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Khinchin weak law for integrable IID variables
Statement
If are IID real random variables and , then, with and , Consequently in probability.
Facts & Assumptions
Zero truncation at a positive level: For a real random variable and a deterministic level , its zero truncation is The threshold event is measurable because is measurable and is Borel; its indicator and the product are measurable by thm-arithmetic-and-lattice-operations-preserve-measurability. Thus is a real random variable as in def-random-element-and-real-random-variable. It equals at both cutoff endpoints and is zero outside the interval. Since , for every its absolute th moment is at most . This is not clipping to the endpoints.
IID finite-variance weak law: Let be IID square-integrable real random variables, with and . For , and in and in probability. Also for .
Measurable coordinatewise functions preserve independence: Let be an independent family of random elements . For each , let be measurable. Then the family is independent.
Dominated convergence: Let and be measurable complex-valued functions such that almost everywhere and almost everywhere for a single nonnegative measurable function with . Then , and hence
Markov's inequality for random variables: If is a nonnegative random variable on a probability space and , then
Finite-measure includes into for : Let be a measure space with . 1. If and , then and 2. If and , then and
Linearity, monotonicity, and the modulus bound for expectation: Let be integrable real or complex random variables on one probability space. 1. For scalars , 2. If and are real-valued and almost surely, then 3.
Proof
Given: The objects and hypotheses of the statement.
Fix and set , . The are bounded IID variables: measurable transformations preserve independence and their laws remain equal by inverse images. The finite-variance result gives .
On a probability space the norm is at most the norm. Finite linearity, the triangle inequality and the modulus bound give . Therefore .
For each fixed let in this bound. Then let run through positive integers tending to infinity. The residual is dominated by the integrable and tends pointwise to zero, so dominated convergence makes the remaining bound tend to zero. Finally Markov applied to proves convergence in probability. No division by a moment occurs, so constant or zero variables are included.
Depends on
- Identical distribution and IID families
- Partial sums, row sums and sample means
- Zero truncation at a positive level
- IID finite-variance weak law
- Measurable coordinatewise functions preserve independence
- Dominated convergence
- Markov's inequality for random variables
- Finite-measure $L^r$ includes into $L^p$ for $p < r$
- Linearity, monotonicity, and the modulus bound for expectation
Used by
Nothing in the library uses this result yet.
Dependency tree · two levels
33 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
- Theorem 3.3, first proof, pp. 56–57 (standard reference, not scraped)
- Theorem 2.2.14, pp. 64–65 (standard reference, not scraped)