TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13

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For $n \geq 1$ independent random signs, $P (∣ S ∣ \geq t) \leq 2 \exp (- t^{2} / (2 n))$ for $t > 0$

Statement

Let $ε_{0}, \dots, ε_{n - 1}$ be mutually independent uniform random signs, where $n \geq 1$ , and put $S = \sum_{i < n} ε_{i}$ . For every $t > 0$ , $P (∣ S ∣ \geq t) \leq 2 \exp ⁣ (- \frac{t^{2}}{2 n}) .$

Facts & Assumptions

Given: Independent random signs, $n \geq 1$ , their sum $S$ , and $t > 0$ .

[L1]

Markov gives $P (Y \geq a) \leq E [Y] / a$ for nonnegative $Y$ and $a > 0$ (Markov's inequality on a finite probability space).

[L2]

The MGF of an independent finite sum is the product of the individual MGFs (The moment generating function of a finite sum of independent variables is the product of their moment generating functions).

[L3]

A uniform sign satisfies $E [\exp (u ε)] \leq \exp (u^{2} / 2)$ (For a uniform random sign $ε$ , $E [e^{t ε}] \leq e^{t^{2} / 2}$ ).

[L4]

The exponential is strictly increasing (The exponential function is strictly increasing).

[L5]

Probability of a finite union is at most the sum of the probabilities (The finite union bound).

[L6]

Mutual independence is the factorization of all joint attained-value probabilities (Pairwise and mutual independence of finite-valued random variables).

Proof

technique · direct

1.1

For $u > 0$ , strict monotonicity gives ${S \geq t} = {\exp (u S) \geq \exp (u t)}$ , so [L1] gives $P (S \geq t) \leq \exp (- u t) E [\exp (u S)]$ .

L1L4

1.2

By [L2] and [L3], $E [\exp (u S)] \leq \exp (n u^{2} / 2)$ .

L2L3algebra

2.1

Choose $u = t / n > 0$ . Substitution in steps 1.1 and 1.2 gives $P (S \geq t) \leq \exp (- t^{2} / (2 n))$ .

step 1.1step 1.2choosealgebra

3.1

Negation merely relabels the two attained values of each sign, so the joint-value factorization in [L6] shows that the variables $- ε_{i}$ are again mutually independent uniform signs. Step 2.1 applied to $- S$ gives the same bound for $P (S \leq - t)$ .

step 2.1L6algebra

4.1

Since ${∣ S ∣ \geq t} = {S \geq t} \cup {S \leq - t}$ , [L5] and steps 2.1 and 3.1 give the result. The excluded boundary $t = 0$ would only give the valid but uninformative bound $1 \leq 2$ .

step 2.1step 3.1L5∎

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Sources

J. Matousek and J. Vondrak, The Probabilistic Method, Section 7.1 (standard reference, not scraped)
Y. Zhao, MIT 18.218 Probabilistic Method in Combinatorics, Section 4.1 (standard reference, not scraped)

For n≥1n≥1 independent random signs, P(∣S∣≥t)≤2exp⁡(−t2/(2n))P(∣S∣≥t)≤2exp(−t2/(2n)) for t>0t>0