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✓ 6 results · all verified · 0 also independently AI-judged
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Statistical, resampling, simulation, and numerical error ledger — Examples

1 · Prerequisites

2 · Summary

Examples

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

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

Sampling and finite-bootstrap errors for a Bernoulli mean

Example

Let X1,X2 be IID observations in {0,1} with P(Xi=1)=3/4 and P(Xi=0)=1/4. Take each observation as one resampling unit in the finite measurable space ({0,1},2{0,1}), and define T2(x1,x2):=(x1+x2)/2 on {0,1}2. The map is measurable because its domain has the full power-set sigma-algebra. Suppose the observed ordered sample is (0,1), an event of probability 3/16>0. The population mean is θ=3/4, while the data estimate is T2(0,1)=1/2. Fix these data and use the size-two empirical-law bootstrap of Size-n bootstrap from the empirical law. Each resample chooses two indices independently and uniformly from {1,2}, with replacement. Its conditional sample-mean law has values 0,1/2,1 with probabilities 1/4,1/2,1/4. In four independent resamples, suppose the ordered index pairs are (1,1),(1,2),(2,2),(2,2), giving means 0,1/2,1,1. This finite realization lets us compare the data-sampling error, the exact conditional bootstrap target, and finite-B simulation error. It makes no bootstrap-validity claim.

Facts & Assumptions

Given: The stated IID two-point observation law, observed sample (0,1), finite unit space and map T2, bootstrap procedure, and four realized resamples. Prove the conditional bootstrap law and error separation.

[F1]

The error ledger uses a declared scalar target and explicit successive reference values, with each stage error equal to the difference of adjacent references. (Successive statistical and computational error targets)

[F2]

The total realized error along the finite reference chain is the sum of its signed stage errors. (Pathwise error decomposition along a reference chain)

[F3]

Identical distribution and IID families: IID means a common law and mutual independence.

[F4]

Size-n bootstrap from the empirical law: for a fixed tuple of length n≥1, draw n indices independently and uniformly from [n], with replacement, and apply the declared measurable statistic to the resampled tuple.

[F5]

Empirical-law bootstrap draws are conditionally IID: conditional on the fixed observed tuple, the resampled coordinates are IID with common law equal to its empirical measure.

[F6]

For B≥2 IID finite-variance replicate outputs, [B(B−1)]−1∑b(Zb−Z‾B)2 is unbiased for the variance of their average. (Unbiased variance estimate from independent computation replicates)

[A1]

The observations are IID with the stated two-point law; the target is their mean θ=3/4.

[A2]

The observed ordered tuple is (0,1) in the finite measurable unit space ({0,1},2{0,1}); each entry is one resampling unit and T2(x1,x2)=(x1+x2)/2 is measurable on the full power-set sample space.

[A3]

Four independent repetitions of the stated bootstrap algorithm produce the listed ordered index pairs.

Verification

1.1givenA1A2algebra

By [A1], the observed event (X1,X2)=(0,1) has probability (1/4)(3/4)=3/16>0. By [A2], its statistic is T2(0,1)=1/2, so the realized data-sampling error is 1/2−3/4=−1/4.

1.2givenA2F4F5constructalgebra

By [F4] and [F5], the four equally likely ordered pairs of bootstrap indices have sample means 0,1/2,1/2,1, respectively. Therefore the conditional bootstrap mean is 1/2 and its conditional variance is (1/4)(1/2)2+(1/2)02+(1/4)(1/2)2=1/8.

2.1givenA3F4step 1.2algebra

The four stated means average to (0+1/2+1+1)/4=5/8, so the finite-B simulation error relative to the exact conditional bootstrap mean is 5/8−1/2=1/8. Each ordered index pair has conditional probability 1/4; independence across the four repetitions makes this listed realization have probability (1/4)4=1/256>0.

2.2F3F6step 1.2A3algebra

Under the conditional resampling law, the four replicate means are IID with variance 1/8 by step 1.2 and [A3]; [F6] gives variance 1/(4⋅8)=1/32 for their average. This is computation-randomization variance conditional on the data, distinct from the data-sampling error in step 1.1.

3.1F1F2step 1.1step 1.2step 2.1algebra∎

The chain 3/4,1/2,1/2,5/8,5/8 has successive errors −1/4,0,1/8,0. Their sum is −1/8=5/8−3/4, so the realized total error is exactly the separate sampling and finite-B errors in steps 1.1 and 2.1.

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Exact-orbit and finite-B permutation p-value resolution

Example

Let Ω=G={e,a,a2} with multiplication a3=e, so aiaj=ai+j (mod 3). The group acts on itself by left multiplication, the null law is uniform on G, and T(e)=3, T(a)=2, T(a2)=1. For the observed point e, the full-orbit inclusive upper-tail rank is one out of three. Separately, define a finite-B reported value for this example by adjoining the observed identity and using p+:=1+∑b=1B1{T(gbe)≥T(e)}B+1. Take B=4 and the realized sampled transformations (g1,g2,g3,g4)=(a2,a,e,a2). This example calculates the two resolutions and their realized difference; it does not establish validity, unbiasedness, or a general error bound for the finite-B formula.

Facts & Assumptions

Given: The group, action, statistic, invariant uniform null law, and finite list of sampled transformations in the Example. Prove the exact-orbit and realized finite-B calculations.

[F1]

A finite-group randomization setup consists of a finite probability space, finite group action, statistic, and null laws invariant under that action. (Finite-group randomization setups)

[F2]

The inclusive orbit upper-tail rank counts group elements with T(gx)≥T(x) and divides by ∣G∣. (Orbit upper-tail rank p-values)

[F3]

Under an invariant null, the full-orbit inclusive rank p-value is superuniform. (Finite-group orbit-rank validity)

[F4]

For a finite scalar reference chain, the total realized error is the sum of its signed adjacent-stage errors. (Pathwise error decomposition along a reference chain)

[A1]

The operation aiaj=ai+j (mod 3) is associative with identity e=a0 and inverses a3−i; for each g∈G, left multiplication x↦gx is a bijection of G.

[A2]

The finite-B formula is the explicitly defined calculation in the Example; no validity property is assumed for it.

Verification

1.1givenA1F1construct

Each map x↦gx permutes G, so the uniform law is invariant and the data form the finite-group setup of [F1].

1.2givenA2algebra

The sampled statistic values are 1,2,3,1, so the inclusive count among the four sampled transforms is one and p+=(1+1)/5=2/5.

2.1givenF2F3step 1.1algebra

Exactly one of e,a,a2 has statistic at least T(e)=3, so [F2] gives the exact-orbit rank p0(e)=1/3; [F3] supplies its superuniformity under this invariant null.

3.1A2F4step 1.2step 2.1algebra∎

The finite-B output lies on the grid {1/5,2/5,3/5,4/5,1} because its count is an integer from zero through four. Set m=1, r0=p0(e), and r1=r~=p+; then the one-stage error is 2/5−1/3=1/15 by [F4].

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Euler discretization bias and path-to-path variance

Example

Let X0 take values in {1,2} with probability 1/2 each, and consider the random-initial-value ODE x′(t)=x(t) on 0≤t≤1. Its exact endpoint is X1=eX0, and the target is θ=E[X12]. Define the one-step explicit Euler rule at h=1 by X1h=X0+hX0=2X0. For an integer M≥2, independently simulate initial values X0(1),…,X0(M) with this law, set Yr:=(2X0(r))2, and return Y‾M:=M−1∑r=1MYr.

Facts & Assumptions

Given: The ODE, initial-value law, one-step Euler rule, and independent path simulations specified in the Example. Compute the discretized-target bias and the path-simulation variance.

[F1]

A scalar error ledger uses successive same-scale references and defines each stage error as the adjacent difference. (Successive statistical and computational error targets)

[F2]

The total error along a finite scalar reference chain is the sum of its signed stage errors. (Pathwise error decomposition along a reference chain)

[F3]

For M≥2 IID finite-variance replicate outputs with variance τ2, the variance of their average is τ2/M. (Unbiased variance estimate from independent computation replicates)

[F4]

For real a<b, a real function continuous on [a,b] and differentiable on (a,b) satisfies f(b)−f(a)=f′(c)(b−a) for some c∈(a,b). (The mean value theorem, as the case g(x)=x of Cauchy's: for f continuous on [a,b] with a<b and differentiable on (a,b) there is c∈(a,b) with f(b)−f(a)=f′(c)(b−a))

[F5]

e=exp⁡(1) for exp⁡(x)=∑k=0∞xk/k!. (The real exponential function and the number e by a power series)

[F6]

exp⁡ is differentiable with (exp⁡)′=exp⁡. (The exponential function is smooth and (exp⁡)′=exp⁡)

[F8]

IID means common distribution and mutual independence. (Identical distribution and IID families)

[F9]

A function differentiable at a limit point is continuous there. (A function differentiable at c is continuous at c)

[F10]

Applying measurable maps coordinatewise to independent random variables preserves independence. (Measurable coordinatewise functions preserve independence)

[F11]

Products of measurable real-valued functions and their real scalar multiples are measurable. (Arithmetic and lattice operations preserve measurability whenever they are defined)

[A1]

X0 equals 1 and 2 with probabilities 1/2 each; an ODE solution x is defined and differentiable on an open interval containing [0,1], satisfies x′=x, and has initial value x(0)=X0.

[A2]

The discretization is the explicitly defined one-step rule X1h=X0+hX0 at h=1; the M simulated initial values are IID copies of X0.

Verification

1.1givenA1F4F5F6F7F9algebra

By [F5] and [F6], x∗(t)=X0exp⁡(t) solves the stated initial-value problem. For any solution x, put f(t)=exp⁡(−t) and u(t)=f(t)x(t). By [F6] and [F7], f′(t)=−f(t). For h≠0 with t+h∈[0,1], u(t+h)−u(t)h=f(t+h)x(t+h)−x(t)h+x(t)f(t+h)−f(t)h. As h→0, continuity of f and differentiability of x,f make the right side tend to f(t)x′(t)+x(t)f′(t)=0 because x′=x. Thus u′=0 on (0,1). The solution x is continuous at 0,1 by [F9], so u is continuous on [0,1]; [F4] gives u(1)=u(0)=X0. The same derivative calculation applied to the continuous product v(t)=f(t)exp⁡(t) gives v′=0 on (0,1); [F4] then yields exp⁡(−1)e=v(1)=v(0)=1, since exp⁡(0)=1 by [F5]. Multiplying u(1)=X0 by e therefore gives x(1)=eX0. By [F5] and E[X02]=(1+4)/2=5/2, the exact target is θ=e2E[X02]=(5/2)e2.

1.2givenA1A2F5algebra

The Euler update gives X1h=2X0, so the exact-arithmetic discretized target is θh=E[4X02]=10. Since e=∑k=0∞1/k!>2, its bias is 10−(5/2)e2<0.

1.3givenA1A2F8algebra

By [F8] and [A2], a single simulated squared Euler endpoint is 4X02, equal to 4 or 16 with equal probability. Its mean is 10 and its variance is ((4−10)2+(16−10)2)/2=36.

2.1F3F8F10F11step 1.2step 1.3A2algebra

By [F11], z↦4z2 is measurable; by [F8] and [A2] the initial values are IID, so [F10] makes the outputs independent. Their common two-point law is computed in step 1.3, hence they are IID. Then [F3] gives variance 36/M for their average, while the discretization bias in step 1.2 remains 10−(5/2)e2 for every M.

3.1F1F2step 1.1step 1.2step 2.1algebra∎

For a realized path average Y‾M, the ledger chain θ,θh,Y‾M gives Y‾M−θ=(θh−θ)+(Y‾M−θh) by [F1] and [F2], separating discretization bias from path-simulation error.

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

Replicate variance for randomized quadrature

Example

Fix the observed data and one randomized quadrature rule for an exact integral I. Suppose its independent full-set randomizations produce IID estimates with mean I and finite variance, and the three realized outputs are Z1=2/5, Z2=1/2, and Z3=7/10. Compute the replicate mean and the variance estimate for that mean. This calculation does not assert a Student interval or estimate variability across new data samples.

Facts & Assumptions

Given: Fixed data and exact integral target I, with IID randomized-rule outputs of mean I and finite variance. The observed outputs are 2/5,1/2,7/10.

[F1]

For R≥2 IID replicate outputs, [R(R−1)]−1∑r(Zr−Z‾R)2 is unbiased for the variance of their average. (Unbiased variance estimate from independent computation replicates)

[F2]

A computation stage is recorded as the difference between its exact reference target and its realized output, with all values on the same scalar scale. (Successive statistical and computational error targets)

Verification

1.1givenalgebra

The replicate mean is Z‾3=(2/5+1/2+7/10)/3=8/15.

2.1givenstep 1.1algebra

The squared deviations from 8/15 sum to (2/5−8/15)2+(1/2−8/15)2+(7/10−8/15)2=7/150.

3.1F1F2step 1.1step 2.1algebra∎

The replicate variance estimate for the mean is (3⋅2)−1(7/150)=7/900 by [F1]. With the data fixed, this estimate concerns only the independent rule-randomization law, not data-sampling variation. The realized computation error in the ledger is Z‾3−I=8/15−I.

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Dot-product cancellation under normal-range rounding

Example

In radix-two binary64 arithmetic with precision 53 and round-to-nearest, ties-to-even, compute the left-to-right dot product of x=(1016,1,−9999999999999998) and y=(1,1,1). The unit roundoff is u=2−53. All three products and all intermediate sums are in the normal range. Define the dot-product condition ratio for this nonzero exact result by κ:=∑j=13∣xjyj∣∣x⋅y∣. This example checks an absolute roundoff bound and illustrates why a bound small relative to the sum of term magnitudes need not mean small relative error in the result when cancellation makes κ large.

Facts & Assumptions

Given: The vectors, binary64 round-to-nearest ties-to-even arithmetic, unit roundoff u=2−53, and the left-to-right evaluation order. Check the exact result, computed result, normal-range hypotheses, and gamma bound.

[F1]

If n≥1, nu<1, and every exact product xjyj, rounded product pj:=fl⁡(xjyj), and intermediate sum lies in the normal range, the computed left-to-right dot product satisfies ∣fl⁡(x⋅y)−x⋅y∣≤γn∑j∣xjyj∣, where γn=nu/(1−nu). (The computed floating-point dot product obeys componentwise and normwise error bounds)

[F2]

The floating-point model uses relative error at most u only on its stated normal-range operation domain; exact zero is handled separately. (The standard relative floating-point model and unit roundoff)

[F3]

The final numerical error is efp=r~−rm for the error ledger. (Successive statistical and computational error targets)

[F4]

The total error along a finite scalar reference chain is the sum of its signed adjacent-stage errors. (Pathwise error decomposition along a reference chain)

[A1]

In the binade [253,254), a radix-two format with precision 53 has spacing 2. Since 253<1016<254, 1016 and 1016+2 are adjacent representable values. The lower significand is 1016/2=5000000000000000, an even integer, whereas the upper significand is odd; therefore the exact midpoint 1016+1 rounds to 1016 under ties-to-even.

Verification

1.1givenalgebra

The exact dot product is 1016+1−(1016−2)=3, and each product by a component of y is exact.

1.2givenA1F2algebra

By [A1], the first rounded partial sum is fl⁡(1016+1)=1016; the next exact sum is 1016−(1016−2)=2, which is represented exactly. The exact intermediate sums are 1016+1 and 2, and the rounded partial sums are 1016 and 2. Together with the exact products of magnitudes 1016,1,1016−2, all lie between the binary64 smallest and largest normal magnitudes stated in [F2].

2.1givenF1step 1.1step 1.2algebra

Here 3u<1, γ3=3/(253−3), and ∑j∣xjyj∣=19999999999999999. The bound is 3(19999999999999999)/(253−3)>1, since its numerator exceeds its denominator; this is consistent with the computed error ∣2−3∣=1.

3.1F3F4step 1.1step 1.2algebra∎

The condition ratio is 19999999999999999/3, while the relative forward error is 1/3. In the ledger take m=0, θ=3, and r~=2; [F3] gives the sole error efp=−1, and [F4] identifies it with the total error. No sampling or simulation stage occurs.

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Perfectly correlated error components

Example

Let U equal −1 or 1 with probability 1/2 each, and set V=U. Both component errors are centered and have variance one, but they are perfectly correlated.

Facts & Assumptions

Given: A two-point probability space with U∈{−1,1} equally likely and V=U, interpreted as two successive signed stage errors. Compute the mean-squared sum and compare it with the variance-only sum.

[F1]

The mean-squared sum of square-integrable components includes twice each pairwise covariance. (MSE of a sum of error components includes covariance terms)

[F2]

An error ledger records each stage error as the difference of adjacent scalar references. (Successive statistical and computational error targets)

Verification

1.1givenF2construct

Set the first reference to 0, the next to U, and the returned value to U+V; by [F2] the two signed stage errors are exactly U and V.

1.2givenalgebra

Since U equals −1 or 1 with equal probability and V=U, EU=EV=0 and Var⁡(U)=Var⁡(V)=1.

2.1givenstep 1.2algebra

Because V=U, Cov⁡(U,V)=E[UV]=1 and E[(U+V)2]=4.

3.1F1step 1.2step 2.1algebra∎

[F1] gives the same value, 1+1+2(1)=4, whereas adding only the two component variances gives 2; the omitted covariance term changes the result.

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