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✓ 5 results · all verified · 0 also independently AI-judged

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Resampling schemes and the sampling-unit contract — Examples

1 · Prerequisites

2 · Summary

Objective

The examples compute an exact conditional bootstrap distribution for a two-point sample mean, enumerate delete-two means on four units, and compare them with size-two subsampling. For the sample mean, complementation makes the two six-point laws equal; their reported outputs remain different, because the delete-two procedure also returns a single uniform average and no variance scaling. A cluster record is resampled as a whole. A stationary binary time-series pair refutes the claim that pointwise IID bootstrap draws reproduce the original joint law. A two-unit Fisher sharp-null design distinguishes known assignment randomization from pointwise bootstrap resampling. None of these calculations alone supplies a general bootstrap-validity theorem.

Development

The sample-mean calculation enumerates all four ordered size-two bootstrap samples. The four-unit example enumerates six delete-two complement means and the six uniformly weighted two-unit subsets, identifying their complement bijection and the shared finite law. The cluster example applies the resampling-unit contract to vector records. In the stationary dependent-pair counterexample, the original covariance is $1 / 8$ while the conditional pointwise-bootstrap covariance is $0$ . In the assignment example, the known design law has support ${- 2, 2}$ , whereas the empirical bootstrap law also has mass at $0$ .

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passOpen item page →

The bootstrap law of a two-point sample mean

Example

Let $X_{1}, X_{2}$ be IID with $P (X_{i} = 0) = P (X_{i} = 2) = 1 / 2$ , and suppose the observed ordered sample is $(0, 2)$ . The size-two statistic is the sample mean ${\overline{X}}_{2} = (X_{1} + X_{2}) / 2$ from Partial sums, row sums and sample means.

Facts & Assumptions

Given: A possible observed realization $(0, 2)$ from two IID real observations with common law assigning mass $1 / 2$ to each of $0$ and $2$ .

[F1]

Identical distribution and IID families: IID observations have a common law and are mutually independent.

[F2]

Empirical measure of observed units: the empirical measure assigns $P_{n} (A) = \frac{1}{n} \sum_{i} 1 {x_{i} \in A}$ to each measurable set $A$ .

[F3]

Partial sums, row sums and sample means: ${\overline{X}}_{n} = n^{- 1} \sum_{i = 1}^{n} X_{i}$ .

[F4]

Size-n bootstrap from the empirical law: a size- $n$ replicate draws $n$ indices independently and uniformly from the observed indices and returns the corresponding ordered tuple.

[F5]

Empirical-law bootstrap draws are conditionally IID: conditional on fixed observations, each bootstrap entry has law $P_{n}$ and the entries are mutually independent.

Verification

1.1givenF1F2algebra

Under the stated IID law the tuple $(0, 2)$ has probability $(1 / 2) (1 / 2) = 1 / 4$ by [F1], so it is an admissible observed realization. For this fixed tuple, [F2] gives $P_{2} ({0}) = P_{2} ({2}) = 1 / 2$ .

1.2F4F5algebra

By [F4] and [F5], each of the four ordered bootstrap pairs $(0, 0), (0, 2), (2, 0), (2, 2)$ has conditional probability $1 / 4$ .

2.1F3step 1.2algebra∎

Applying the sample-mean formula [F3] to these four pairs gives values $0, 1, 1, 2$ , respectively. Thus the conditional bootstrap mean has probabilities $1 / 4, 1 / 2, 1 / 4$ at $0, 1, 2$ . This is the exact conditional resampling law for this fixed sample and is not a claim of general bootstrap validity.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passOpen item page →

Delete-d replicates and subsampling on four units

Example

Take four observed real units $z = (1, 2, 4, 7)$ and use $T_{m} (u_{1}, \dots, u_{m}) = m^{- 1} \sum_{j = 1}^{m} u_{j}$ as the statistic at every nonempty subsample size. For delete-two jackknife replicates, remove each two-unit subset in turn and average the six size-two complement means. For subsampling, draw a two-unit subset uniformly without replacement and compute its mean.

Facts & Assumptions

Given: The ordered tuple $(1, 2, 4, 7)$ and the sample-mean statistic $T_{m} (u_{1}, \dots, u_{m}) = m^{- 1} \sum_{j = 1}^{m} u_{j}$ for $m = 2, 4$ .

[F1]

Delete-d jackknife replicates and their uniform average: the delete- $d$ procedure evaluates the declared size- $(n - d)$ statistic on every complement of a $d$ -subset and returns the uniform average of the replicates; it defines no variance scaling.

[F2]

Uniform subsampling without replacement: the subsampling procedure selects a uniformly distributed subset of the declared size without replacement and applies the corresponding statistic.

[F3]

Partial sums, row sums and sample means defines ${\overline{X}}_{m} = m^{- 1} \sum_{j = 1}^{m} X_{j}$ for real random variables on one probability space. Evaluating this finite-sum identity at an outcome gives the tuple formula declared in Given.

Verification

1.1givenF1F3algebra

The six deleted pairs ${1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}$ leave complement means $11 / 2, 9 / 2, 3, 4, 5 / 2, 3 / 2$ , respectively.

2.1F1F3step 1.1algebra

Their uniform average is $(11 / 2 + 9 / 2 + 3 + 4 + 5 / 2 + 3 / 2) / 6 = 7 / 2$ , equal to the original sample mean; [F1] supplies no multiplier or variance interpretation for that average.

3.1F1F2F3step 1.1step 2.1algebra∎

The six two-unit subsets have means $3 / 2, 5 / 2, 4, 3, 9 / 2, 11 / 2$ , each with probability $1 / 6$ by [F2]. The complement of every two-subset again has size two, and taking the complement twice returns the original subset; hence complementation bijects the deleted pairs with the sampled pairs. Therefore the size-two subsampling law has the same six equally weighted values as the delete-two replicate law in this example. The jackknife output in step 2.1 is their average, a single number, and is not a variance estimate.

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Resampling complete cluster records

Example

Suppose each of three clusters is recorded as a two-measurement vector, with records $C_{1} = (1, 0)$ , $C_{2} = (1, 1)$ , and $C_{3} = (0, 0)$ . The declared design treats the entire vector as the sampling unit, so a bootstrap draw selects one complete vector at a time.

Facts & Assumptions

Given: Three observed cluster records and a design that declares each vector indivisible for the target analysis.

[F1]

Sampling units and the resampling-unit contract: a resampling unit is a complete unit record or predeclared group selected together; unit compatibility does not assert validity.

[F2]

Size-n bootstrap from the empirical law: each bootstrap draw selects an observed unit record by an independently uniform index.

[F3]

Empirical-law bootstrap draws are conditionally IID: conditional on the fixed unit records, the selected records are IID from their empirical measure.

[F4]

Empirical measure of observed units: each of the three indexed cluster records has empirical mass $1 / 3$ .

Verification

1.1givenF1

The declared records are $C_{1}, C_{2}, C_{3}$ , each a complete ordered pair of measurements, so [F1] makes the vector rather than either coordinate the resampling unit.

1.2F2F3F4algebra

For the index tuple $(2, 2, 1)$ , [F2] returns $(C_{2}, C_{2}, C_{1}) = ((1, 1), (1, 1), (1, 0))$ . By [F2], the three independent index draws each assign probability $1 / 3$ to their specified index, so this index tuple has conditional probability $3^{- 3} = 1 / 27$ . The empirical record law assigns mass $1 / 3$ to each of these three distinct cluster records by [F4], and [F3] identifies the record draws as conditionally IID.

2.1F1F2step 1.2∎

Every bootstrap tuple is formed by selecting records from ${C_{1}, C_{2}, C_{3}}$ , so every selected two-measurement vector remains intact. This verifies the unit contract in this finite example; it makes no claim that resampling clusters is valid for every target or design.

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passOpen item page →

Pointwise bootstrap need not preserve pair dependence

Statement refuted

The conditional pointwise size- $n$ bootstrap reproduces the original joint sampling law for every possibly dependent $n$ -tuple of observed units.

Facts & Assumptions

Given: Consecutive observations $(X_{1}, X_{2})$ from a stationary binary Markov chain with invariant law $(1 / 2, 1 / 2)$ and transition probabilities $3 / 4$ to stay at the current state and $1 / 4$ to switch, the observed pair $x = (X_{1}, X_{2})$ , and the size-two pointwise empirical bootstrap.

[F1]

Empirical measure of observed units: the empirical law is $P_{2} = \frac{1}{2} (δ_{x_{1}} + δ_{x_{2}})$ , with multiplicity when values repeat.

[F2]

Size-n bootstrap from the empirical law: conditional bootstrap entries are obtained from independent uniform draws of the observed indices.

[F3]

Empirical-law bootstrap draws are conditionally IID: conditional on fixed observations, the bootstrap entries are independent with common law $P_{2}$ .

[F4]

Moments, variance, and covariance on a probability space: for square-integrable real random variables, covariance is $E [(X - E X) (Y - E Y)]$ .

Counterexample

The transition rule and invariant initial law give the finite joint law $P (0, 0) = P (1, 1) = 3 / 8, P (0, 1) = P (1, 0) = 1 / 8.$

1.1givenF4algebra

The transition matrix is $(\begin{array}{cc} 3 / 4 & 1 / 4 \\ 1 / 4 & 3 / 4 \end{array})$ , whose columns sum to one, so $(1 / 2, 1 / 2)$ is invariant; multiplying this initial law by the transition probabilities gives the displayed four joint masses. Both marginals are Bernoulli with mean $1 / 2$ , while $P (X_{1} = 0, X_{2} = 0) = 3 / 8 \neq 1 / 4 = P (X_{1} = 0) P (X_{2} = 0)$ , so the observations are dependent. Also $E [X_{1} X_{2}] = 3 / 8$ , and [F4] gives original covariance $3 / 8 - (1 / 2) (1 / 2) = 1 / 8$ .

1.2F1F2F3F4algebra

Fix any realized pair $x = (x_{1}, x_{2})$ and put $m = (x_{1} + x_{2}) / 2$ . By [F1]–[F3], each bootstrap coordinate has conditional mean $m$ , and the two coordinates are conditionally independent; hence $E^{*} [X_{1}^{*} X_{2}^{*}] = m^{2}$ and [F4] gives ${Cov}^{*} (X_{1}^{*}, X_{2}^{*}) = 0$ .

2.1step 1.1step 1.2algebra∎

Both the original pair and the bootstrap pair are bounded, so equality of their joint laws would force equality of their covariances. The original covariance is $1 / 8$ but the conditional bootstrap covariance is $0$ , so the laws differ and the statement is false. This finite counterexample concerns exact joint-law reproduction only; it does not decide asymptotic validity for any particular statistic.

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Assignment randomization and bootstrap are different laws

Example

Take two fixed units with potential outcomes $y = (0, 2)$ under the Fisher sharp null, and suppose the observed assignment is $z^{(1)} = (1, 0)$ . The allowed assignment vectors are $z^{(1)} = (1, 0)$ and $z^{(2)} = (0, 1)$ , each with probability $1 / 2$ . Predeclare the statistic as treated outcome minus untreated outcome. Compare its law under the known assignment mechanism with its law when the two observed outcomes are pointwise bootstrapped with replacement and the observed assignment $z^{(1)}$ is held fixed in the contrast.

Facts & Assumptions

Given: Two fixed units, observed assignment $z^{(1)} = (1, 0)$ , the two allowed assignments with equal design probabilities, potential outcomes fixed at $(0, 2)$ under every assignment, and the predeclared treated-minus-untreated score.

[F1]

Finite randomized treatment-assignment models: units and potential outcomes are fixed, the assignment has the known finite probability law, and the assignment is the only randomness in the design.

[F2]

Fisher sharp null and predeclared test orderings: under the Fisher sharp null the response vector is constant over allowed assignments, and the score is fixed in advance; upper-tail ordering includes ties.

[F3]

Fisher sharp-null randomization p-values are valid: under every such finite sharp-null design, the inclusive weighted upper-tail rank p-value is superuniform using the actual assignment weights.

[F4]

Size-n bootstrap from the empirical law: pointwise bootstrap samples observed unit values with independent uniform index draws.

[F5]

Empirical-law bootstrap draws are conditionally IID: conditional on the fixed data, the bootstrap entries are IID from the empirical law.

[F6]

Finite-population average treatment effect and weak null: the weak zero-average-effect null restricts only the average unit effect, while the Fisher sharp null requires every unit effect to be zero.

[F7]

Sampling units and the resampling-unit contract: the conditional bootstrap law is a law on resampled data records, while a Bayesian posterior is a conditional law on parameter values under a specified prior and joint model.

[F8]

Valid p-values: a p-value statistic is valid for a null hypothesis when it is superuniform under every null law.

[F9]

Empirical measure of observed units: the empirical law assigns each indexed observation mass $1 / n$ , counting repetitions by index.

Verification

1.1givenF1F2construct

Define $T (z, y) = \sum_{i : z_{i} = 1} y_{i} - \sum_{i : z_{i} = 0} y_{i}$ . The sharp-null schedule fixes $y = (0, 2)$ , so $T (z^{(1)}, y) = - 2$ and $T (z^{(2)}, y) = 2$ .

2.1F1F2F3F8step 1.1algebra

The design randomization law therefore assigns probability $1 / 2$ to each of $- 2$ and $2$ . Its inclusive upper-tail p-value is $1$ at $z^{(1)}$ and $1 / 2$ at $z^{(2)}$ , a superuniform distribution under the assignment law, so [F8] identifies it as valid for this null; [F3] also gives the general validity statement.

3.1F4F5F9step 1.1step 2.1algebra

The observed unit values are $(0, 2)$ , so [F9] gives empirical mass $1 / 2$ to each. By [F4] and [F5], an ordered size-two bootstrap pair takes $(0, 0), (0, 2), (2, 0), (2, 2)$ with probabilities $1 / 4$ each. The same difference score has bootstrap law with masses $1 / 2$ at $0$ and $1 / 4$ at each of $- 2, 2$ , which differs from the design law in step 2.1.

4.1F1F4F5F6F7step 2.1step 3.1∎

The two laws have different state spaces and sampling mechanisms: one resamples assignments under a known design, the other resamples observed unit records conditional on those records. By [F7], a Bayesian posterior, when defined, is instead a conditional law on parameter values under a specified prior and joint model; this example supplies no such prior or posterior. By [F6], its sharp-null calculation also asserts no validity under the weaker average-effect null.

Resampling schemes and the sampling-unit contract — Examples

1 · Prerequisites

2 · Summary

Objective

Development

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

The bootstrap law of a two-point sample mean

Example

Facts & Assumptions

Verification

Delete-d replicates and subsampling on four units

Example

Facts & Assumptions

Verification

Resampling complete cluster records

Example

Facts & Assumptions

Verification

Pointwise bootstrap need not preserve pair dependence

Statement refuted

Facts & Assumptions

Counterexample

Assignment randomization and bootstrap are different laws

Example

Facts & Assumptions

Verification

Sources