Sampling Algorithms for Conditional Poisson Processes

VLVZ was proposed for anyanti-hereditaryconditioning setE, defined as. Definition 2.4. E isanti-hereditaryifx∈ E ⇒x∪ {ξ} ∈ E, for anyξ /∈x.

Definition 2.5. E islattice(or has the lattice property) ifx, y ∈ E ⇒x∩y∈ E andx∪y∈ E. This lattice structure becomes crucial when employing Corollary 2.1 in order to establish strict stochastic dominance (cf. Theorem 2.4 & Lemma 2.6). Letθbe a random mapθ :_R2

e →R2e which

takesx∈ _R2

e and returns some (random) sub-configuration ofx. Forx ∈R2e refer toθx ≡ θ(x)as

a thinningofx sinceθx ⊆ x. Henceforth θ will represent anindependent p-thinning (cf. Section

1.1.9), so that θx is the configuration obtained by independently retaining each point of x with

probabilityp. Letn(x)be the number of points inxandπ_λE the distribution of a Poisson(λ)which is conditioned to belong to someE.

A number of algorithms are now presented that have the objective of sampling from π_λE, for someλ > 0 andE a lattice anti-hereditary conditioning set. The basis of all the algorithms is the generation first of adominatingprocess of a higher intensityκ > λ, which is then thinned in some way to obtain a process of lower intensity. Processes of higher intensity are more likely to belong to

E (see the discussion in Van Lieshout & Van Zwet 2001); hence letκbe large enough so that it is feasible to use Rejection sampling to draw fromπE_κ and setp= λ_κ. Define two dominating processes

DandD0by: D∼πE_κ andD0is obtained via: Seti= 1.

whilei6= 0:

drawD∼πE_κ and sampleθD.

ifθD ∈ E: seti= 0;else: seti=i+ 1.

returnD0 =D.

Thus the distribution ofD0 is a weighted version ofπE_κ since we acceptD0 = D ∼ πE_κ with proba- bilityP[θD ∈ E |D], which is given by:

P[θD ∈ E |D] = X x⊆d P[θD =x|D=d]1{x∈E}= X x⊆d pn(x)(1−p)n(d)−n(x)1{x∈E}

sinceθis an independentp-thinning. Three algorithms, which are variants of the same basic idea (a two-step rejection sampling), and the VLVZ algorithm are now described below:

Algorithm 2.1 (Rejection/Thinning (RT)). Seti= 1.

whilei6= 0:

drawD∼πE_κ and sampleθD.

ifθD ∈ E: seti= 0;else: seti=i+ 1.

returnXRT =θD.

Algorithm 2.2 (Iterated Rejection/Thinning (IRT)). Seti= 1; drawD∼πE_κ.

whilei6= 0: sampleθD.

ifθD ∈ E: seti= 0;else: seti=i+ 1.

returnXIRT _=θ D.

Algorithm 2.3 (Modified Iterated Rejection/Thinning (Mod)). Seti= 1; drawD0.

whilei6= 0: sampleθD0.

ifθD0 ∈ E: seti= 0;else: seti=i+ 1. returnXM od_=θ

D0. Algorithm 2.4 (VLVZ).

DrawD∼πE_κ.

Define a conditional spatial birth-and-death processΦ˜ with the following dynamics: the birth rate is ₁₋p_p for each point inD\Φ˜;

unit death rate per point inΦ˜;

births are generated fromD\Φ˜ and always accepted; deaths are accepted only ifE is not violated.

Apply CFTP to obtain an exact sample from the equilibrium distribution ofΦ˜. returnXV LV Z, the output of the CFTP procedure.

Remark2.3. Observe the analogy between the random variablesN, N0 (X, X0)of Section 2.2.1 and the processes D, D0 XIRT, XM od. In Section 2.2.1 N0 is obtained fromN via an accept reject mechanism where the acceptance probability is_P[θN ∈ E |N]; hereD0 is similarly obtained from

Dwith acceptance probability_P[θD ∈ E |D]. MoreoverX, X0 are conditional thinnings ofN, N0

andXIRT, XM od are those ofD, D0respectively.

The difference between IRT and modified IRT is in the generation of the dominating processes D

andD0 respectively. Except for trivialE implementation of modified IRT will be computationally burdensome; however the principle of the algorithm enables derivation of some distributional re- sults. The difference between RT and IRT is that in RT a newdominating process is generated at every iteration, while in IRT θ is applied tosame dominating process. Thus the output of IRT is aconditional thinningof the dominating processD. For VLVZ the equilibrium distribution of the

conditional processΦ˜ depends onDand detailed balance calculations (cf. Eq. 1.25) show that the equilibrium (conditional) distribution ofΦ˜, given a realization ofD, is that ofθDbut conditioned to

belong toE, ie. _PXV LV Z =x|D∝_P[θD =x|D]1{x∈E}.

Remark2.4. Only the RT and modified IRT algorithms actually produce samples from the required target distributionπ_λE; the other two will generally not, as the following Lemmas show.

Lemma 2.2. The modified IRT algorithm is equivalent to the RT algorithm, in that they output samples with the same distribution.

Proof. This can be seen via conditional expectation; supposegis a measurable function. For RTD

is the dominating process,θD the process produced by applyingθtoDandXRT the output. Then

Eg XRT= E g(θD)1{θD∈E} P[θD ∈ E] (2.4) where_P[θD ∈ E] = R dP[θD ∈ E |D=d]dπ E κ(d). Iff andf 0

are the densities (with respect to a unit rate Poisson) ofDandD0 respectively, thenf0 ∝f _P[θD ∈ E |D], as the distribution ofD0is

that ofDbut weighted by_P[θD ∈ E |D]. ThusE[E[g(θD0)|D0]] =_E h E[g(θD)|D]×P [θD∈E|D] P[θD∈E] i . Since the output of modified IRT,XM od_{, is a conditional thinning ofD}0

, obtained by iteratively samplingθD0 until it satisfiesE:

PXM od =x|D0= P[θD 0 =x|D0] P[θD0 ∈ E |D0] 1{x∈E}; so that _Eg XM od=_EEg XM od |D0=_E " Eg(θD0)1_{θ D0∈E}|D 0 P[θD0 ∈ E |D0] # =_E " Eg(θD)1{θD∈E}|D P[θD ∈ E |D] × P[θD ∈ E |D] P[θD ∈ E] # = E Eg(θD)1{θD∈E}|D P[θD ∈ E] = E g(θD)1{θD∈E} P[θD ∈ E] . (2.5)

The third equality follows asD0 is a weighted realization ofD. Now compare Eqs.(2.4 & 2.5). Lemma 2.3. The IRT and RT do not, in general, produce samples from the same distribution. Proof. This can also be seen via conditional expectation. If XIRT is the output of IRT, then the distribution ofXIRT givenDis that ofθD but weighted so that

PXIRT =x|D = P[θD =x|D]

P[θD ∈ E |D]

1{x∈E} thus _Eg XIRT=_EEg XIRT|D=_E

" Eg(θD)1{θD∈E}|D P[θD ∈ E |D] # . (2.6) 48

Comparing Eqs. (2.4 & 2.6)one can see that the difference lies in the outermost expectation: in IRT one takes the outer expectation of the ratio while in RT it is the ratio of the outer expectations. The two expressions will only be equal if _P[θD ∈ E |D] does not depend on D; but D ∈ E by

construction, so that _P[θD ∈ E |D] will generally depend on D (unless p = 1 in which case it

would be a very trivial thinning!).

Lemma 2.4. The VLVZ algorithm is equivalent to the IRT algorithm.

Proof. LetXV LV Z _{denote the output from VLVZ andX}IRT _{that from IRT. The dominating process}

for either isD, andθD is obtained by applyingθtoD. Detailed balance calculations (Van Lieshout

& Van Zwet 2001) show that the distribution ofXV LV Z_{, given D, is simply that ofθ}

D conditioned

to belong toE. The distribution of XIRT_{, given D, is that of θ}

D but conditioned to belong toE.

SinceXV LV Z |DandXIRT |Dagree in distribution, so willXV LV Z andXIRT.

Thus VLVZ produces biased samples asXV LV Z _andXRT _{do have the same distribution.}