> [!theorem] > > Let $X$ be a non-negative [[Random Variable|random variable]]. Then for any $a > 0$, > $ > \bp\bracs{X \ge a} \le \frac{\ev(X)}{a} > $ > For any strictly increasing function $\phi: \real \to [0, \infty)$, > $ > \bp\bracs{X \ge t} = \bp\bracs{\phi(X) \ge \phi(t)} \le \frac{\ev(\phi(X))}{\phi(t)} > $ > > > *Proof*. By writing the [[Probability|probability]] as an [[Expectation|expectation]], > $ > \bp(X \ge a) = \ev{(\chi_{[a, \infty)}(X))} \ge \ev(a \cdot \chi_{X \ge a}) = a\bp\bracs{X \ge a} > $