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Ordinary and self-normalized importance sampling — Examples

1 · Prerequisites

2 · Summary

These examples compute ordinary and self-normalized estimates on a two-point space, exhibit a full-support light-tailed proposal whose ordinary importance estimator has infinite variance, and show that a finite sample can have ESS equal to its full size while missing a region with positive target probability.

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 →

Ordinary and self-normalized importance sampling on a two-point space

Example

Let $X = {a, b}$ with counting measure, unnormalized target density $\tilde{p} (a) = \tilde{p} (b) = 1$ , proposal probabilities $Q (a) = 3 / 4$ and $Q (b) = 1 / 4$ , and integrand $f = 1_{{b}}$ . Then $Z_{p} = 2$ , the target probabilities are $π (a) = π (b) = 1 / 2$ , and $θ = \int f d π = 1 / 2$ . The proposal has full target support.

Facts & Assumptions

Given: The finite space, target, proposal, integrand, and normalizer in the Example.

[F1]

Ordinary importance sampling: For each proposal draw, ordinary importance sampling uses $Y_{i} = f (X_{i}) \tilde{p} (X_{i}) / (Z_{p} q (X_{i}))$ and returns their sample mean.

[F2]

Self-normalized importance sampling: When the sum of the nonnegative weights $W_{i} = \tilde{p} (X_{i}) / q (X_{i})$ is positive, the self-normalized estimate is $\sum_{i} W_{i} f (X_{i}) / \sum_{i} W_{i}$ .

[F3]

Ordinary importance sampling is unbiased: Under the stated support and integrability assumptions, the ordinary importance sampling estimate has expectation $\int f d π$ .

Verification

Given: Use one proposal draw $X \sim Q$ and the definitions in [F1] and [F2].

1.1givenF3algebra

The target normalizer is $Z_{p} = \tilde{p} (a) + \tilde{p} (b) = 2$ , so $π (a) = π (b) = 1 / 2$ and $θ = E_{π} f = 1 / 2$ . Since both proposal probabilities are positive and $X$ is finite, the support and integrability assumptions of [F3] hold.

2.1F1F3step 1.1algebra

The ordinary weighted value is $0$ at $a$ and $2$ at $b$ . Its expectation is $(3 / 4) 0 + (1 / 4) 2 = 1 / 2 = θ$ , as also guaranteed by [F3].

3.1F2step 1.1algebra∎

The self-normalized weights are $4 / 3$ at $a$ and $4$ at $b$ , so the single-draw ratio in [F2] equals $f (X)$ . Its expectation is $Q (X = b) = 1 / 4$ , and its bias for $θ$ is $1 / 4 - 1 / 2 = - 1 / 4$ . This verifies that self-normalization need not be unbiased at finite sample size.

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

A light-tailed proposal can give ordinary importance sampling infinite variance

Statement refuted

Full proposal support and finite target variance of $f$ are sufficient for ordinary importance sampling to have finite estimator variance.

Facts & Assumptions

Given: The claim in “Statement refuted.”

[F1]

Ordinary importance sampling: Ordinary importance sampling averages $Y_{i} = f (X_{i}) \tilde{p} (X_{i}) / (Z_{p} q (X_{i}))$ over IID proposal draws.

[F2]

Change of variables for expectation: For a nonnegative measurable function of a random element with law $Q$ , expectation equals its integral against $Q$ .

[F3]

Ordinary importance sampling has finite variance and a CLT under a weighted second moment: Its finite-variance formula assumes $E_{Q} [Y_{1}^{2}] < \infty$ and makes no finite-variance claim when that moment is infinite.

[F4]

Moments, variance, and covariance on a probability space: For square-integrable real $Y$ , $Var (Y) = E [(Y - E Y)^{2}]$ ; finite variance therefore requires a finite centered second moment.

[F5]

Ordinary importance sampling is unbiased: Under its support and integrability hypotheses, the ordinary importance sampling estimate has expectation $\int f d π$ .

[F6]

Monotonicity and nonnegative homogeneity of the nonnegative integral: If nonnegative measurable $u \leq v$ , then $\int u d μ \leq \int v d μ$ , including infinite integrals.

[F7]

Integrating against a density agrees with integrating the product: for a nonnegative measurable density $r$ and nonnegative measurable $h$ , $\int h d (r d μ) = \int h r d μ$ , including infinite values.

Counterexample

1.1givenalgebra

On $R$ with Lebesgue measure, take target density $p (x) = \tilde{p} (x) = \frac{1}{2} e^{- ∣ x ∣}$ , proposal density $q (x) = e^{- 2 ∣ x ∣}$ , and $f (x) = 1$ . Direct integration gives $\int p = \frac{1}{2} (\int_{- \infty}^{0} e^{x} d x + \int_{0}^{\infty} e^{- x} d x) = 1$ and $\int q = 2 \int_{0}^{\infty} e^{- 2 x} d x = 1$ . Also $q (x) > 0$ for every $x$ , while $f = 1$ has target mean $1$ and variance $0$ .

2.1F1F2F3step 1.1algebraF7

Use [F7] with density $r = q$ for integration against $Q$ . For every $x$ , the importance weight is $W (x) = p (x) / q (x) = \frac{1}{2} e^{∣ x ∣}$ . By [F2], $E_{Q} W = \int_{R} p (x) d x = 1$ and $E_{Q} [W^{2}] = \int_{R} \frac{1}{4} e^{2 ∣ x ∣} e^{- 2 ∣ x ∣} d x = \int_{R} \frac{1}{4} d x = \infty$ . The proposal has full target support, but the finite-second-moment hypothesis in [F3] fails.

3.1F1F4F5F6step 2.1algebra∎

By [F5], the ordinary estimate ${\overset{ˉ}{W}}_{n} = n^{- 1} \sum_{i} W_{i}$ has mean $1$ . For each finite $n$ , nonnegativity gives ${\overset{ˉ}{W}}_{n} \geq W_{1} / n$ , hence $W_{1}^{2} / n^{2} \leq {\overset{ˉ}{W}}_{n}^{2}$ . Monotonicity [F6] and step 2.1 imply $E_{Q} [{\overset{ˉ}{W}}_{n}^{2}] \geq n^{- 2} E_{Q} [W_{1}^{2}] = \infty$ . If its centered second moment were finite, the inequality $x^{2} \leq 2 (x - 1)^{2} + 2$ would imply $E_{Q} [{\overset{ˉ}{W}}_{n}^{2}] < \infty$ , a contradiction. Thus the centered second moment is infinite, so the estimate does not have finite variance under [F4]. This refutes the sufficiency claim.

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

A high observed weight ESS can miss target mass

Statement refuted

If a finite importance sample has weight-based effective sample size $n$ , then the sample has visited every region to which the target assigns positive probability.

Facts & Assumptions

Given: The claim in “Statement refuted.”

[F1]

Weight-based effective sample size: For nonnegative weights with positive sum, ${ESS}_{w} = (\sum_{i} w_{i})^{2} / \sum_{i} w_{i}^{2}$ .

[F2]

Self-normalized importance sampling: The input $X_{1}, \dots, X_{n}$ is IID from the proposal law $Q$ , and the unnormalized importance weight at a draw is $W = \tilde{p} (X) / q (X)$ when $q (X) > 0$ .

Counterexample

1.1givenF2algebra

Equip $X = {a, b}$ with counting measure $μ$ , set $\tilde{p} (a) = \tilde{p} (b) = 1$ , and let $q (a) = 99 / 100$ and $q (b) = 1 / 100$ . Then $Z_{p} = 2$ and $π (b) = 1 / 2$ , and the proposal law $Q (d x) = q (x) μ (d x)$ has full target support. For any $n \geq 1$ , take the IID $Q$ -sample $X_{1}, \dots, X_{n}$ from [F2]. The event $E_{n} = {X_{1} = \dots = X_{n} = a}$ has probability $(99 / 100)^{n} > 0$ .

2.1F1F2step 1.1algebra∎

On $E_{n}$ , all observed weights equal $W (a) = 100 / 99 > 0$ . By [F1], ${ESS}_{w} = (n W (a))^{2} / (n W (a)^{2}) = n$ , while no sample is at $b$ and the target assigns $b$ probability $1 / 2$ . Thus the claimed certification fails on an event of positive probability.