Standard-Deviation Normalization Proof

In the paper's words

Let (Ω, F, P)be a probability space, and let Xbe a real-valued random variable defined on this space, representing a physical observable. Assume Xhas a well-defined mean μ = E[X]and finite, positive variance σ² = E[(X - μ)²] > 0. Let Mbe a statistical manifold representing a family of probability distributions \p(x|θ)\_θ ∈ Θ, where θare parameters.

The abstract, verbatim, and its opening sentence is the claim. Every line below is quoted from the text.

Numbers it reports

No sentence of this paper reports a figure with a unit or a quantity word, so none is shown.

How it is done

1. Role of Standard-Deviation Normalization in a Scale-Invariant Statistical Framework Author: Rowan Brad Quni-Gudzinas Affiliation: QNFO Contact: rowan.quni@qnfo.org ORCID: 0009-0002-4317-5604 ISNI: 0000 0005 2645 6062 DOI: https://papers.qnfo.org/papers/standard-deviation-normalization-proof/ Publication Date: 2025-09-28 Version: 1.0

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2. Theorem (Fundamental Role of Standard-Deviation Normalization).

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What would falsify it

1. Let $(\Omega, \mathcal{F}, P)$be a probability space, and let $X$be a real-valued random variable defined on this space, representing a physical observable.

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2. 2: The Standardized Variable Z is Invariant under Scaling Consider a global scale transformation of the observable $X$: $$ X \mapsto X' = \lambda X, \quad \lambda \gt 0. $$ The mean and standard deviation transform as: $$ \mu' = \lambda \mu, \quad \sigma' = \lambda \sigma. $$ The standardized variable transforms as: $$ Z' = \frac{X' - \mu'}{\sigma'} = \frac{\lambda X - \lambda \mu}{\lambda \sigma} = \frac{X - \mu}{\sigma} = Z. $$ Thus, $Z$is invariant under global scaling of the original variable $X$.

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3. The standard deviation $\sigma$acts as the natural unit for measuring the variability of an observable $X$, consistent with the epistemic principle that knowledge is limited by inherent fluctuations.

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Where the claim stops

1. Assume $X$has a well-defined mean $\mu = \mathbb{E}[X]$and finite, positive variance $\sigma^2 = \mathbb{E}[(X - \mu)^2] \gt 0$.

   from Given

2. Assume the family includes location-scale distributions of the form $p(x|\mu, \sigma) = \sigma^{-1} f((x-\mu)/\sigma)$, where $f$is a fixed probability density.

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Where to read

1. Given90 words · 1 min
2. Goal33 words · 1 min
3. Derivation743 words · 3 min
4. Conclusion118 words · 1 min