> [!theorem] > > Let $X$ be a [[Random Variable|random variable]] with $X \in L^1$, then > $ > \bp\bracs{\abs{X - \ev(X)} \ge t} \le \frac{\var{X}}{t^2} > $ > More familiarly, if $\sigma = \sqrt{\var{X}}$ is the [[Standard Deviation|standard deviation]], then > $ > \bp\bracs{\abs{X - \ev(X)} \ge k\sigma} \le \frac{1}{k^2} > $ > *Proof*. By [[Markov's Inequality]]. > [!theorem] > > Let $\seqf{X_i} \subset L^1$ be a family of [[Correlation|uncorrelated]] random variables. Let $S_n = \sum_{i = 1}^nX_i$, and $s_n = \ev(S_n)$, then for all $t \ge 0$, > $ > \bp\bracs{\abs{S_n - s_n} \ge t} \le \frac{\sum_{i = 1^n}\var{X_i}}{t^2} > $ > *Proof*. Denote $x_i = \ev(X_i)$, then > $ > \begin{align*} > \var{S_n} &= \ev\braks{(S_n - s_n)^2} \\ > &= \ev\braks{\paren{\sum_{i = 1}^n(X_i - x_i)}^2} \\ > &= \sum_{i = 1}^n\ev\braks{(X_i - x_i)^2} + \sum_{1 \le i < j \le n}\cov{X_i, X_j} \\ > &= \sum_{i = 1}^n\var{X_i} > \end{align*} > $