Kallenberg's L² Bound (Lemma 1.2) #
The L² distance bound for weighted averages of an exchangeable sequence — the
"Lemma 1.2" of Kallenberg (2005). Self-contained algebraic statement; this file
is not used by the rest of the ViaL2 proof (which proceeds through
l2_contractability_bound to avoid the Exchangeable hypothesis), but it is the
clean form of the original Kallenberg argument and is retained here for
documentation.
Main results #
kallenberg_L2_bound: the L² distance bound, with full Exchangeable hypothesis
References #
- Kallenberg (2005), Probabilistic Symmetries and Invariance Principles, Chapter 1, Lemma 1.2
Kallenberg's L² Approach (Lemma 1.2 + Second Proof) #
This section implements Kallenberg's "second proof" of de Finetti's theorem using elementary L² bounds. The key is Lemma 1.2: for exchangeable sequences, weighted averages satisfy a simple variance bound that makes Cesàro averages Cauchy in L².
No ergodic theory is used - only:
- Exchangeability → constant pairwise second moments
- Algebraic identity for variance of weighted sums
- Completeness of L²
This is the lightest-dependency route to de Finetti.
References:
- Kallenberg (2005), Probabilistic Symmetries, Chapter 1, pp. 27-28
- Lemma 1.2 (L² bound for exchangeable weighted sums)
- "Second proof of Theorem 1.1" (the L² route to de Finetti)
Kallenberg's L² bound (Lemma 1.2) - Core of the elementary proof.
For an exchangeable sequence and centered variables Z_i := f(X_i) - E[f(X_1)], the L² distance between any two weighted averages satisfies:
‖∑ p_i Z_i - ∑ q_i Z_i‖²_L² ≤ C_f · sup_i |p_i - q_i|
where C_f := E[(Z_1 - Z_2)²].
Key application: For uniform block averages of length n, ‖A_{m,n} - A_{m',n}‖_L² ≤ √(C_f/n)
making the family {A_{m,n}}_m Cauchy in L² as n→∞.
Proof: Pure algebra + exchangeability:
- Expand ‖∑ c_i Z_i‖² = ∑ c_i² E[Z_i²] + ∑_{i≠j} c_i c_j E[Z_i Z_j]
- By exchangeability: E[Z_i²] = E[Z_1²], E[Z_i Z_j] = E[Z_1 Z_2] for i≠j
- For c_i = p_i - q_i (differences of probability weights): ∑ c_i = 0
- Algebraic bound: ∑ c_i² ≤ (∑|c_i|) · sup|c_i| ≤ 2 · sup|c_i|
- Substitute and simplify to get the bound
This is exactly Kallenberg's Lemma 1.2. No ergodic theory needed!
Why this proof uses l2_contractability_bound instead of kallenberg_L2_bound #
The Circularity Problem:
The de Finetti theorem we're proving establishes: Contractable ↔ Exchangeable
contractable_of_exchangeable(✓ proved in Contractability.lean): Exchangeable → Contractablecesaro_to_condexp_L2(this file): Contractable → Exchangeable (via conditionally i.i.d.)
Since we're trying to prove Contractable → Exchangeable, we cannot assume exchangeability in this proof - that would be circular!
Why kallenberg_L2_bound requires exchangeability:
kallenberg_L2_bound needs Exchangeable μ Z to establish uniform second moments:
- E[Z_i²] = E[Z_0²] for all i (uniform variance)
- E[Z_i Z_j] = E[Z_0 Z_1] for all i≠j (uniform pairwise covariance)
Exchangeability gives this via permutation invariance: swapping indices doesn't change the distribution.
Why contractability is insufficient for kallenberg_L2_bound:
Contractability only tells us about increasing subsequences:
- For any increasing k : Fin m → ℕ, the subsequence (Z_{k(0)}, ..., Z_{k(m-1)}) has the same distribution as (Z_0, ..., Z_{m-1})
This is weaker than exchangeability:
- ✓ We know (Z_0, Z_1) has same distribution as (Z_1, Z_2), (Z_2, Z_3), etc.
- ✗ We DON'T know (Z_0, Z_1) has same distribution as (Z_1, Z_0) - contractability doesn't give permutation invariance!
However: contractability DOES give uniform covariance!
Even though contractability ≠ exchangeability, contractability is sufficient for:
- E[Z_i²] = E[Z_0²] for all i (from the increasing subsequence {i})
- E[Z_i Z_j] = E[Z_0 Z_1] for all i<j (from the increasing subsequence {i,j})
This is exactly the covariance structure needed by l2_contractability_bound from
L2Helpers.lean, which doesn't assume full exchangeability.
Note: By the end of this proof, we'll have shown Contractable → Exchangeable, so contractable sequences ARE exchangeable. But we can't use that equivalence while proving it - that would be begging the question!