ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14

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The fair-coin strong law as a shift average

Example

Assume the Axiom of Countable Choice. For a binary sequence $x$ and the first-coordinate observable $f (x) = x_{0}$ ,

$A_{n} f (x) = \frac{1}{n} \sum_{k = 0}^{n - 1} x_{k} .$

Thus the fair-coin shift theorem is precisely the almost-sure convergence of the empirical proportion of heads to $1 / 2$ .

Facts & Assumptions

Given: Countable choice, binary sequence space, its left shift $σ$ , and $f (x) = x_{0}$ .

[F1]

Binary cylinders prescribe finitely many coordinates (Binary-sequence cylinders and fair-coin content).

[F2]

The fair-coin frequency theorem says $n^{- 1} \sum_{k < n} x_{k} \to 1 / 2$ almost surely (Fair-coin frequency strong law).

Verification

technique · direct coordinate calculation

1.1

The $k$ th iterate of the left shift satisfies $(σ^{k} x)_{0} = x_{k}$ . Therefore $f (σ^{k} x) = x_{k}$ .

givenalgebra

2.1

Summing step 1.1 for $0 \leq k < n$ and dividing by $n$ gives $A_{n} f (x) = n^{- 1} \sum_{k < n} x_{k}$ .

step 1.1algebra

3.1

Applying [F2] to the right side proves $A_{n} f (x) \to 1 / 2$ almost surely. The observable is the cylinder indicator $1_{{x : x_{0} = 1}}$ from [F1], so this is the usual heads-frequency strong law written dynamically. Countable choice is inherited from [F2].

F1F2step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

16 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

Charles Walkden, Ergodic Theory lecture notes (standard reference, not scraped)