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IID Monte Carlo integration and error — Examples

1 · Prerequisites

2 · Summary

Examples

Gaussian integrand: exact Monte Carlo moments and asymptotic error checks the finite-sample mean and variance formulas and the asymptotic normal error for a Gaussian integrand. Finite mean does not give a finite-variance Monte Carlo error exhibits a finite mean with an infinite second moment: the sample mean remains consistent, while there is no finite population standard error σ/n and the finite-variance CLT does not apply. For a finite real sample of size n≥2, the empirical sample standard deviation can still be computed, but it has no finite-variance guarantee in this example.

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 →

Gaussian integrand: exact Monte Carlo moments and asymptotic error

Example

Assume AC and let X have law N(μ,σ2) with σ>0. Use the identity integrand g(x)=x, whose target is E[g(X)]=μ, and take IID outputs (Yi)i≥1 with this law. Then E[Yˉn]=μ and Var⁡(Yˉn)=σ2/n; the estimated standard error is relatively consistent, and n(Yˉn−μ)⇒N(0,σ2).

Facts & Assumptions

Given: AC, σ>0, and a real IID sequence (Yi)i≥1 with common law N(μ,σ2).

[A1]

The Axiom of Choice: AC is the axiom that every family of nonempty sets has a choice function.

[F1]

Location-scale normal laws have their stated mean and variance: Under AC, a real random variable with law N(m,σ2) for σ≥0 has mean m and variance σ2.

[F2]

Unbiasedness and exact variance of the IID Monte Carlo mean: For integrable IID outputs, E[Yˉn]=μ; with a finite second moment, Var⁡(Yˉn)=σ2/n.

[F3]

Consistency of the IID Monte Carlo sample variance: With finite second moment, sn2→σ2 almost surely, and for σ2>0, se^n/(σ/n)→1 almost surely.

[F4]

Lindeberg–Lévy limit for the IID Monte Carlo mean: Under AC and 0<σ2<∞, n(Yˉn−μ)⇒N(0,σ2).

[F5]

IID Monte Carlo sample mean, variance, and standard error: se^n=sn/n for n≥2 under the finite-second-moment contract.

Verification

1.1A1F1F2given

By [F1], each output has mean μ and variance σ2, so its first absolute moment and second moment are finite. Applying [F2] gives E[Yˉn]=μ and Var⁡(Yˉn)=σ2/n.

1.2A1F1F3F5given

The finite-second-moment hypotheses of [F3] hold by [A1] and [F1]. Thus sn2→σ2 almost surely and, using the plug-in definition [F5], se^n/(σ/n)→1.

2.1A1F1F4given∎

The sequence has positive finite variance and AC is assumed, so [F4] applies and gives n(Yˉn−μ)⇒N(0,σ2).

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

Finite mean does not give a finite-variance Monte Carlo error

Example

Assume the Axiom of Countable Choice. Let F(x)=0 for x<1 and F(x)=1−x−3/2 for x≥1. The CDF-to-law theorem gives a probability law with CDF F. Let Y be the identity random variable under this law, and let (Yi)i≥1 be any real IID sequence with this common law. Then E[Y1]=3 but E[Y12]=+∞. Thus the sample mean is strongly and weakly consistent, while no finite population standard error σ/n is available. For a finite sample of real outputs the empirical quantity sn/n can still be computed, but this example supplies no finite-variance interpretation for it.

Facts & Assumptions

Given: Countable Choice, the displayed function F, and a real IID sequence (Yi)i≥1 with the probability law constructed from F.

[A1]

The Axiom of Countable Choice (ACω): Countable Choice says every countable family of nonempty sets has a choice function.

[F1]

Continuity and derivatives of positive-base real powers: For rational α, x↦xα is continuous and differentiable on (0,∞) with derivative αxα−1.

[F2]

Monotonicity of r↦ar and of a↦ar: For positive rational r, a↦ar is strictly increasing on positive reals.

[F3]

Probability laws correspond to distribution functions: Under Countable Choice, every nondecreasing right-continuous F with limits 0 and 1 at the two infinities is the CDF of a unique Borel probability measure on R.

[F4]

Random elements and real random variables: A real random variable is a measurable map to R; the identity map on a Borel probability space is measurable.

[F5]

Expectation of a nonnegative or integrable random variable: A nonnegative random variable may have extended expectation E[Y]=∫Y dP.

[F6]

Layer-cake formulas for random variables: For nonnegative Y, E[Y]=∫0∞P(Y>t) dt, with infinity allowed.

[F7]

For 0 < p < infinity, the layer-cake formula computes the integral of |f|^p from the distribution function: For measurable Y and p>0, E[∣Y∣p]=p∫0∞tp−1P(∣Y∣>t) dt, with infinity allowed.

[F8]

Truncated integrals of rational powers: For rational p≠1 and 0<A<B, ∫ABt−p dt=(B1−p−A1−p)/(1−p).

[F9]

A nonnegative improper Riemann integral on a half-line agrees with the Lebesgue integral: Under Countable Choice, if a nonnegative function is Riemann integrable on compact intervals and its improper Riemann integral is finite, that value equals its Lebesgue integral on the half-line.

[F10]

A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral: Under Countable Choice, a bounded Riemann-integrable function on a compact interval has the same Lebesgue and Riemann integrals.

[F11]

Continuous functions on Euclidean spaces are Borel measurable: Under Countable Choice, continuous maps between Euclidean spaces are Borel measurable.

[F12]

Arithmetic and lattice operations preserve measurability whenever they are defined: Products of measurable real functions are measurable.

[F13]

Monotone convergence for the integral: Integrals of an increasing sequence of nonnegative measurable functions increase to the integral of their pointwise limit.

[F14]

Strong and weak consistency of the IID Monte Carlo mean: For a real IID sequence with finite first absolute moment, its Monte Carlo sample mean converges in probability and almost surely to the common mean.

[F15]

IID Monte Carlo sample mean, variance, and standard error: The empirical sn/n is a statistic for real samples; its interpretation as an estimate of a finite population standard error is under the finite-second-moment condition.

[F16]

Change of variables for expectation: Expectations of measurable functions of a random variable are integrals against its law, including extended expectations for nonnegative functions.

[F17]

A continuous function on [a,b] is Riemann integrable, by Heine-Cantor and Riemann's criterion: A continuous real function on a compact interval is bounded and Riemann integrable there.

Verification

1.1A1F1F2F3F4given

For x≥1, x↦x3/2 is strictly increasing by [F2], so its positive reciprocal x−3/2 is decreasing; it is continuous by [F1]. Thus the displayed F is nondecreasing and right-continuous, including at x=1 where its two formulas agree. For x≥1, [F2] gives x1/2≥1, hence 0≤x−3/2=x−1x−1/2≤x−1. Since x−1→0, F(x)→1 as x→+∞; also F(x)→0 as x→−∞. By [A1] and [F3], F defines a Borel probability law μ. Since F(1)=0, μ((−∞,1])=0 and hence μ([1,∞))=1. Restricting μ to [1,∞) gives a probability space on which the identity random variable Y(x)=x is measurable by [F4], has CDF F, and satisfies Y≥1.

2.1F1F3F5F6F8F9step 1.1

If 0≤t<1, F(t)=0 and hence P(Y>t)=1. For t≥1, the CDF gives P(Y>t)=1−F(t)=t−3/2. By the nonnegative expectation definition [F5] and layer-cake formula [F6], E[Y]=1+∫1∞t−3/2 dt. The function t−3/2 is nonnegative and continuous on every compact [1,R] by [F1]. For R>1, [F8] gives ∫1Rt−3/2 dt=2(1−R−1/2)→2; [F9] identifies this finite improper integral with its Lebesgue value. Therefore E[Y]=3.

3.1F1F5F7F8F10F11F12F13F17step 2.1algebra

Since Y≥1 on the restricted space, P(∣Y∣>t)=P(Y>t). By the nonnegative expectation definition [F5] and [F7] with p=2, E[Y2]=1+2∫1∞t−1/2 dt. Extend h(t)=t−1/2 from [1,∞) by h(t)=1 for t<1; it is continuous at 1 and on both pieces, so [F11] makes it Borel. Since [1,N] is Borel, [F12] makes each h1[1,N] measurable. For each integer N≥2, [F1] gives continuity of t−1/2 on [1,N], so [F17] makes it bounded and Riemann integrable there; [F8] and [F10] then give ∫[1,N]t−1/2 dλ=2(N−1). These values diverge because the natural numbers are unbounded. The truncations increase pointwise to t−1/21[1,∞), so [F13] gives ∫[1,∞)t−1/2 dλ=+∞. Hence E[Y2]=+∞.

4.1F11F16givenstep 2.1step 3.1

The given sequence has common law μ. By [F11], the functions y↦∣y∣ and y↦y2 are Borel; applying [F16] to these nonnegative functions gives E∣Y1∣=E∣Y∣=3 and E[Y12]=E[Y2]=+∞. Thus Y1 is integrable, and [F16] applied to the identity gives E[Y1]=E[Y]=3.

5.1givenF14F15step 4.1∎

By step 4.1, E∣Y1∣=3<∞, so [F14] proves strong and weak consistency of Yˉn. Step 4.1 shows E[Y12]=+∞, so the finite-second-moment condition in [F15] fails and no finite population standard error σ/n is available. For each realized sample of size n≥2, the empirical formula remains a statistic by [F15].

Sources