Chebyshev's inequality
/ CHEB-i-shev /
How far from its average can a random variable wander? If a process is tightly controlled, big swings should be rare. Chebyshev's inequality makes this quantitative using just one extra piece of information beyond the mean: the variance, which measures typical spread. It answers 'how unlikely is a deviation of k standard deviations?' for any distribution at all.
The statement: for any random variable X with mean mu and variance sigma^2, and any k > 0, P(|X - mu| >= k sigma) <= 1 / k^2. So being at least 2 standard deviations from the mean has probability at most 1/4; at least 3, at most 1/9; at least 10, at most 1/100. The proof is just Markov's inequality wearing a disguise: apply Markov to the non-negative variable (X - mu)^2, whose mean is exactly sigma^2, and the threshold (k sigma)^2 turns E[(X-mu)^2]/(k sigma)^2 into sigma^2/(k^2 sigma^2) = 1/k^2. That is the whole trick.
Chebyshev's great virtue is universality — it needs no shape assumption, only a finite variance — which is also why it is loose for nice distributions (for a normal, 3 sigma is about 0.3 percent, far below Chebyshev's 1/9). Its starring role is proving the weak law of large numbers: the variance of an average of n independent terms shrinks like 1/n, so Chebyshev forces the average to concentrate at the mean as n grows. It is the first rigorous bridge from variance to concentration.
A machine fills bottles to mean 500 ml with standard deviation 5 ml, shape unknown. What fraction lie outside 480-520 ml (4 sigma wide, so 2 sigma each side)? Chebyshev: P(|X - 500| >= 10) <= 1/2^2 = 1/4. At most a quarter — no distribution shape can do worse.
Variance alone caps the chance of straying k standard deviations: at most 1/k^2.
Chebyshev only bites for k > 1 (for k <= 1 the bound 1/k^2 >= 1 says nothing), and it needs a finite variance — for heavy-tailed laws with infinite variance it does not apply.