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A skewed finite-replicate law defeats a nominal Student interval

Example

Assume AC. Use the nested uniform scramble of the base- $2$ net ${0, 1 / 2}$ and $f (x) = 1_{[0, 1 / 100)} (x)$ . For two independent full-set replicates, calculate the exact coverage of the usual two-sided $95 %$ Student interval based on the sample variance, namely ${\overline{Z}}_{2} \pm c \sqrt{{\hat{V}}_{2}}$ with $c$ the $97.5$ th percentile of $t_{1}$ . This checks the finite- $R$ coverage caveat.

Facts & Assumptions

Given: AC, the net and scramble, the fixed target $I = \int_{0}^{1} f (x) d x$ , and two independent scrambles.

[F1]

$(λ, t, m, d)$ -nets in base $b$ : The input net has one point in each half interval.

[F2]

Nested uniform scrambling of a finite point set: Conditional on the first permutation, the two point tails are independent uniform tails in their assigned halves.

[F3]

Independent RQMC replicates estimate randomization variance: For $R = 2$ , ${\hat{V}}_{2} = \frac{1}{2} \sum_{r = 1}^{2} (Z_{r} - {\overline{Z}}_{2})^{2}$ estimates the replicate-mean variance.

[F4]

Student t and Fisher F laws: $t_{1}$ is the law of $G / \sqrt{Q_{1}}$ , with independent standard normal $G$ and $Q_{1} \sim χ_{1}^{2}$ .

[F5]

Chi-square laws, including zero degrees: $Q_{1}$ is the square of a standard normal, so $G^{2}$ and the independent $Q_{1}$ have the same $χ_{1}^{2}$ law.

[F6]

Positive-degree chi-square laws have a positive density and regular CDF: A positive-degree chi-square variable is positive almost surely and has a continuous density.

[F7]

Student t and Fisher F CDFs are continuous and strictly increasing on their supports: Under AC, $t_{1}$ has a continuous, symmetric, strictly increasing CDF, so its $0.975$ quantile is unique.

[A1]

The Axiom of Choice: AC is used by the stated Student t law and its CDF result [F4, F7].

[F8]

Expectation of a nonnegative or integrable random variable: Expectation is integration against the probability law.

Verification

1.1F1F2F8givenalgebra

The point in $[0, 1 / 2)$ is uniform there by [F2], while the point in $[1 / 2, 1)$ never enters $[0, 1 / 100)$ . Thus one rule estimate is $Z = \frac{1}{2} B$ , where $B$ is Bernoulli with success probability $(1 / 100) / (1 / 2) = 1 / 50$ . Consequently [F8] gives $E Z = 1 / 100 = I$ , and $Z$ has a skewed two-point law.

2.1F3step 1.1algebra

For two independent estimates, if both $B$ values agree then ${\hat{V}}_{2} = 0$ by [F3], so the interval is the singleton ${0}$ or ${1 / 2}$ and misses $I = 1 / 100$ . This event has probability $(49 / 50)^{2} + (1 / 50)^{2} = 2402 / 2500$ .

2.2A1F3F4F5F6F7step 1.1algebra

If the two $B$ values differ, then ${\overline{Z}}_{2} = 1 / 4$ and ${\hat{V}}_{2} = 1 / 16$ , so the interval half-width is $c / 4$ with $c$ as in the Example. By [F4]–[F6], $T_{1}$ can be represented as $G / \sqrt{Q_{1}}$ with $G^{2}$ and $Q_{1}$ independent and identically distributed; their continuous law gives $P (G^{2} = Q_{1}) = 0$ . Exchangeability then gives $P (G^{2} < Q_{1}) = P (Q_{1} < G^{2}) = 1 / 2$ , so $P (∣ T_{1} ∣ < 1) = 1 / 2$ . Symmetry and continuity of the CDF [F7] imply $F_{t_{1}} (1) = 3 / 4$ , hence its $0.975$ quantile satisfies $c > 1$ . Therefore the interval half-width $c / 4 > 1 / 4 > ∣ 1 / 4 - 1 / 100 ∣ = 6 / 25$ , and it contains $I$ . This event has probability $2 (49 / 50) (1 / 50) = 98 / 2500$ .

3.1step 2.1step 2.2algebra∎

The resulting two-sided interval coverage is $98 / 2500 = 0.0392$ , far below $0.95$ . The replicate variance formula remains unbiased; this finite- $R$ example shows that unbiased variance estimation alone does not guarantee accurate Student interval coverage.

Depends on

Used by

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Sources

Art B. Owen, Monte Carlo Book: The Quasi-Monte Carlo Parts, Chapter 17 §17.1 and §17.4 (standard reference, not scraped)