ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

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Two-stage queue overflow and a redundant split at equal work

Example

Consider a two-period single-server queue with no service during these two periods and independent Bernoulli arrivals $B_{1}, B_{2}$ , each with probability $1 / 2$ . Capacity is one, so overflow occurs when both arrivals occur. Let $A_{1} = {B_{1} + B_{2} \geq 1}$ and $A_{2} = {B_{1} + B_{2} = 2}$ . Then $P (A_{1}) = 3 / 4$ , $P (A_{2} ∣ A_{1}) = 1 / 3$ , and $P (A_{2}) = 1 / 4$ .

Under unit cost per conditional draw, use two draws at each of the two levels for work $4$ . Compare with an augmented chain having two repeated $A_{1}$ levels before $A_{2}$ , and one draw at each of four levels, also work $4$ . The added levels have conditional probability one but the estimator has larger relative variance.

Facts & Assumptions

Given: The two independent Bernoulli arrivals, nested events $A_{1} \supseteq A_{2}$ , unit draw cost, and the two stated independent-splitting designs.

[F1]

Independent fixed-level splitting: The estimator is a product of independent sample means of level indicators, with $N$ conditional draws per level.

[F2]

Independent splitting estimator moments and work: Its relative variance is $\prod_{ℓ} (1 + (1 - p_{ℓ}) / (N p_{ℓ})) - 1$ , and unit-cost work for $L$ levels is $N L$ .

Verification

technique · direct

1.1givenalgebra

The four arrival outcomes are equiprobable. Exactly three have at least one arrival and exactly one has two arrivals; hence $P (A_{1}) = 3 / 4$ , $P (A_{2}) = 1 / 4$ , and $p_{2} = (1 / 4) / (3 / 4) = 1 / 3$ .

2.1F1F2step 1.1algebra

The two designs use the product estimator specified in [F1]. With two levels and $N = 2$ , [F2] gives relative variance $(1 + \frac{1 / 4}{2 (3 / 4)}) (1 + \frac{2 / 3}{2 (1 / 3)}) - 1 = \frac{7}{6} \cdot 2 - 1 = \frac{4}{3} .$ The work is $2 \cdot 2 = 4$ .

2.2F2step 1.1algebra

In the augmented chain, the conditional probabilities are $3 / 4, 1, 1, 1 / 3$ , and $N = 1, L = 4$ . The two unit-probability factors in [F2] equal one, while the other factors are $4 / 3$ and $3$ ; the relative variance is $4 - 1 = 3$ , with work $1 \cdot 4 = 4$ .

3.1step 2.1step 2.2algebra∎

Since $3 > 4 / 3$ , equal work does not make the redundant-level design equally precise. Its unit-probability stages add cost without reducing the one-draw randomness left in the two nontrivial levels.

Depends on

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Sources

Gino Biondini, An Introduction to Rare Event Simulation and Importance Sampling, Handbook of Statistics 33, Chapter 2, §8 (standard reference, not scraped)
Adam M. Johansen, Pierre Del Moral, and Arnaud Doucet, Sequential Monte Carlo Samplers for Rare Events, §§2–3 (standard reference, not scraped)