LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-12

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Chebyshev bounds the majority error of repeated Bernoulli trials

Statement

Let $k \geq 1$ , let $X_{1}, \dots, X_{k}$ be independent Bernoulli random variables with $P (X_{i} = 1) \geq 2 / 3$ for every $i$ , and put $S : = X_{1} + \dots + X_{k}$ . Then

$P ⁣ (S \leq \frac{k}{2}) \leq \frac{9}{k} .$

Thus the probability that a majority vote is wrong decays like $O (1 / k)$ .

Facts & Assumptions

Given: an integer $k \geq 1$ and independent Bernoulli variables $X_{1}, \dots, X_{k}$ with $P (X_{i} = 1) \geq 2 / 3$ .

[L1]

Expectation is linear over finite sums (Expectation is linear for every finite family of random variables, without any independence hypothesis).

[L2]

Bernoulli mean and variance satisfy $E [X_{i}] = p_{i}$ and $Var (X_{i}) = p_{i} (1 - p_{i}) \leq 1 / 4$ (A Bernoulli $(p)$ variable has mean $p$ and variance $p (1 - p)$ ; a binomial $(n, p)$ variable has mean $n p$ and variance $n p (1 - p)$ ).

[L3]

Variance adds for pairwise-independent finite families (Variance adds for every finite pairwise-independent family).

[L4]

Chebyshev's inequality bounds $P (∣ Y - E [Y] ∣ \geq t)$ by $Var (Y) / t^{2}$ (Chebyshev's inequality on a finite probability space).

Proof

technique · direct

1.1L1L2L3given

Let $μ : = E [S]$ . By [L1] and [L2], $μ = \sum_{i = 1}^{k} E [X_{i}] \geq 2 k / 3$ . Also [L2] and [L3] give $Var (S) = \sum_{i = 1}^{k} Var (X_{i}) \leq k / 4$ .

2.1L4step 1.1algebra

If $S \leq k / 2$ , then since $μ \geq 2 k / 3$ we have $μ - S \geq k / 6$ , so ${S \leq k / 2} \subseteq {∣ S - μ ∣ \geq k / 6}$ . Applying [L4] with $Y = S$ and $t = k / 6$ yields $P (S \leq k / 2) \leq Var (S) / (k / 6)^{2} \leq (k / 4) / (k^{2} / 36) = 9 / k$ .

3.1step 2.1∎

The event $S \leq k / 2$ is exactly the event that at most half the trials are correct, so the same bound controls majority-vote error.

Depends on

Used by

Dependency tree · two levels

18 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Eric Blais, Models of Computation, 14. Randomized Computation (standard reference, not scraped)
Sanjeev Arora and Boaz Barak, Computational Complexity: A Modern Approach (standard reference, not scraped)