Helper: Sub-Gaussian MGF properties from boundedness #
For bounded mean-zero random variables, derive the Hoeffding MGF bound and
exponential integrability using Mathlib's hasSubgaussianMGF_of_mem_Icc_of_integral_eq_zero.
Hoeffding's Inequality — Sum Form #
The sum form states: for independent mean-zero bounded random variables
Xᵢ ∈ [aᵢ, bᵢ], we have P(∑ Xᵢ > t) ≤ exp(-2t²/∑(bᵢ-aᵢ)²).
This is the core result from which the sample-mean form follows.
Hoeffding's inequality (upper tail, sum form).
For independent mean-zero Xᵢ ∈ [aᵢ, bᵢ] a.s., and t > 0:
P(∑ᵢ Xᵢ > t) ≤ exp(-2t² / ∑ᵢ(bᵢ - aᵢ)²).
Hoeffding's inequality (lower tail, sum form).
For independent mean-zero Xᵢ ∈ [aᵢ, bᵢ] a.s., and t > 0:
P(∑ᵢ Xᵢ < -t) ≤ exp(-2t² / ∑ᵢ(bᵢ - aᵢ)²).
Hoeffding's Inequality — Sample Mean Form (Theorem 1.9) #
The sample-mean form: for independent Xᵢ ∈ [aᵢ, bᵢ] (not necessarily centered),
P(X̄ - E(X̄) > t) ≤ exp(-2n²t²/∑(bᵢ-aᵢ)²).
This follows from the sum form by noting that X̄ - E(X̄) > t iff ∑(Xᵢ - E[Xᵢ]) > nt,
and the centered variables Yᵢ = Xᵢ - E[Xᵢ] are mean-zero, bounded, and independent.
Theorem 1.9 — Hoeffding's inequality (combined).
For independent random variables Xᵢ ∈ [aᵢ, bᵢ] a.s. with aᵢ < bᵢ,
the centered sum satisfies both tail bounds:
P(∑ᵢ Xᵢ > t) ≤ exp(-2t² / ∑ᵢ(bᵢ - aᵢ)²)
P(∑ᵢ Xᵢ < -t) ≤ exp(-2t² / ∑ᵢ(bᵢ - aᵢ)²)
where the Xᵢ are assumed to be centered (mean zero).
For general (uncentered) variables, apply this to Yᵢ = Xᵢ - E[Xᵢ].
Hoeffding's Inequality — Sample Mean Form (Theorem 1.9 proper) #
The textbook states Hoeffding's inequality for the sample mean:
for independent Xᵢ ∈ [aᵢ, bᵢ] (not necessarily centered),
P(X̄ - E(X̄) > t) ≤ exp(-2n²t²/∑(bᵢ-aᵢ)²)
where X̄ = (1/n)∑ᵢ Xᵢ.
The proof centers the variables: Yᵢ = Xᵢ - E[Xᵢ] are mean-zero, bounded in
[aᵢ - E[Xᵢ], bᵢ - E[Xᵢ]] (so bᵢ - aᵢ is preserved), and independent.
Since X̄ - E(X̄) > t ↔ ∑Yᵢ > nt, we apply the sum form with threshold nt.
Hoeffding's inequality (upper tail, sample mean form).
For independent Xᵢ ∈ [aᵢ, bᵢ] a.s. with n > 0:
P(X̄ - E[X̄] > t) ≤ exp(-2n²t² / ∑ᵢ(bᵢ - aᵢ)²),
where X̄ = (1/n)∑ Xᵢ, i.e., X̄ - E[X̄] = (∑ Xᵢ)/n - (∑ E[Xᵢ])/n.
Hoeffding's inequality (lower tail, sample mean form).
For independent Xᵢ ∈ [aᵢ, bᵢ] a.s. with n > 0:
P(X̄ - E[X̄] < -t) ≤ exp(-2n²t² / ∑ᵢ(bᵢ - aᵢ)²).
Theorem 1.9 — Hoeffding's inequality (sample mean form).
Let X₁,...,Xₙ be n independent random variables with Xᵢ ∈ [aᵢ, bᵢ] a.s.
Let X̄ = (1/n) ∑ᵢ Xᵢ. Then for any t > 0:
P(X̄ - E[X̄] > t) ≤ exp(-2n²t² / ∑ᵢ(bᵢ - aᵢ)²)
P(X̄ - E[X̄] < -t) ≤ exp(-2n²t² / ∑ᵢ(bᵢ - aᵢ)²)
Here X̄ - E[X̄] is expressed as (∑ Xᵢ)/n - (∑ E[Xᵢ])/n.