Cantelli's inequality
Dependencies:
- Random variable
- Expected value of a random variable
- Variance of a random variable
- Linearity of expectation
- Markov's bound
$\newcommand{\E}{\operatorname{E}}$ $\newcommand{\Var}{\operatorname{Var}}$ Let $X$ be a real-valued random variable. Let $t \ge 0$ and $\Var(X) = \sigma^2$. Then \[ \Pr(X \ge \E(X) + t) \le \frac{\sigma^2}{\sigma^2 + t^2}, \] \[ \Pr(X \le \E(X) - t) \le \frac{\sigma^2}{\sigma^2 + t^2}. \]
This is called Cantelli's inequality, or the one-sided Chebyshev inequality.
Proof
Let $Z = X - \E(X)$. Then $\E(Z) = 0$ and $\Var(Z) = \E(Z^2) = \sigma^2$.
Let $u = \sigma^2/t$. Then \begin{align} & \Pr(X \ge \E(X) + t) \\ &= \Pr(Z \ge t) \\ &\le \Pr((Z+u)^2 \ge (t+u)^2) \\ &\le \frac{\E((Z+u)^2)}{(t+u)^2} \tag{by Markov's bound} \\ &= \frac{\sigma^2 + u^2}{(t+u)^2} \\ &= \frac{\sigma^2}{\sigma^2 + t^2}. \end{align}
This proves the first part of Cantelli's inequality. To prove the second part, let $Y = \E(X) - X$, and apply Cantelli's inequality to $Y$. Since $\E(Y) = 0$ and $\Var(Y) = \sigma^2$, we get
\[ \Pr(X \le \E(X) - t) = \Pr(Y \ge t) \le \frac{\sigma^2}{\sigma^2 + t^2}. \]
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