\leq \mathbb{E}[(t-X) \mathbb{I}_{X<t}]$ , where $\mathbb{I}$ is the indicator function.

From the Cauchy-Schwarz inequality,

$t^2 \leq \mathbb{E}[(t-X)^2] \mathbb{E}[{I}_{X<t}^{2}]$ ,

$=\mathbb{E}(t-X)^2\mathbb{P}\{X<t\}$

$=(Var(X)+t^2)\mathbb{P}\{X<t\}.$

Indeed the theorem claim follows.

One-sided Chebyshev Inequality (Chebyshev-Cantelli Inequality)