Markov’s Inequality
Markov’s inequality provides a bound on the upper-tail probability of non-negative random variables. The bound is expressed with the expectation of the random variable.
[Markov’s Inequality]
Let X be a non-negative random variable. Then, for any \lambda > 0,
Pr[ X \geq \lambda ] \leq \frac{E[X]}{\lambda}.
Equivalently, for any k > 1,
Pr[ X \geq k E[X] ] \leq \frac{1 }{k}.
Proof.
Define a random variable Y such that
Y = \begin{cases}
\lambda & X\geq \lambda, \\
0 & X< \lambda. \\
\end{cases}
Then [latex]E[X] \geq E[Y] \geq \lambda\cdot Pr[X\geq \lambda][/latex].
Markov's inequality only bounds the upper-tail probability. The following is a similar bound on the lower-tail probability if the random variable is bounded above.
[Lemma]
Let X be a random variable over [0,1]. Then, for any positive \(k < 1[/latex],
\[
Pr[ X < k E[X] ] \leq \frac{1 - E[X]}{1 - k E[X]}.
\]
The following are some applications of Markov's inequality.
[Borel-Cantelli Lemma]
Let [latex]E_1, E_2, \ldots\) be a sequence of events such that \(\sum_n Pr[E_n] <\infty[/latex]. Then the probability that no more than [latex]k[/latex] of the events happen is at least
\[
1 - \frac{ \sum_{n} Pr[E_n] }{ M }.
\]
This implies that, with probability 1, at most finitely many events happen.
Proof.
For each event [latex]E_i\), define binary random variable X_i such that X_i = 1 if E_i hold, and X_i = 0 otherwise. Then E[X_i] = Pr[E_i], and, by linearity of expectation,
E[\sum_i X_i ] = \sum_i E[ X_i ] = \sum_i Pr[ E_i ].
The result follows by applying Markov’s inequality on \sum_i X_i.
[Chebyshev’s Inequality]
Let X be a random variable. Then, for any \lambda > 0,
Pr[ |X-\mathbb{E}(X)| \geq \lambda ] \leq \frac{ Var[X] }{\lambda^2}.
Equivalently, for any k > 1,
Pr[ |X-\mathbb{E}(X)| \geq k Var[X] ] \leq \frac{1}{k^2}.
Chebyshev’s inequality follows by applying Markov’s inequality on the random variable (X – E[X])^2, and the fact that Var[X] = E [ (X – E[X])^2 ].