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A finite ABC-SMC tolerance schedule and its mixture weights

Example

Let $Θ = {0, 1}$ have the uniform prior and $Y = {0, 1, 2}$ . The simulator has $K (0, {0}) = K (0, {1}) = \frac{1}{2}$ and $K (0, {2}) = 0$ , while $K (1, {0}) = \frac{1}{4}$ , $K (1, {1}) = \frac{1}{2}$ , and $K (1, {2}) = \frac{1}{4}$ . Take $S (0) = S (1) = 0$ , $S (2) = 1$ , and $s_{0} = 0$ . Let $ϵ_{0} = \frac{3}{2}$ and $ϵ_{1} = \frac{1}{2}$ . At stage $0$ every simulated datum meets the tolerance. Condition on the possible stage-0 population $(θ_{1}, θ_{2}) = (0, 0)$ with weights $(\frac{1}{2}, \frac{1}{2})$ . At stage $1$ use a perturbation kernel with $Q_{1} (0, {0}) = \frac{99}{100}$ and $Q_{1} (0, {1}) = \frac{1}{100}$ , and set the same row from $θ = 1$ so $Q_{1}$ is fully specified.

Verification

Given: The model, tolerances, previous weighted population, and perturbation kernel specified in the Example.

[F1] Augmented ABC tolerance target: the tolerance target has unnormalized measure $Π (d θ) K (θ, d y) 1_{A_{ϵ}} (y)$ and normalizer $Z_{ϵ}$ .

[F2] Sequential ABC importance-resampling with tolerance populations: the proposal mixture is the weighted mixture of perturbation kernels, and the incremental weight is the tolerance indicator times $d Π / d ϕ_{t}^{N}$ .

[F3] Sequential ABC Radon–Nikodym weights target each tolerance law: under the mixture support condition, the conditional expected unnormalized weighted proposal law is $γ_{t} (d θ, d y) = Π (d θ) K (θ, d y) 1_{A_{ϵ_{t}}} (y)$ , whose total mass is $Z_{ϵ_{t}}$ . Dividing this measure by $Z_{ϵ_{t}}$ gives the current normalized tolerance target; this does not assert unbiasedness of a self-normalized finite-particle estimate.

1.1F1givenalgebra

At $ϵ_{0} = 3 / 2$ , all simulated summaries are accepted, so $Z_{ϵ_{0}} = 1$ ; at $ϵ_{1} = 1 / 2$ , acceptance probabilities are $1$ and $3 / 4$ at the two parameter values, giving $Z_{ϵ_{1}} = 7 / 8$ and target parameter masses $(4 / 7, 3 / 7)$ .

1.2F2givenalgebra

For the previous population $(0, 0)$ , $ϕ_{1}^{N} ({0}) = 99 / 100$ and $ϕ_{1}^{N} ({1}) = 1 / 100$ , so it dominates the uniform prior and the RN ratios are $r_{1}^{N} (0) = (1 / 2) / (99 / 100) = 50 / 99$ and $r_{1}^{N} (1) = (1 / 2) / (1 / 100) = 50$ .

2.1F1F2F3step 1.1step 1.2algebra∎

A proposal matches tolerance with probability $(99 / 100) \cdot 1 + (1 / 100) \cdot (3 / 4) = 399 / 400$ ; its expected weight contributions at $θ = 0, 1$ are $(99 / 100) (1) (50 / 99) = 1 / 2$ and $(1 / 100) (3 / 4) (50) = 3 / 8$ , totaling $7 / 8$ . Their ratio is $(4 / 7, 3 / 7)$ , the population target marginal by [F3]; these are normalized expected unnormalized weights, not a realized self-normalized finite-particle estimate.

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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, §5.2 and Appendix, Algorithm 4 (standard reference, not scraped)
Benjamin M. Turner and Thomas Van Zandt, A tutorial on approximate Bayesian computation, §3.2 (standard reference, not scraped)