There are several famous inequalities in the theory of probability. The simplest one is the Markov inequality:

\[\Pr(|X|\ge a) \le \frac{E[|X|]}{a},\]for any random variable \(X\) and real number \(a>0\).

If we consider the random variable \((X-E[X])^2\), substitute into the Markov inequality, we have

\[\begin{align} \Pr((X-E[X])^2\ge a^2) &\le \frac{E[(X-E[X])^2]}{a^2} = \frac{Var[X]}{a^2} \\ \Pr(|X-E[X]|\ge a) &\le \frac{Var[X]}{a^2}. \end{align}\]By substituting \(a=k\sigma\) where \(Var[X]=\sigma^2\), we have the Chebyshev inequality

\[\Pr(|X-E[X]| \ge k\sigma) \le \frac{1}{k^2}.\]which it is required that \(k>0\) and \(\sigma\) is non-zero and finite.

Two other useful but more complicated probability inequalities are about deviation from mean. The Chernoff bound says that, for \(n\) i.i.d. Bernoulli random variables \(X_1,X_2,\cdots,X_n\), each having the probability \(p>\frac{1}{2}\), the probability of having more than \(n/2\) occurrences among the \(n\) of them is

\[\sum_{j=\lfloor n/2\rfloor+1}^{n} \binom{n}{j} p^j (1-p)^{n-j} \ge 1-\exp\big(-2n(p-\frac{1}{2})^2\big).\]While these \(n\) Bernoulli random variable shall produce the expectation of \(np\) occurrences, the probability of deviation from \(np\) is bounded by the Hoefding’s inequality, saying that for \(\epsilon > 0\), the probability of no less than \(n(p+\epsilon)\) occurrences is

\[P_{\ge n(p+\epsilon)} \le \exp(-2\epsilon^2 n),\]and the probability of no more than \(n(p-\epsilon)\) occurrences is

\[P_{\le n(p-\epsilon)} \le \exp(-2\epsilon^2 n),\]so the probability of having \(k\) occurrences, which \(k\in[n(p-\epsilon),n(p+\epsilon)]\), is

\[P_{\in[n(p-\epsilon),n(p+\epsilon)]} > 1-2\exp(-2\epsilon^2 n).\]Hoefding’s inequality can be generalized so that, for \(X_i\in[a_i,b_i]\) a.s. and \(X_1,X_2,\cdots,X_n\) are independent, we have the empirical mean and its expectation

\[\begin{align} \bar{X} &= \frac{1}{n}(X_1+X_2+\cdots+X_n) \\ E[\bar{X}] &= \frac{1}{n}(E[X_1]+E[X_2]+\cdots+E[X_n]), \end{align}\]then for \(t>0\) we have

\[\begin{align} \Pr[\bar{X}-E[\bar{X}]\ge t] &\le \exp(-\frac{2t^2n^2}{\sum_{i=1}^n (b_i-a_i)^2}) \\ \Pr[E[\bar{X}]-\bar{X}\ge t] &\le \exp(-\frac{2t^2n^2}{\sum_{i=1}^n (b_i-a_i)^2}). \end{align}\]