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ABC acceptance probability vanishes with the tolerance

Example

Let the parameter space be the singleton ${*}$ with its unit prior and let the simulator return $Y \sim Uniform [0, 1]$ , independently of the parameter. Use the independence proposal $Q = H$ , meaning that each proposal draws the singleton parameter and a fresh uniform datum. Take $S (y) = y$ , observation $s_{0} = \frac{1}{2}$ , and absolute distance. For $0 \leq ϵ \leq \frac{1}{2}$ , the tolerance event is $[\frac{1}{2} - ϵ, \frac{1}{2} + ϵ]$ ; for $ϵ > \frac{1}{2}$ it contains the full simulator support $[0, 1]$ . For positive tolerance use the explicit Metropolis–Hastings density representatives constructed below; acceptance at individual starting states can depend on this choice.

Verification

Given: A uniform simulator on $[0, 1]$ , the stated summary and observation, and a tolerance $ϵ \geq 0$ .

[F1] Augmented ABC tolerance target: the prior-predictive tolerance-event probability is $Z_{ϵ}$ , and the indicator target is defined only when $Z_{ϵ} > 0$ .

[F2] ABC Metropolis–Hastings on simulator-augmented states: a proposed augmented state is simulated and retained only according to the fixed-tolerance Metropolis–Hastings accepted-flow rule.

[F3] Metropolis–Hastings kernel: the generic procedure accepts with the density of the common symmetric part of the forward and transposed proposal flows relative to the forward flow and otherwise makes a self-transition. The procedure chooses measurable representatives $f = d μ / d (μ + μ^{T})$ and $g = d μ^{T} / d (μ + μ^{T})$ with $f + g = 1$ everywhere, and uses $α = \min {1, g / f}$ when $f > 0$ and $α = 1$ when $f = 0$ .

1.1F1givenalgebra

For $0 \leq ϵ \leq \frac{1}{2}$ , the tolerance interval has length $2 ϵ$ , so the singleton prior gives $Z_{ϵ} = 2 ϵ$ ; for $ϵ > \frac{1}{2}$ it covers all of $[0, 1]$ , so $Z_{ϵ} = 1$ .

2.1F1F2F3step 1.1algebra

Fix $ϵ > 0$ and write $h (x)$ for the tolerance indicator of the datum in the augmented state $x$ . Relative to $H (d x) H (d y)$ , the forward and transposed flows have densities $h (x) / Z_{ϵ}$ and $h (y) / Z_{ϵ}$ . Choose $f (x, y) = h (x) / (h (x) + h (y))$ and $g (x, y) = h (y) / (h (x) + h (y))$ when $h (x) + h (y) > 0$ , and $f = g = 1 / 2$ otherwise. These are measurable, sum to one everywhere, and multiplying them by the density $(h (x) + h (y)) / Z_{ϵ}$ of $μ + μ^{T}$ recovers the two flow densities; hence they are valid representatives for [F3].

3.1F1F2F3step 1.1step 2.1algebra∎

At every admissible current state $x$ , $h (x) = 1$ , so these representatives give $f (x, y) > 0$ , $g (x, y) / f (x, y) = h (y)$ and thus $α (x, y) = h (y)$ . Consequently the acceptance probability is $\int h (y) H (d y) = Z_{ϵ}$ , equal to $2 ϵ \to 0$ for $0 < ϵ \leq \frac{1}{2}$ and to $1$ for $ϵ > \frac{1}{2}$ . At $ϵ = 0$ , $Z_{0} = 0$ and [F1] assigns no target. The pointwise acceptance calculation uses the specified representatives, rather than arbitrary modifications on flow-null sets.

Depends on

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Sources

Jarno Lintusaari, Michael U. Gutmann, Ritabrata Dutta, Samuel Kaski, and Jukka Corander, Fundamentals and Recent Developments in Approximate Bayesian Computation, §§4–5 (standard reference, not scraped)
Benjamin M. Turner and Thomas Van Zandt, A tutorial on approximate Bayesian computation, §3.1 (standard reference, not scraped)