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✓ 6 results · all verified · 2 also independently AI-judged

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Control variates, antithetics, and stratification — Examples

1 · Prerequisites

2 · Summary

These finite and elementary examples calculate when a known-mean control removes variance, when a control has no effect, how a monotone antithetic pair cancels exactly, and how proportional stratification compares with an ordinary sample mean. A two-observation regression example also verifies that fitting a control coefficient on the production sample need not preserve unbiasedness; the pilot example checks conditional unbiasedness when the coefficient is independent of production data.

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 →

A perfect-correlation control removes all variance

Example

Let $X_{i}$ be IID uniform on $[0, 1]$ , take $f (x) = h (x) = x$ , and use the known control mean $θ = 1 / 2$ . The fixed coefficient $β = 1$ gives an exact estimate with no Monte Carlo variance.

Facts & Assumptions

Given: The uniform law on $[0, 1]$ , $f = h = id$ , and the known mean $θ = 1 / 2$ .

[F1]

Unbiasedness and optimal coefficients for fixed controls: The scalar minimizing coefficient is $Cov (f (X), h (X)) / Var (h (X))$ when $Var (h (X)) > 0$ .

[F2]

Control-variate estimators: ${\hat{μ}}_{β} = n^{- 1} \sum_{i} [f (X_{i}) - β (h (X_{i}) - θ)]$ .

Verification

Given: The specified $f, h, θ$ and IID $X_{i} \sim Unif [0, 1]$ .

1.1F1algebra

Direct integration gives $E X = 1 / 2$ and $Var (X) = \int_{0}^{1} (x - 1 / 2)^{2} d x = 1 / 12 > 0$ , so [F1] gives $β^{*} = 1$ .

2.1F2step 1.1given∎

For every sample value, $X_{i} - (X_{i} - 1 / 2) = 1 / 2$ . By [F2], ${\hat{μ}}_{1} = 1 / 2$ identically, hence its variance is zero and its expectation is the target $1 / 2$ .

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

An uncorrelated control has no variance benefit

Example

Let $A$ and $B$ be independent fair signs, let $X = (A, B)$ , and set $f (A, B) = A$ , $h (A, B) = B$ , and $θ = 0$ . The optimal fixed control coefficient is zero, so this control does not reduce variance.

Facts & Assumptions

Given: The four equiprobable outcomes $(A, B) \in {- 1, 1}^{2}$ .

[F1]

Unbiasedness and optimal coefficients for fixed controls: The scalar optimum is covariance divided by control variance when that variance is positive.

[F2]

Control-variate estimators: The fixed-coefficient estimate subtracts $β (\overset{ˉ}{h} - θ)$ from $\overset{ˉ}{f}$ .

Verification

Given: $f = A$ , $h = B$ , $E A = E B = 0$ , and the four outcomes each have probability $1 / 4$ .

1.1givenalgebra

Summing the four table values gives $E f = E h = 0$ , $Var (h) = 1$ , $Var (f) = 1$ , and $Cov (f, h) = E [A B] = 0$ .

2.1F1F2step 1.1∎

By [F1], $β^{*} = 0$ and the minimum variance of one corrected draw equals $Var (f) = 1$ . Thus [F2] reduces to the ordinary sample mean, with variance $1 / n$ ; this control gives no reduction.

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

A monotone uniform integrand has a zero-variance antithetic pair

Example

Take $d = 1$ , $T (u) = u$ , $f (x) = x$ , and $U \sim Unif [0, 1]$ . The antithetic pair average is constant, while the ordinary mean of two independent draws has positive variance.

Facts & Assumptions

Given: $U$ uniform on $[0, 1]$ , $T (u) = u$ , and $f (x) = x$ .

[F1]

Antithetic randomized sampling: A pair averages $f (T (U))$ and $f (T (1 - U))$ .

[F2]

Antithetic variance and a monotone sufficient condition: The antithetic variance is $(σ^{2} + Cov (Y, \tilde{Y})) / (2 n)$ ; it is no larger than the IID comparator exactly when the covariance is nonpositive, and is strictly smaller exactly when the covariance is negative.

Verification

Given: The one-dimensional uniform input and identity sampler/integrand.

1.1givenalgebra

Direct integration gives $E U = E (1 - U) = 1 / 2$ , $Var (U) = 1 / 12$ , and $E [U (1 - U)] = 1 / 6$ , so $Cov (U, 1 - U) = 1 / 6 - 1 / 4 = - 1 / 12$ .

2.1F1F2step 1.1algebra∎

For each pair, $(U + (1 - U)) / 2 = 1 / 2$ , so [F1] gives a constant estimator of variance zero. By [F2], the comparator variance for two independent uniform draws is $(1 / 12) / 2 = 1 / 24$ , hence the reduction is strict.

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

Proportional stratification reduces variance for a uniform mean

Example

For $P = Unif [0, 1]$ and $f (x) = x$ , split into the two half-intervals and draw one point independently from each. This is proportional allocation with total sample size two and gives variance $1 / 96$ , compared with $1 / 24$ for two ordinary IID draws.

Facts & Assumptions

Given: $A_{1} = [0, 1 / 2)$ , $A_{2} = [1 / 2, 1]$ , $p_{1} = p_{2} = 1 / 2$ , and $n_{1} = n_{2} = 1$ .

[F1]

Stratified Monte Carlo estimator: ${\hat{μ}}_{s t r a t} = p_{1} {\overset{ˉ}{f}}_{1} + p_{2} {\overset{ˉ}{f}}_{2}$ .

[F2]

Stratified estimator unbiasedness, variance, and proportional allocation: The fixed-stratum variance is $\sum_{h} p_{h}^{2} σ_{h}^{2} / n_{h}$ and proportional allocation compares it with the ordinary IID mean of the same total size.

Verification

Given: Independent uniform draws from each half-interval and one ordinary IID comparator sample of size two.

1.1F1F2givenalgebra

Each conditional uniform law has mean $1 / 4$ or $3 / 4$ and variance $(1 / 2)^{2} / 12 = 1 / 48$ . Substitution into the estimator definition [F1] and variance formula [F2] gives $Var ({\hat{μ}}_{s t r a t}) = 2 (1 / 4) (1 / 48) = 1 / 96$ .

2.1givenstep 1.1algebra∎

The ordinary sample mean has variance $Var (U) / 2 = (1 / 12) / 2 = 1 / 24$ . Since $1 / 96 < 1 / 24$ , this specified proportional allocation reduces variance by a factor of four.

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

A same-sample fitted coefficient can bias the estimator

Example

Let $X_{i}$ be IID uniform on ${- 1, 0, 1}$ , let $f (x) = x^{2}$ , $h (x) = x$ , and use $θ = E h (X) = 0$ . For $n = 2$ , fit the scalar coefficient on the same two observations by $\hat{β} = \overset{}{Cov} (h, f) / \overset{}{Var} (h)$ when the sample variance is positive, and set it to $0$ when the sample variance is zero. The resulting control estimator is biased.

Facts & Assumptions

Given: The three-point uniform law, sample size two, the specified $f, h, θ$ , and the zero-denominator convention.

[F1]

Same-sample coefficient remainder is of order $O_{P} (n^{- 1})$ : The same-sample coefficient is the inverse sample control covariance times the sample cross-covariance when defined, with a zero fallback when singular.

[F2]

Same-sample coefficient remainder is of order $O_{P} (n^{- 1})$ defines the plug-in estimator pointwise; for a scalar fitted coefficient it is $\overset{ˉ}{f} - \hat{β} (\overset{ˉ}{h} - θ)$ .

Verification

Given: Each of the nine ordered pairs $(X_{1}, X_{2}) \in {- 1, 0, 1}^{2}$ has probability $1 / 9$ .

1.1F1F2givenalgebra

On the three equal pairs, the fallback coefficient is zero and the estimates are $1, 0, 1$ . On the two ordered pairs $(- 1, 1)$ and $(1, - 1)$ , the fitted slope is zero and each estimate is $1$ . On each of the four ordered adjacent pairs $(- 1, 0), (0, - 1), (0, 1), (1, 0)$ , the fitted slope is respectively $- 1, - 1, 1, 1$ , and direct substitution into [F2] gives estimate $0$ .

2.1givenstep 1.1algebra∎

Exactly four of the nine equally likely pairs yield estimate $1$ and the other five yield $0$ , so the expected estimate is $4 / 9$ . The target is $E f (X) = \frac{1}{3} (1 + 0 + 1) = 2 / 3$ , hence this same-sample fitted-coefficient estimator is biased.

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

A random pilot coefficient remains conditionally unbiased

Example

Let the production draws be IID uniform on $[0, 1]$ , take $f (x) = x^{2}$ , $h (x) = x$ , and $θ = 1 / 2$ . From an independent pilot draw $V \sim Unif [0, 1]$ , set the random coefficient $\hat{β} = 2 V$ . The control estimator remains conditionally unbiased given the pilot.

Facts & Assumptions

Given: The pilot variable $V$ is independent of all production draws and $\hat{β} = 2 V$ has finite first moment.

[F1]

An independent pilot coefficient preserves conditional unbiasedness: An integrable pilot-measurable coefficient independent of the production sample gives conditional expectation $E_{P} f$ .

[F2]

Control-variate estimators: For a fixed scalar coefficient $b$ , the estimator is $\overset{ˉ}{f} - b (\overset{ˉ}{h} - θ)$ .

Verification

Given: The stated production and pilot laws.

1.1givenalgebra

Production integration gives $E_{P} f = \int_{0}^{1} x^{2} d x = 1 / 3$ and $E_{P} h = \int_{0}^{1} x d x = 1 / 2 = θ$ ; also $E ∣ \hat{β} ∣ = E [2 V] = 1 < \infty$ .

2.1F1F2givenstep 1.1algebra∎

Set $P = σ (V)$ . The independent pilot makes $P$ independent of the sigma-algebra of the entire production sample, and $\hat{β} = 2 V$ is $P$ -measurable and integrable by step 1.1. The functions $f, h$ are bounded and hence integrable. The random-coefficient estimator defined in [F1] is $\overset{ˉ}{f} - 2 V (\overset{ˉ}{h} - 1 / 2)$ , extending the fixed-coefficient expression in [F2]. Therefore [F1] gives $E [{\hat{μ}}_{\hat{β}} ∣ P] = 1 / 3$ almost surely.