Poisson point processes

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Suppose {Ft}\left\{ \mathcal{F}_{t}\right\} is a filtration satisfying the usual conditions and Pt1P_{t}^{1} and Pt2P_{t}^{2} are Poisson processes with respect to {Ft}\left\{ \mathcal{F}_{t}\right\} with parameters λ1,λ2\lambda_{1}, \lambda_{2}, respectively. Suppose Pt1+Pt2P_{t}^{1}+P_{t}^{2} is a Poisson process with parameter λ1+λ2\lambda_{1}+\lambda_{2}. Prove that P1P^{1} and P2P^{2} are independent processes.

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Suppose {Ft}\left\{ \mathcal{F}_{t}\right\} is a filtration satisfying the usual conditions, PtP_{t} is a Poisson process with respect to {Ft}\left\{ \mathcal{F}_{t}\right\}, and WtW_{t} is a Brownian motion with respect to {Ft}\left\{ \mathcal{F}_{t}\right\}. Show that if Wt+PtW_{t}+P_{t} has stationary and independent increments, then PP and WW are independent processes.

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If HH is as in (18.1) and NN is defined by (18.2), show that NN is a martingale, Nc=0N^{c}=0, and [N]t[N]_{t} is given by (18.3).

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Suppose {As,0<s<∞}\left\{ A_{s}, 0<s<\infty \right\} is a collection of subsets of S\mathcal{S} such that λ(As)→∞\lambda \left(A_{s}\right) \rightarrow \infty as s→∞s \rightarrow \infty. Show that Nt(As)/λ(As)N_{t}\left(A_{s}\right) / \lambda \left(A_{s}\right) converges to tt uniformly over finite intervals, where the convergence is in probability.

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Suppose {As,0<s<∞}\left\{ A_{s}, 0<s<\infty \right\} is a collection of subsets of S\mathcal{S} such that Ar⊂AsA_{r} \subset A_{s} if r≤sr \leq s and λ(As)→∞\lambda \left(A_{s}\right) \rightarrow \infty as s→∞s \rightarrow \infty. Show that for each tt,

sup⁡u≤t∣Nu(As)λ(As)−u∣ \sup _{u \leq t}\left|\frac{N_{u}\left(A_{s}\right)}{\lambda \left(A_{s}\right)}-u\right|

tends to zero almost surely as s→∞s \rightarrow \infty.

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Let S\mathcal{S} be a metric space and λ\lambda a σ\sigma-finite measure on S\mathcal{S}. Construct a Poisson point process which has λ\lambda as the corresponding measure.

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Let Ptj,j=1,2,…P_{t}^{j}, j=1,2, \ldots be independent Poisson processes with parameter βj\beta_{j}. Let Xt=∑j=1∞ajPtjX_{t}=\sum_{j=1}^{\infty } a_{j} P_{t}^{j}, where aja_{j} is a sequence such that XtX_{t} is finite, a.s. For A⊂R\{0}A \subset \mathbb {R} \backslash \left\{ 0\right\}, define Nt(A)N_{t}(A) to be the number of times before time tt that XX has a jump whose size is in AA :

Nt(A)=∑s≤t  1A(Xs−Xs−) N_{t}(A)=\sum _{s \leq t} \; \mathbb {1}_{A}\left(X_{s}-X_{s-}\right)

Prove that NtN_{t} is a Poisson point process and determine λ\lambda.

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Let δ>0\delta >0 and Aδ={f∈E:sup⁡t∣f(t)∣>δ}A_{\delta }=\left\{ f \in \mathcal{E}: \sup_{t}\left|f(t)\right|>\delta \right\}. Let S1=inf⁡{t:Kt(Aδ)=1}S_{1}=\inf \left\{ t: K_{t}\left(A_{\delta }\right)=1\right\} and S2=inf⁡{t>S1:Kt(Aδ)=2}S_{2}=\inf \left\{ t>S_{1}: K_{t}\left(A_{\delta }\right)=2\right\}. Thus S1S_{1} and S2S_{2} are the times the first and second excursions in AδA_{\delta } have been completed. Let Y1(t)Y_{1}(t) be the excursion completed at time S1S_{1} and define Y2(t)Y_{2}(t) similarly. To be more precise, if R1=sup⁡{t<S1:Wt=0}R_{1}=\sup \left\{ t<S_{1}: W_{t}=0\right\}, then Y1(s)=WR1+sY_{1}(s)=W_{R_{1}+s} if s≤S1−R1s \leq S_{1}-R_{1} and Y1(s)Y_{1}(s) is equal to 0 for all s≥S1−R1s \geq S_{1}-R_{1}.

Prove that Y1Y_{1} and Y2Y_{2} are independent.

Hint: Use the strong Markov property at time S1S_{1}.

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Let Nt(A)N_{t}(A) be a Poisson point process with respect to the measure space (S,m)(\mathcal{S}, m) and let As,s>0A_{s}, s>0, be an increasing sequence of subsets of S\mathcal{S} with m(As)→∞m\left(A_{s}\right) \rightarrow \infty as s→∞s \rightarrow \infty. Does

Nt(As)−m(As)m(As) \frac{N_{t}\left(A_{s}\right)-m\left(A_{s}\right)}{\sqrt{m\left(A_{s}\right)}}

converge weakly with respect to D[0,1]D[0,1] as s→∞s \rightarrow \infty ? What is the limit?

This can be applied to get central limit theorems for the number of downcrossings of a Brownian motion, for example.

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Let XX be a Lévy process with Lévy measure mm. Prove that if AA and BB are disjoint closed sets, then

Ex[∑s≤t  1A(Xs−)  1B(Xs)]=Ex[∫0t  1A(Xs)m(B−Xs)ds] \mathbb {E}_{x}\left[\sum _{s \leq t} \; \mathbb {1}_{A}\left(X_{s-}\right) \; \mathbb {1}_{B}\left(X_{s}\right)\right]=\mathbb {E}_{x}\left[\int _{0}^{t} \; \mathbb {1}_{A}\left(X_{s}\right) m\left(B-X_{s}\right) d s\right]

for each xx, where B−y={z−y:z∈B}B-y=\left\{ z-y: z \in B\right\}. This is the Lévy system formula in the case of Lévy processes. There is an analogous formula for Hunt processes.

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Here is an application of the existence theorem in which TT is not a subset of the line. Let (N,N,ν)(N, \mathscr {N}, \nu ) be a measure space, and take TT to consist of the N\mathscr {N}-sets of finite ν\nu-measure. The problem is to construct a generalized Poisson process, a stochastic process [XA:A∈T]\left[X_{A}: A \in T\right] such that (i) XAX_{A} has the Poisson distribution with mean ν(A)\nu (A) and (ii) XA1,…,XAnX_{A_{1}}, \ldots , X_{A_{n}} are independent if A1,…,AnA_{1}, \ldots , A_{n} are disjoint. Hint: To define the finite-dimensional distributions, generalize this construction: For A,BA, B in TT, consider independent random variables Y1,Y2,Y3Y_{1}, Y_{2}, Y_{3} having Poisson distributions with means ν(A∩Bc),ν(A∩B),ν(Ac∩B)\nu \left(A \cap B^{c}\right), \nu (A \cap B), \nu \left(A^{c} \cap B\right), take μA,B\mu_{A, B} to be the distribution of (Y1+Y2,Y2+Y3)\left(Y_{1}+Y_{2}, Y_{2}+Y_{3}\right).

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Example 6. Let λ\lambda be a finite measure (λ≢0\lambda \not\equiv 0) defined on some measurable space (S,B)(\mathrm{S}, \mathscr {B}). Let Y,X1,X2,…Y, X_{1}, X_{2}, \ldots be independent random elements on a probability space (Ω,F,P)(\Omega , \mathscr {F}, \mathbb {P}) such that YY is a real Poisson random variable with parameter λ(S)\lambda (\mathrm{S}), and X1,X2,…X_{1}, X_{2}, \ldots are independent identically distributed F∣B\mathscr {F} \mid \mathscr {B}-measurable variables for which PX1(B)=λ(B)/λ(S)\mathbb {P}_{X_{1}}\left(B\right) = \lambda (B) / \lambda (\mathrm{S}) for B∈BB \in \mathscr {B}. Define on B×Ω\mathscr {B} \times \Omega the function Z(B,ω)Z(B, \omega ) by setting

Z(B,ω)=∑j=1Y(ω)1B(Xj(ω))(14) Z(B, \omega ) = \sum _{j = 1}^{Y(\omega )} \mathbf{1}_{B}\left(X_{j}(\omega )\right) \tag {14}

(Z(B,ω)=0Z(B, \omega ) = 0 when Y(ω)=0Y(\omega ) = 0). Formula (14) defines the so-called Poisson random measure with intensity measure (or leading measure) λ\lambda.

Let λ\lambda be a σ\sigma-finite measure on some measurable space ( S,B\mathrm{S}, \mathscr {B} ) (S=⋃m=1∞Sm\left(\mathrm{S} = \bigcup_{m = 1}^{\infty } \mathrm{S}_{m}\right., where Sm∈B,Sm∩Sk=∅\mathrm{S}_{m} \in \mathscr {B}, \mathrm{S}_{m} \cap \mathrm{S}_{k} = \varnothing for m≠km \neq k and 0<λ(Sm)<∞0 < \lambda \left(\mathrm{S}_{m}\right) < \infty, m,k∈Nm, k \in \mathbb {N} ). In this case the construction of the Poisson random measure (14) is easily modified as follows.

On some probability space (Ω,F,P)(\Omega , \mathscr {F}, \mathbb {P}) construct a countable collection of independent variables

Y(1),X1(1),X2(1),…,Y(m),X1(m),X2(m),… Y^{(1)}, X_{1}^{(1)}, X_{2}^{(1)}, \ldots , Y^{(m)}, X_{1}^{(m)}, X_{2}^{(m)}, \ldots

such that Y(m)Y^{(m)} has the Poisson distribution with parameter λm=λ(Sm)\lambda_{m} = \lambda \left(\mathrm{S}_{m}\right) and Law⁡(Xj(m))=λ(⋅)/λ(Sm)\operatorname {Law}\left(X_{j}^{(m)}\right) = \lambda (\cdot ) / \lambda \left(\mathrm{S}_{m}\right) on Bm=B∩Sm;m,j∈N\mathscr {B}_{m} = \mathscr {B} \cap \mathrm{S}_{m} ; m, j \in \mathbb {N}. Let Zm(B,ω)Z_{m}(B, \omega ) for B∈BmB \in \mathscr {B}_{m} be defined according to (14) with YY replaced by Y(m)Y^{(m)} and XjX_{j} by Xj(m)X_{j}^{(m)}. Set

Z(B,ω)=∑m=1∞Zm(B∩Sm,ω),B∈B,ω∈Ω(54) Z(B, \omega ) = \sum _{m = 1}^{\infty } Z_{m}\left(B \cap \mathrm{S}_{m}, \omega \right), \quad B \in \mathscr {B}, \quad \omega \in \Omega \tag {54}

Note that Z(B,ω)Z(B, \omega ) may also take infinite values.

Prove that if λ(B)=∞\lambda (B) = \infty for some B∈BB \in \mathscr {B}, then Z(B)=∞Z(B) = \infty a.s., and if λ(B)<∞\lambda (B) < \infty, then Z(B)<∞Z(B) < \infty a.s.

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Example 6. Let λ\lambda be a finite measure (λ≢0\lambda \not\equiv 0) defined on some measurable space (S,B)(\mathrm{S}, \mathscr {B}). Let Y,X1,X2,…Y, X_{1}, X_{2}, \ldots be independent random elements on a probability space (Ω,F,P)(\Omega , \mathscr {F}, \mathbb {P}) such that YY is a real Poisson random variable with parameter λ(S)\lambda (\mathrm{S}), and X1,X2,…X_{1}, X_{2}, \ldots are independent identically distributed F∣B\mathscr {F} \mid \mathscr {B}-measurable variables for which PX1(B)=λ(B)/λ(S)\mathbb {P}_{X_{1}}\left(B\right) = \lambda (B) / \lambda (\mathrm{S}) for B∈BB \in \mathscr {B}. Define on B×Ω\mathscr {B} \times \Omega the function Z(B,ω)Z(B, \omega ) by setting

Z(B,ω)=∑j=1Y(ω)1B(Xj(ω))(14) Z(B, \omega ) = \sum _{j = 1}^{Y(\omega )} \mathbf{1}_{B}\left(X_{j}(\omega )\right) \tag {14}

(Z(B,ω)=0Z(B, \omega ) = 0 when Y(ω)=0Y(\omega ) = 0). Formula (14) defines the so-called Poisson random measure with intensity measure (or leading measure) λ\lambda.

Let λ\lambda be a σ\sigma-finite measure on some measurable space ( S,B\mathrm{S}, \mathscr {B} ) (S=⋃m=1∞Sm\left(\mathrm{S} = \bigcup_{m = 1}^{\infty } \mathrm{S}_{m}\right., where Sm∈B,Sm∩Sk=∅\mathrm{S}_{m} \in \mathscr {B}, \mathrm{S}_{m} \cap \mathrm{S}_{k} = \varnothing for m≠km \neq k and 0<λ(Sm)<∞0 < \lambda \left(\mathrm{S}_{m}\right) < \infty, m,k∈Nm, k \in \mathbb {N} ). In this case the construction of the Poisson random measure (14) is easily modified as follows.

On some probability space (Ω,F,P)(\Omega , \mathscr {F}, \mathbb {P}) construct a countable collection of independent variables

Y(1),X1(1),X2(1),…,Y(m),X1(m),X2(m),… Y^{(1)}, X_{1}^{(1)}, X_{2}^{(1)}, \ldots , Y^{(m)}, X_{1}^{(m)}, X_{2}^{(m)}, \ldots

such that Y(m)Y^{(m)} has the Poisson distribution with parameter λm=λ(Sm)\lambda_{m} = \lambda \left(\mathrm{S}_{m}\right) and Law⁡(Xj(m))=λ(⋅)/λ(Sm)\operatorname {Law}\left(X_{j}^{(m)}\right) = \lambda (\cdot ) / \lambda \left(\mathrm{S}_{m}\right) on Bm=B∩Sm;m,j∈N\mathscr {B}_{m} = \mathscr {B} \cap \mathrm{S}_{m} ; m, j \in \mathbb {N}. Let Zm(B,ω)Z_{m}(B, \omega ) for B∈BmB \in \mathscr {B}_{m} be defined according to (14) with YY replaced by Y(m)Y^{(m)} and XjX_{j} by Xj(m)X_{j}^{(m)}. Set

Z(B,ω)=∑m=1∞Zm(B∩Sm,ω),B∈B,ω∈Ω(54) Z(B, \omega ) = \sum _{m = 1}^{\infty } Z_{m}\left(B \cap \mathrm{S}_{m}, \omega \right), \quad B \in \mathscr {B}, \quad \omega \in \Omega \tag {54}

Note that Z(B,ω)Z(B, \omega ) may also take infinite values.

(see [32; p. 370]). Let B∈BB \in \mathscr {B} be such that 0<λ(B)<∞0 < \lambda (B) < \infty, where λ\lambda is a σ\sigma-finite measure on (S,B)(\mathrm{S}, \mathscr {B}). Let B=⋃q=1nBqB = \bigcup_{q = 1}^{n} B_{q}, where Bq∈B,Bj∩Bq=∅B_{q} \in \mathscr {B}, B_{j} \cap B_{q} = \varnothing for q≠j(q,j=1,…,n)q \neq j(q, j = 1, \ldots , n). Prove that for any k1,…,kn∈Z+k_{1}, \ldots , k_{n} \in \mathbb {Z}_{+} and k=∑q=1nkqk = \sum_{q = 1}^{n} k_{q} the following equality holds

P(Z(B1)=k1,…,Z(Bn)=kn∣Z(B)=k)==k!k1!⋯kn!(λ(B1)λ(B))k1⋯(λ(Bn)λ(B))kn \begin{aligned} & \mathbb {P}\left(Z\left(B_{1}\right) = k_{1}, \ldots , Z\left(B_{n}\right) = k_{n} \mid Z(B) = k\right) = \\ & = \frac{k !}{k_{1} ! \cdots k_{n} !}\left(\frac{\lambda \left(B_{1}\right)}{\lambda (B)}\right)^{k_{1}} \cdots \left(\frac{\lambda \left(B_{n}\right)}{\lambda (B)}\right)^{k_{n}} \end{aligned}
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Define m(B)=E[μ(B,⋅)]m(B) = \mathbb {E}\left[\mu (B, \cdot )\right] for all B∈B(R+)B \in \mathscr {B}\left(\mathbb {R}_{+}\right), where μ(B,⋅)\mu (B, \cdot ) appears in (85).

Suppose that on some probability space (Ω,F,P)(\Omega , \mathscr {F}, \mathbb {P}) we are given a sequence of random variables {τk}k⩾1\left\{ \tau_{k}\right\}_{k \geqslant 1} such that for ω∈Ω0\omega \in \Omega_{0}, where P(Ω0)=1\mathbb {P}\left(\Omega_{0}\right) = 1,

0<τ1<τ2<… and τn→∞ as n→∞. (84) 0 < \tau _{1} < \tau _{2} < \ldots \text{ and } \tau _{n} \rightarrow \infty \text{ as } n \rightarrow \infty \text{. } \tag {84}

Definition 11. The random point measure μ:B(R+)×Ω→Z‾+=Z+∪{∞}\mu : \mathscr {B}\left(\mathbb {R}_{+}\right) \times \Omega \rightarrow \overline{\mathbb {Z}}_{+} = \mathbb {Z}_{+} \cup \left\{ \infty \right\} is given by the formula

μ(B,ω)=∑k=1∞δτk(ω)(B),B∈B(R+),ω∈Ω(85) \mu (B, \omega ) = \sum _{k = 1}^{\infty } \delta _{\tau _{k}(\omega )}(B), \quad B \in \mathscr {B}\left(\mathbb {R}_{+}\right), \quad \omega \in \Omega \tag {85}

where the sequence {τk}k⩾1\left\{ \tau_{k}\right\}_{k \geqslant 1} is described above, and δx\delta_{x} is the Dirac measure.

Will the function m(⋅)m(\cdot ) be a measure on B(R+)\mathscr {B}\left(\mathbb {R}_{+}\right)? Can a Poisson process with a locally finite intensity measure mm be represented in the form (86)?

Definition 12. The random point process Y={Y(t),t⩾0}Y = \left\{ Y(t), t \geqslant 0\right\}, where Y(0)=0Y(0) = 0, is introduced for t>0t > 0 as follows:

Y(t,ω)=μ((0,t],ω)=∑k=1∞1(0,t](τk(ω))(86) Y(t, \omega ) = \mu ((0, t], \omega ) = \sum _{k = 1}^{\infty } \mathbf{1}_{(0, t]}\left(\tau _{k}(\omega )\right) \tag {86}

where the random point measure μ\mu is given by formula (85).

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Let N={N(t),t⩾0}N = \left\{ N(t), t \geqslant 0\right\} be a Poisson process with mean measure m=m(⋅)m = m(\cdot ), determined by the sequence {τk}k⩾1\left\{ \tau_{k}\right\}_{k \geqslant 1} that specifies the jump points of NN. Let {ηk}k⩾1\left\{ \eta_{k}\right\}_{k \geqslant 1} be a sequence of independent identically distributed real random variables with Law⁡(ηk)=P~\operatorname {Law}\left(\eta_{k}\right) = \widetilde{\mathbb {P}}. Take D∈B(R)D \in \mathscr {B}(\mathbb {R}) such that P~(D)>0\widetilde{\mathbb {P}}(D) > 0. Prove that the "thinned process" Y(t,D),t⩾0Y(t, D), t \geqslant 0, defined by formula (88), is a Poisson process with mean measure m(⋅)P~(D)m(\cdot ) \widetilde{\mathbb {P}}(D).

Definition 11. The random point measure μ:B(R+)×Ω→Z‾+=Z+∪{∞}\mu : \mathscr {B}\left(\mathbb {R}_{+}\right) \times \Omega \rightarrow \overline{\mathbb {Z}}_{+} = \mathbb {Z}_{+} \cup \left\{ \infty \right\} is given by the formula

μ(B,ω)=∑k=1∞δτk(ω)(B),B∈B(R+),ω∈Ω(85) \mu (B, \omega ) = \sum _{k = 1}^{\infty } \delta _{\tau _{k}(\omega )}(B), \quad B \in \mathscr {B}\left(\mathbb {R}_{+}\right), \quad \omega \in \Omega \tag {85}

where the sequence {τk}k⩾1\left\{ \tau_{k}\right\}_{k \geqslant 1} is described above, and δx\delta_{x} is the Dirac measure.

Definition 12. The random point process Y={Y(t),t⩾0}Y = \left\{ Y(t), t \geqslant 0\right\}, where Y(0)=0Y(0) = 0, is introduced for t>0t > 0 as follows:

Y(t,ω)=μ((0,t],ω)=∑k=1∞1(0,t](τk(ω))(86) Y(t, \omega ) = \mu ((0, t], \omega ) = \sum _{k = 1}^{\infty } \mathbf{1}_{(0, t]}\left(\tau _{k}(\omega )\right) \tag {86}

where the random point measure μ\mu is given by formula (85).

Formula (88):

Y(t,D;ω)={0 if Y(t,ω)=0∑k=1Y(t)1D(ηk) if Y(t)⩾1(88) Y(t, D ; \omega ) = \begin{cases} 0 & \text{ if } Y(t, \omega ) = 0 \\ \sum _{k = 1}^{Y(t)} \mathbf{1}_{D}\left(\eta _{k}\right) & \text{ if } Y(t) \geqslant 1 \end{cases} \tag {88}

where the process YY is given by formula (86).

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References

[Bas11] Bass R.F. Stochastic Processes. Cambridge University Press, 2011.
[Bil95] Billingsley P. Probability and Measure. 3rd ed., Wiley-Interscience, 1995.
[БуШи05] Булинский А.В., Ширяев А.Н. Теория случайных процессов. ФИЗМАТЛИТ, 2005.