 Markov's inequality

For a nonnegative random variable $Y$ and threshold $a>0$, Markov's inequality givesThe useful case has finite expected value; an infinite right side gives no finite information. The bound also holds if $Y$ is nonnegative almost surely. Combining it with the elementary bound one gives $\mathbb{P}(Y\ge a)\le \min \{1,\mathbb{E}Y/a\}$. Nonnegativity is essential: negative values could otherwise cancel the contribution of large positive values to the expected value.

The idea is an average budget. Every outcome where $Y\ge a$ contributes at least $a$ times its probability to $\mathbb{E}Y$, and the other outcomes contribute nonnegative amounts. Formally, the indicator function gives $Y\ge a{1}_{\{Y\ge a\}}$; taking the expected value proves the bound. ^[ref].

For example, let $Y$ take $0$, $1$, $2$, $5$ with respective probabilities $0.5$, $0.2$, $0.2$, $0.1$. Its expected value is $1.1$. At $a=2$, the tail probability is $0.3$, while the bound is $1.1/2=0.55$. At $a=5$, the tail probability is $0.1$, while the bound is $0.22$. A different probability distribution with the same expected value, putting masses $0.78$ and $0.22$ at $0$ and $5$, attains the bound $0.22$ exactly at $a=5$.

Markov's inequality is useful when a moment is easy to compute but a full probability distribution is difficult to determine. A nonnegative count with expected value less than one must have some positive probability of being zero, giving the first moment method. Applying the same bound to a squared centered random variable gives Chebyshev's inequality. Applying it to an exponential function of a random variable gives an exponential Markov bound, which can exploit more than a single moment.