1.5 Markov’s Inequality
Theorem 1.5.1 (Markov’s inequality). If \(X\geq 0\) and \(a>0\), then \(P(X\geq a)\leq \frac {E(X)}{a}\).
Proof. For continuous \(X\) with density \(f\), \[E(X) = \int _0^ax f(x)\,dx + \int _a^\infty xf(x)\,dx\geq \int _a^\infty xf(x)\,dx\geq a\int _a^\infty f(x)\,dx = aP(X\geq a) .\] The discrete case is the same with sums. □
Applied to \((X - \mu )^2\) with \(a = k^2\), Markov’s inequality gives Chebyshev’s inequality \(P(|X - \mu |\geq k)\leq \frac {\sigma ^2}{k^2}\).
Example 1.5.2. The number of items a factory produces in a week has mean \(500\). What can be said about the probability that this week’s production is at least \(1000\)? If the variance is also known to be \(100\), what can be said about the probability that production is between \(400\) and \(600\)?
Solution. By Markov’s inequality, \(P(X\geq 1000)\leq \frac {500}{1000} = \frac 12\). By Chebyshev’s inequality with \(k = 100\), \(P(|X - 500|\geq 100)\leq \frac {100}{100^2} = 0.01\), so \(P(400<X<600)\geq 0.99\). The bounds hold whatever the distribution; knowing the variance makes them far sharper.
Questions on this section
Stuck on something here? Ask below and it stays attached to this topic.