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Stationary means and sample autocovariance — Examples

1 · Prerequisites

2 · Summary

The finite examples separate pairwise uncorrelated data from mutual independence and compare the two sample-ACVF denominator conventions. A three-observation vector gives an exact negative Toeplitz quadratic form for the pair-count-normalized version. No asymptotic or central-limit claim is attached to either calculation.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Pairwise uncorrelated data can be jointly dependent

Example

Let $Ω = {(- 1, - 1, 1), (- 1, 1, - 1), (1, - 1, - 1), (1, 1, 1)}$ with each outcome assigned probability $1 / 4$ . Let $X_{1}, X_{2}, X_{3}$ be the three coordinate random variables. This finite data vector is pairwise uncorrelated, but it is not mutually independent.

Facts & Assumptions

Given: The four-outcome uniform probability space $Ω$ above and its three coordinate random variables.

[F1]

A finite probability space assigns nonnegative outcome weights summing to one; events are subsets of outcomes with probability the sum of their weights (Finite probability spaces, outcome weights, events, and event probabilities).

[F2]

A finite-valued real random variable is a function on the finite outcome space (Real random variables on finite probability spaces and their finite distributions).

[F3]

Expectation on a finite probability space is the weighted sum (Expectation of a real random variable on a finite probability space).

[F4]

Covariance is the expectation of the centered product (Variance, standard deviation, and covariance on a finite probability space).

[F5]

Mutual independence of finite-valued random variables requires factorization of every joint point probability (Pairwise and mutual independence of finite-valued random variables).

Verification

technique · enumerate the four outcomes

1.1F1F2F3givenalgebra

Each coordinate takes the value $1$ twice and $- 1$ twice. By [F1]–[F3], $E [X_{i}] = 0$ and $E [X_{i}^{2}] = 1$ for each $i \in {1, 2, 3}$ .

2.1F3F4step 1.1algebra

For each of the three pairs, the four products $X_{i} X_{j}$ contain two $1$ values and two $- 1$ values. Thus $E [X_{i} X_{j}] = 0$ ; using the means from step 1.1 in [F4] gives $Cov (X_{i}, X_{j}) = 0$ whenever $i \neq j$ .

3.1F1F5step 1.1algebra∎

The event ${X_{1} = 1, X_{2} = 1, X_{3} = 1}$ contains only the outcome $(1, 1, 1)$ , so its probability is $1 / 4$ . Each event ${X_{i} = 1}$ has probability $1 / 2$ by step 1.1. Mutual independence would give joint probability $(1 / 2)^{3} = 1 / 8$ by [F5], contradicting $1 / 4$ . Therefore the variables are not mutually independent.

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Pair-count-normalized sample ACVF need not be positive semidefinite

Statement refuted

For every finite real data vector, the pair-count-normalized sample ACVF ${\tilde{γ}}_{n}$ has nonnegative Toeplitz quadratic forms.

Facts & Assumptions

Given: The pair-count-normalized sample ACVF definition for a finite data vector of length $n \geq 1$ .

[F1]

For $∣ h ∣ < n$ , the pair-count-normalized rescaling divides the centered lag-product sum by $n - ∣ h ∣$ and is zero outside that lag range (Autocovariance and sample autocovariance).

Counterexample

1.1F1givenalgebra

Take $n = 3$ and data $(x_{1}, x_{2}, x_{3}) = (1, 0, 0)$ . Its sample mean is $1 / 3$ , so its centered values are $(2 / 3, - 1 / 3, - 1 / 3)$ . By [F1], ${\tilde{γ}}_{3} (0) = 2 / 9$ , ${\tilde{γ}}_{3} (1) = - 1 / 18$ , and ${\tilde{γ}}_{3} (2) = - 2 / 9$ ; negative-lag values equal the corresponding positive-lag values.

2.1step 1.1algebra∎

In the Toeplitz matrix on indices $0, 1, 2$ , the diagonal entries are $2 / 9$ , the first off-diagonal entries are $- 1 / 18$ , and the outer off-diagonal entries are $- 2 / 9$ by step 1.1. For $v = (1, \frac{1}{2}, 1)^{T}$ , the quadratic form is $v^{T} T v = (2 / 9) (1 + 1 / 4 + 1) + 2 (- 1 / 18) (1 / 2 + 1 / 2) + 2 (- 2 / 9) = - 1 / 18 < 0$ . A positive-semidefinite Toeplitz form must be nonnegative on every real vector, so this data vector refutes the statement.