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A finite sample can fail a criterion that rejects a positive-probability sequence

Physical scope

A hypothetical finite-sampling analysis of the declared two-path arrangement: for a coherent or which-path preparation whose recorded law is fixed, and under the declared independent-and-identically-distributed $N$ -trial sampling model, every declared fringe criterion that rejects at least one outcome sequence of positive probability fails with positive probability, and the empirical frequency of a label concentrates around its recorded probability. The indistinguishability conclusion for the phase-zero coherent and weighted which-path laws assumes that phase zero is admissible. The item is a conditional consequence of the declared preparation and detector models, the declared sampling model in the sense of P7, and published probability suppliers with local calculations. It asserts no measured counts, no p-value, no confidence level, no universal failure probability, and no empirical conclusion about nature; a finite sample without a visible fringe pattern is not a falsification of the coherent model, and a sample with a visible pattern is not a proof of it.

Statement

Adopt the declared two-path preparation model The declared two-path preparation model and its ideal detection bins and the declared recorded detector and sampling model The declared recorded detector and sampling model of the two-path arrangement with finite label set $A$ satisfying $0 \notin A$ , recorded law $ν$ (a probability law on $A \cup {0}$ , with $0$ the distinct no-record label), and the declared $N$ -trial sampling model of clause (d) of that definition; the identification of repetitions with that model is the additional statistical assumption of clause (ii) of the preparation and apparatus postulate P7 (Preparation and apparatus postulate P7: declared preparations, isolation and sampling assumptions). Fix a preparation $ρ$ , which may be a coherent preparation $ρ_{c o h} (φ)$ or a which-path preparation $ρ_{m i x} (w)$ of the declared model, so that its recorded law is the law $ν$ of Coherent and which-path detection laws of the declared two-path model and Recorded bin laws, loss and conditioning of the declared two-path detector. Let $(O_{1}, \dots, O_{N})$ be the outcome sequence, $S_{N} (a) : = # {k : O_{k} = a}$ and $F_{N} (a) : = S_{N} (a) / N$ the empirical frequency of $a \in A \cup {0}$ . Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Then:

(a) Concentration. For every label $a$ and every $ε > 0$ , $E [F_{N} (a)] = ν (a), Var (F_{N} (a)) = \frac{ν (a) (1 - ν (a))}{N}, P (∣ F_{N} (a) - ν (a) ∣ \geq ε) \leq \frac{ν (a) (1 - ν (a))}{N ε^{2}} .$

(b) Positive failure probability. Let $C \subseteq (A \cup {0})^{N}$ be a declared fringe criterion, and suppose there is a sequence $(a_{1}, \dots, a_{N})$ with $(a_{1}, \dots, a_{N}) \notin C$ and $ν (a_{k}) > 0$ for every $k$ . Then the failure probability of the criterion is $P ((O_{1}, \dots, O_{N}) \notin C) \geq \prod_{k = 1}^{N} ν (a_{k}) > 0.$ In particular, if $ν (a) > 0$ for some label $a$ , then every criterion that rejects the constant sequence $(a, \dots, a)$ has failure probability at least $ν (a)^{N} > 0$ .

(c) No discrimination between coincident laws. If the model is admissible at phase $0$ and the recorded coherent law of phase $0$ and the recorded which-path law of weight $w_{*}$ coincide in the sense of clause (d) of Recorded bin laws, loss and conditioning of the declared two-path detector, then no criterion whatsoever has different probability of acceptance under the two declared preparations; in particular no finite sample distinguishes them with certainty.

(d) Scope. Clauses (a)-(c) are conditional on the declared preparation, detector and sampling models and the cited mathematics. The independence and identical distribution of the trials are declared model assumptions, not consequences of P1-P6; no failure probability, error rate, p-value, confidence level, power, sample size or measured count is computed or asserted, and no conclusion is drawn about any real apparatus. In particular the statements do not assert that a coherent preparation must show a visible fringe pattern in a finite sample, and do not assert that a sample without a visible pattern falsifies the coherent model.

Facts & Assumptions

Given: The declared preparation, detector and sampling models, with the recorded law $ν$ on $A \cup {0}$ ; the outcome sequence $O_{1}, \dots, O_{N}$ and the counts $S_{N} (a)$ , $F_{N} (a)$ ; a declared criterion $C$ and a sequence $(a_{1}, \dots, a_{N})$ as in (b); the Axiom of Countable Choice.

[F1]

Independent trials. The random elements $O_{1}, \dots, O_{N}$ are mutually independent and identically distributed with $P (O_{k} = a) = ν (a)$ for every $a \in A \cup {0}$ (The declared recorded detector and sampling model of the two-path arrangement clause (d), Pairwise and mutual independence of finite-valued random variables).

[F2]

Expectation, variance and concentration suppliers. Expectation is the integral against the probability measure; it is linear for integrable random variables. For a square-integrable real variable $X$ , $Var (X) = E [(X - E X)^{2}]$ ; variance adds across a finite pairwise-independent family, and Chebyshev bounds $P (∣ X - E X ∣ \geq ε) \leq Var (X) / ε^{2}$ (Expectation of a nonnegative or integrable random variable, Linearity, monotonicity, and the modulus bound for expectation, Moments, variance, and covariance on a probability space, Variance adds for every finite pairwise-independent family, Chebyshev's inequality for random variables).

[F3]

Recorded laws of the declared preparations. The recorded probabilities of the coherent and which-path preparations are those of Coherent and which-path detection laws of the declared two-path model clauses (a)-(b) as read by Recorded bin laws, loss and conditioning of the declared two-path detector clauses (a)-(b); its coincidence conclusion is used only when phase $0$ is admissible.

Proof

Proof technique: direct; reduce each label frequency to a sum of pairwise independent Bernoulli indicators, compute its moments and apply Chebyshev, use mutual independence for the selected-sequence event, and compare product laws when the recorded one-trial laws coincide.

Given: The data of the statement and [F1]-[F3].

1.1givenF1algebra

[Indicator marginals and pairwise independence] Fix $a \in A \cup {0}$ , [given, F1, algebra] put $p = ν (a)$ , and define $X_{k} = 1_{{O_{k} = a}}$ . Then $P (X_{k} = 1) = p$ and $P (X_{k} = 0) = 1 - p$ , so $X_{k}$ is Bernoulli $(p)$ by Bernoulli random variables and binomial random variables as sums of independent Bernoulli trials. For distinct $i, j$ , mutual independence of $O_{i}, O_{j}$ , summed over the finite set of outcomes other than $a$ when an indicator is zero, gives $P (X_{i} = 1, X_{j} = 1) = p^{2}, P (X_{i} = 1, X_{j} = 0) = p (1 - p), P (X_{i} = 0, X_{j} = 1) = p (1 - p), P (X_{i} = 0, X_{j} = 0) = (1 - p)^{2} .$ Thus the $X_{k}$ are pairwise independent.

1.2givenF1algebra

[Positive-probability sequences] Let [given, F1, algebra] $E = {O_{k} = a_{k} for every k = 1, \dots, N}$ . By mutual independence [F1], $P (E) = \prod_{k = 1}^{N} ν (a_{k}) > 0$ . Since the chosen sequence is outside $C$ , $E \subseteq {(O_{1}, \dots, O_{N}) \notin C}$ , so the failure probability is at least $\prod_{k} ν (a_{k})$ . The constant sequence is a special case. This proves (b).

2.1givenstep 1.1F2algebra

[Frequency moments and concentration] Since [given, step 1.1, F2, algebra] $F_{N} (a) = N^{- 1} \sum_{k = 1}^{N} X_{k}$ , expectation linearity gives $E F_{N} (a) = p$ . The two-point law gives $Var (X_{k}) = E [(X_{k} - p)^{2}] = p (1 - p)$ . By [F2], $Var (F_{N} (a)) = N^{- 2} Var (\sum_{k = 1}^{N} X_{k}) = \frac{p (1 - p)}{N} .$ Applying Chebyshev from [F2] with threshold $ε > 0$ gives the bound in (a). This proves (a).

3.1givenF1F3algebrastep 1.2step 2.1∎

[No discrimination] Under the coincidence hypothesis of (c), the [given, F1, F3, algebra, step 1.2, step 2.1] recorded probabilities agree for each $a \in A$ . Their no-record probabilities also agree because $ν (0) = 1 - \sum_{a \in A} ν (a)$ . For each sequence in the finite set $(A \cup {0})^{N}$ , its probability under the declared IID model is the product of its $N$ one-trial probabilities; these products therefore agree under the two preparations. Summing over any criterion $C$ shows equal acceptance probabilities, proving (c). Steps 2.1 and 1.2 prove (a)-(b), and (d) records that these are conditional model deductions, not empirical results.

Remarks

A null result is compatible with the coherent model. By (b), for any criterion that would reject the outcome actually realized in some positive-probability sequence, failure has positive probability under the declared coherent model itself; a finite sample without a visible pattern is therefore not by itself a refutation.
A pattern is not a proof. The same computation applies to the which-path model, and (c) shows that when the two recorded laws coincide no criterion distinguishes them. Agreement of a dataset with a declared model is evidence about the combined model, apparatus and sampling assumptions, not a proof of the model.

A finite sample can fail a criterion that rejects a positive-probability sequence

Statement

Facts & Assumptions

Proof

Remarks

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