Markov and Chebyshev Inequalities

Most tail calculations assume you know the whole distribution — you say "returns are normal" and read off a probability. But what if you know almost nothing? What if you have only the mean, or only the mean and the variance, and no idea of the shape? The Markov and Chebyshev inequalities answer exactly this. They give you a hard ceiling on how much probability can hide in the tail, using nothing but a moment or two. The price of assuming so little is that the bounds are loose — but they are true for every distribution, including the fat-tailed ones that break tidier models.

Markov: a mean is enough to bound a tail

If a random quantity $X$ is never negative and has mean $E[X]$, then for any level $a > 0$,

$$P(X \ge a) \le \frac{E[X]}{a} .$$

Here $P(X \ge a)$ is the chance of landing at or above $a$, and $E[X]$ is the average. The intuition is a budget argument: the average is a fixed pot of "size." If too much probability piled up at large values, the average would have to be bigger than it is. So a large tail is capped by how much mean you have to spend. If daily trade volume averages $1{,}000$ lots, then at most $\tfrac{1000}{5000} = 20\%$ of days can see $5{,}000$ lots or more — you needn't know anything else about the distribution.

Chebyshev: add a variance, bound the spread

Markov only sees one side and needs non-negativity. Apply it to the squared distance from the mean and you get something far more useful. With mean $\mu$ and standard deviation $\sigma$, for any $k > 0$,

$$P\bigl(|X - \mu| \ge k\sigma\bigr) \le \frac{1}{k^2} .$$

In words: the chance of landing more than $k$ standard deviations from the mean is at most $1/k^2$, no matter the distribution. Beyond two sigma, at most a quarter of the probability; beyond three sigma, at most a ninth; beyond ten sigma, at most one in a hundred. It's a universal speed limit on how often a variable can stray far from its center.

How loose is it? Chebyshev vs. the normal

The bound is universal precisely because it must cover the worst-behaved distribution, so for a well-behaved one it is far from tight. Compare Chebyshev's ceiling to what the normal actually delivers:

Distance from mean	Chebyshev ceiling	Actual for a normal
beyond $1\sigma$	$\le 100\%$	about $32\%$
beyond $2\sigma$	$\le 25\%$	about $4.6\%$
beyond $3\sigma$	$\le 11.1\%$	about $0.27\%$
beyond $4\sigma$	$\le 6.25\%$	about $0.006\%$

The gap is enormous — Chebyshev allows an $11\%$ three-sigma tail where a normal has a $0.3\%$ one. That is not a flaw; it's the point. The normal earns its tiny tail by assuming a specific thin-tailed shape. Chebyshev makes no such assumption, so it must leave room for the ugliest distribution consistent with that variance.

Worked example: bounding a bad day

A trading desk's daily P&L has mean $0$ and a standard deviation $\sigma$ of $10,000, but you refuse to assume the shape is normal — you've seen the fat tails. What can you say about the chance of a loss worse than $30,000?

That threshold is $k = 3$ standard deviations out. Chebyshev caps both tails combined at $\tfrac{1}{3^2} = \tfrac{1}{9} \approx 11.1\%$. Since a loss is only the lower tail, a common refinement (the one-sided Cantelli inequality) tightens the single-tail bound to $\tfrac{1}{1+k^2} = \tfrac{1}{10} = 10\%$. Either way, you can promise a rigorous ceiling — "no worse than about one such day in nine or ten" — without ever naming the distribution. A normal would claim one in seven hundred, but you don't trust that, and Chebyshev lets you stay honest.

Where it leads

Chebyshev is the engine behind the The Law of Large Numbers: as you average more independent draws, the variance of the average shrinks like $1/n$, so Chebyshev forces the probability of straying from the true mean to zero. It also sits opposite Jensen's Inequality in the toolkit of moment-based reasoning — one bounds tails from a variance, the other bounds averages through curvature.