= Markov's inequality
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{title2=$\mathbb P(Y\geq a)\leq\mathbb E Y/a$}
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= Markov inequality
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For a nonnegative <random variable> $Y$ and threshold $a>0$, <Markov's inequality> gives
$$
\mathbb P(Y\geq a)\leq\frac{\mathbb E Y}{a}.
$$
The 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 $\mathbb P(Y\geq a)\leq\min\{1,\mathbb E Y/a\}$. 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 $Y\geq a$ contributes at least $a$ times its <probability> to $\mathbb E Y$, and the other outcomes contribute nonnegative amounts. Formally, the <indicator function> gives $Y\geq a\mathbf1_{\{Y\geq a\}}$; taking the <expected value> proves the bound. https://ocw.mit.edu/courses/18-175-theory-of-probability-spring-2014/887e96172e82a615dac16fb26702d88b_MIT18_175S14_Lecture8.pdf[Sheffield, MIT 18.175, lecture 8: the expectation argument]{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 $a=5$.
\Image[/markov-average-and-tail.png]
{title=An average bounds how much probability can lie beyond a positive threshold}
{description=Left: the stipulated four-point <probability distribution> has <expected value> $1.1$. Right: its exact <tail probability> $\mathbb P(Y\geq a)$ and the capped <Markov's inequality> bound. Filled markers include an atom at the threshold; open markers show the smaller value immediately after it. The highlighted threshold $a=2$ has actual <probability> $0.3$ and bound $0.55$.}
<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>.
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