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A lighter-tailed proposal has no finite rejection envelope

Example

On $N = {0, 1, 2, \dots}$ with counting measure, put $f (0) = g (0) = 0$ and for $n \geq 1$ define $f (n) = \frac{1}{n (n + 1)}, g (n) = \frac{4}{n (n + 1) (n + 2)} .$ Both are probability masses, but their ratio $f (n) / g (n) = (n + 2) / 4$ for $n \geq 1$ is unbounded. Thus this lighter-tailed proposal has no finite rejection envelope for the target.

Facts & Assumptions

[A1]

The sample space is $N$ with counting measure. The displayed target and proposal masses are nonnegative and vanish at $0$ .

[F1]

Rejection sampling: a finite rejection envelope requires $f (n) \leq c g (n)$ at every point outside a common counting-null set.

[F2]

Counting measure on an arbitrary set, Counting measure is a measure, and Counting-measure integration is a nonnegative series: counting measure assigns each set its cardinality, so its only null set is empty; integration of a nonnegative function against it is the nonnegative series of its point masses.

[F3]

The reals form a totally ordered field and The Cauchy-sequence reals have the least-upper-bound property: $R$ is a complete ordered field.

[F4]

For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ : for every $ε > 0$ some positive integer $m$ satisfies $1 / m < ε$ .

[F5]

Inverses of positives are positive, and reciprocation reverses order and Canonical naturals are positive and strictly increasing: reciprocation reverses positive order, and the canonical naturals are positive and strictly increasing.

[F6]

Limits and Cauchy sequences of reals: a real sequence tends to $x$ when every positive rational tolerance eventually bounds its distance from $x$ .

[F7]

Every complete ordered field is Archimedean: for every real $x$ there is a positive integer $n > x$ .

Verification

Given: The two mass functions in the Example, counting measure on $N$ , and the real number field.

1.1A1F2F3F4F5F6algebra

The partial sums telescope: $\sum_{n = 1}^{N} f (n) = 1 - \frac{1}{N + 1}$ and $\sum_{n = 1}^{N} g (n) = 1 - \frac{2}{(N + 1) (N + 2)}$ . For any positive rational $ε$ , [F4] applied to $ε / 2$ supplies $m$ with $1 / m < ε / 2$ ; for all $N + 1 \geq m$ , [F5] gives $1 / (N + 1) \leq 1 / m < ε / 2$ and $2 / ((N + 1) (N + 2)) \leq 2 / (N + 1) \leq 2 / m < ε$ . Thus both remainders tend to zero by [F6], and [F2] gives total masses $1$ .

2.1A1F1F2F7step 1.1algebra∎

At each $n \geq 1$ , $\frac{f (n)}{g (n)} = \frac{n + 2}{4}$ . For any finite proposed envelope $c > 0$ , [F7] gives a positive integer $n > 4 c$ , so $f (n) / g (n) = (n + 2) / 4 > c$ . Since [F2] says the only counting-null set is empty, no common null exception can remove these points. This contradicts the required envelope in [F1]; the pair has no finite rejection envelope.

Depends on

Used by

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Sources

Art B. Owen, Monte Carlo Theory, Methods and Examples, Chapter 4 §4.7 (standard reference, not scraped)
Computational Statistics with R, Chapter 6 §6.1 (standard reference, not scraped)