Often all we know about a random variable is its mean, or its mean and variance. That is still enough to say something definite about how likely large values are. The two inequalities here hold for every distribution, discrete or continuous, and they are the tools behind the law of large numbers.
Key idea
Markov: if $X \ge 0$, then for every $a > 0$, $$P(X \ge a) \le \frac{E[X]}{a}.$$ Chebyshev: if $X$ has mean $\mu$ and standard deviation $\sigma$, then for every $c > 0$, $$P(\lvert X - \mu \rvert \ge c) \le \frac{\sigma^{2}}{c^{2}}, \quad \text{equivalently} \quad P(\lvert X - \mu \rvert \ge k\sigma) \le \frac{1}{k^{2}}.$$
Why it works
Markov: compare two random variables. Wherever $X \ge a$, the quantity $a \cdot 1_{X \ge a}$ equals $a \le X$; elsewhere it is $0 \le X$, because $X \ge 0$. So $a \cdot 1_{X \ge a} \le X$ always, and taking expected values gives $a\,P(X \ge a) \le E[X]$.
Chebyshev: apply Markov to the nonnegative variable $(X - \mu)^{2}$ with $a = c^{2}$: $$P(\lvert X - \mu \rvert \ge c) = P\left((X - \mu)^{2} \ge c^{2}\right) \le \frac{E[(X - \mu)^{2}]}{c^{2}} = \frac{\sigma^{2}}{c^{2}}.$$
Chebyshev: apply Markov to the nonnegative variable $(X - \mu)^{2}$ with $a = c^{2}$: $$P(\lvert X - \mu \rvert \ge c) = P\left((X - \mu)^{2} \ge c^{2}\right) \le \frac{E[(X - \mu)^{2}]}{c^{2}} = \frac{\sigma^{2}}{c^{2}}.$$
| $k$ | 1.5 | 2 | 3 | 4 |
|---|---|---|---|---|
| Chebyshev: at least this much within $k\sigma$ | $\frac{5}{9} \approx 56\%$ | $\frac{3}{4} = 75\%$ | $\frac{8}{9} \approx 89\%$ | $\frac{15}{16} \approx 94\%$ |
| A normal distribution actually has | 87% | 95% | 99.7% | 99.99% |
Worked examples
Worked example: Only the mean
Waiting times at a clinic average 10 minutes and are never negative. Bound the probability that a wait is at least 40 minutes.
- Markov with $a = 40$: $P(X \ge 40) \le \frac{10}{40} = \frac{1}{4}$.
- So at most a quarter of waits are that long, whatever the distribution.
Answer: at most $\frac{1}{4}$
Worked example: Mean and spread
Test scores have mean 70 and standard deviation 5. Give a lower bound for the fraction of scores strictly between 60 and 80.
- 60 and 80 are $c = 10$ from the mean, which is $k = 2$ standard deviations.
- $P(\lvert X - 70 \rvert \ge 10) \le \frac{25}{100} = \frac{1}{4}$, so at least $\frac{3}{4}$ of scores lie strictly between 60 and 80.
Answer: at least $\frac{3}{4}$
Worked example: How tight is the bound?
For one fair die roll ($\mu = 3.5$, $\sigma^{2} = \frac{35}{12}$), compare Chebyshev's bound on $P(\lvert X - 3.5 \rvert \ge 2.5)$ with the true value.
- Bound: $\frac{35/12}{2.5^{2}} = \frac{35/12}{25/4} = \frac{7}{15} \approx 0.47$.
- True value: only 1 and 6 are that far out, so $P = \frac{2}{6} = \frac{1}{3}$.
Answer: Bound $\frac{7}{15}$, true value $\frac{1}{3}$: correct, not tight.
Common mistake
Reading a bound as the exact probability ("Markov says the chance is $\frac{1}{4}$"), or using Markov's inequality for a variable that can be negative. If $X = \pm 10$ with equal chances, $E[X] = 0$, yet $P(X \ge 5) = \frac{1}{2}$, more than $\frac{0}{5}$. Check: Markov needs $X \ge 0$, and both inequalities give "at most" (or "at least") statements.