Markov Inequality¶

For any non-negative random variable \(X\) and positive \(a\),

\[ \begin{aligned} P(X \geq a) \leq \frac{E[X]}{a} \end{aligned} \]

This means that for a random variable with small mean, the probability of taking large values is small.

This can be proved as follows

\[ \begin{aligned} E[X] &= \int_{0}^{\infty} xp_{X}(x) dx = \int_{0}^{a} xp_{X}(x) dx + \int_{a}^{\infty} xp_{X}(x) dx\newline &\geq \int_{a}^{\infty} xp_{X}(x) dx \geq \int_{a}^{\infty} ap_{X}(x) dx = a\int_{a}^{\infty} p_{X}(x) dx\newline &\geq aP(X \geq a) \end{aligned} \]

Based on experiments with simple distributions (like uniform distribution), it can be verified that the bounds provided by this inequality can be quite loose.