> [!NOTE] Theorem (Markov's Inequality for Non-negative Discrete Real-valued Random Variables) > Let $(\Omega,\mathcal{F},\mathbb{P})$ be a [[Probability Space|probability space]]. Let $X$ be a [[Square-Integrable Discrete Real-Valued Random Variable|square-integrable discrete real-valued random variable]] on $(\Omega,\mathcal{F},\mathbb{P}).$ For all $a>0,$ $\mathbb{P}((X-\mathbb{E}[X])^{2} \geq a^{2}) \leq \frac{\text{Var}(X)}{a^{2}},$where $\mathbb{E}$ denotes [[Expectation of Discrete Real-Valued Random Variable|expectation]], $\text{Var}$ denotes [[Variance of Square-Integrable Discrete Real-Valued Random Variable|variance]] and $|x|$ denotes the [[Absolute value function|absolute value]] of $x.$ **Note**: $(X-\mathbb{E}[X])^{2} \geq a^{2}$ iff $|X-\mathbb{E}[X]|\geq a$ iff $X \geq a+\mathbb{E}[X]$ or $X \leq -a+\mathbb{E}[X].$ **Proof**: Since $(X-\mathbb{E}[X])^{2}$ is a non-negative random variable, we can apply [[Markov's Inequality for Non-negative Integrable Discrete Real-valued Random Variables]] with $x=a^{2}$ to obtain $\mathbb{P}\{ (X- \mathbb{E}[X])^{2}\geq a^{2} \} \leq \frac{\mathbb{E}[(X-\mathbb{E}[X])^{2}]}{a^{2}}$Since $(X-\mathbb{E}[X])^{2} \geq a^{2}\iff |X-\mathbb{E}[X]| \geq a$ this yields $\mathbb{P}\{ |X-\mathbb{E}[X]| \geq a \} \leq \frac{\mathbb{E}[(X-\mathbb{E}[X])^{2}]}{a^{2}} = \frac{\text{Var}(X)}{a^{2}}.$ # Applications