Chebyshev’s inequality

For any distribution with finite variance, the probability of landing $k$ standard deviations away from the mean is bounded by $1/ k^{2}$ :

P (∣ X - μ ∣ \geq k σ) \leq \frac{1}{k ^{2}}

No assumptions beyond finite variance. No symmetry, no shape requirements.

True

P (∣ X - μ ∣ \geq k σ)

= 0.0455

Chebyshev

1/ k^{2}

= 0.2500

Bound / truth ratio: 5.5× — tight only for distributions that pack their variance far from the mean.

Distribution (all σ = 1) k (std devs) 2.0

The two-sided tail probability is bounded above by

1/ k^{2}

. Chebyshev is distribution-free — so it has to accommodate the worst case and usually overshoots.

What to notice

At k = 2, the bound says ≤ 25%. For the Normal it’s actually 4.6% — loose by a factor of five. For a heavy-tailed rescaled t(3) it’s ~12%. The gap shrinks as distributions get heavier, because Chebyshev has to cover the worst case.
Pathological distributions can make Chebyshev tight. A two-point distribution that puts probability $1/ (2 k^{2})$ on each of $μ \pm k σ$ and the rest at the mean actually achieves the bound. Chebyshev is the price of distribution-freeness.
Below k = 1 the bound is trivial — it exceeds 1, so it tells you nothing. One-sided versions (Cantelli) tighten this somewhat.

Why it matters

Chebyshev is the shortest path from “finite variance” to a weak form of the Law of Large Numbers. Apply it to the sample mean $\overset{ˉ}{X}_{n}$ , whose variance is $σ^{2} / n$ , and you get:

P (∣ \overset{ˉ}{X}_{n} - μ ∣ \geq ϵ) \leq \frac{σ ^{2}}{n ϵ ^{2}}

which goes to zero as $n \to \infty$ . That’s convergence in probability — no other ingredient needed.

Proof

Apply Markov’s inequality to $(X - μ)^{2}$ :

P (∣ X - μ ∣ \geq k σ) = P ((X - μ)^{2} \geq k^{2} σ^{2}) \leq \frac{E [( X - μ ) ^{2} ]}{k ^{2} σ ^{2}} = \frac{1}{k ^{2}}

The squaring trick converts the one-sided non-negativity requirement of Markov into a two-sided concentration bound — the template for nearly every concentration inequality that followed.