One-sided Chebyshev Inequality (Chebyshev-Cantelli Inequality)

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.

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