Repetition
Laplace functional
Point Processes
Repetition
Laplace functional
General Poisson process
Consider a sample space Ω and anonther space E ⊂ Rd for
some d ≥ 1
A sequence of random variables Xn : Ω → E for n ∈ N is
defined
Let E be the Borel σ-algebra on E
Given E ⊂ Rd and a measure µ, the point processes N is PRM(µ)
(Poisson random measure (µ)) if
N(A) is Poisson distributed with mean µ(A) for all A ∈ E
For any A ∈ E, let
{An }kn=1 disjoint ⇒ {N(An )}kn=1 independent
N(A) = N(ω, A)
X
Xn (A)
=
n∈N
=
X
1(X
1 n ∈ A)
n∈N
Stochastic processes III, lecture 3
Repetition
Stochastic processes III, lecture 3
Laplace functional
Repetition
Laplace functional
The Laplace functional
Let {Ij ; 1 ≤ j ≤ k} be a (finite) collection of bounded
rectangles in E ⊂ Rd
The finite
P dimensional distributions of the point process
N = n∈N Xn are the collection of multivariate mass
functions:
P(N(Ij ) = nj ; 1 ≤ j ≤ k},
for k, n1 , n2 · · · nk ∈ N.
Proposition 4.7.1: The finite dimensional distributions of the
point process N uniquely determine the distribution of N
Let B+ , be the non-negative and bounded functions on E .
For f ∈ B+ , and a point process N define
Z
X
N(f ) :=
f (x)dN(x) :=
f (Xn )
x∈E
The Laplace functional of N is defined as
h
i
−N(f )
ΨN (f ) := E e
The proof is skipped in this course and follows from basic
measure theory
Stochastic processes III, lecture 3
n∈N
Stochastic processes III, lecture 3
Repetition
Laplace functional
Repetition
Laplace functional
The Poisson process
If {Ij ; 1 ≤ j ≤ k} is a collection of bounded rectangles in
k
X
E ⊂ Rd and f (x) =
λj1(x
1 ∈ Ij ) for λi ≥ 0, then
Let µ be a measure on E ⊂ Rd , where E has Borel σ-algebra
E
Recall: a point process is a Poisson process with mean
measure µ (PRM) on E if
j=1
h Pk
i
ΨN (f ) = E e − j=1 λj N(Ij ) ,
1
For A ∈ E
(
which is the joint Laplace transform of (N(Ij ); 1 ≤ j ≤ k)
P(N(A) = k) =
This implies ΨN (f ) determines the joint distribution of
(N(Ij ); 1 ≤ j ≤ k)
2
Propositions 4.7.1 and 4.7.2 give that the Laplace functional
uniquely determines the distribution of N
Repetition
Laplace functional
The distribution of PRM(µ) is uniquely determined by
For A ∈ E
(
P(N(A) = k) =
{An }kn=1 disjoint subsets of E ⇒ {N(An )}kn=1 independent
We want to show that the properties above determine a
unique point process and we want to identify the form of the
Laplace functional
Repetition
Laplace functional
e −µ(A) (µ(A))k
k!
0
if µ(A) < ∞,
if µ(A) = ∞.
First we prove that the form of the Laplace functional follows from
the properties of the Poisson process
If f (x) = λ1
1(x ∈ A), where λ > 0 then N(f ) = λN(A), and
N(A) ∼ Poisson(µ(A))
This gives
h
i
h
i
ΨN (f ) := E e −N(f ) = E e −λN(A) =
{An }kn=1 disjoint subsets of E ⇒ {N(An )}kn=1 independent
Furthermore, the point process N is PRM(µ) if and only if its
Laplace functional is of the form
Z
ΨN (f ) = exp[− (1 − e −f (x) )µ(dx)],
f ∈ B+
E
=
∞
X
e −µ(A) (µ(A))k
k!
k=0
e −λk = exp[−(1 − e −λ )µ(A)] =
Z
(1 − e −λ )1
1(x ∈ A)µ(dx)] =
Z
Z
−λ1
1(x∈A)
= exp[− (1−e
)µ(dx)] = exp[− (1−e −f (x) )µ(dx)]
= exp[−
E
E
Stochastic processes III, lecture 3
if µ(A) < ∞,
if µ(A) = ∞.
Proof
Theorem (Theorem 4.8.2)
2
0
Stochastic processes III, lecture 3
Stochastic processes III, lecture 3
1
e −µ(A) (µ(A))k
k!
Stochastic processes III, lecture 3
E
Repetition
Laplace functional
Repetition
P
Let f (x) = kn=1 λn1(x
1 ∈ An ), where λn ≥ 0 and {An }kn=1
disjoint subsets of E
This gives
h
i
h Pk
i
ΨN (f ) := E e −N(f ) = E e − n=1 λn N(An ) =
Z
k
k
h
i Y
Y
−λn N(An )
−λn1(x∈A
1
n)
=
E e
=
exp − (1 − e
)µ(dx) =
n=1
= exp[−
For f ∈ B+ , let
n
n2
X
i −1 i
i −1
1(f
1 (x) ∈ [ n , n )) + n1
1(f (x) ≥ n)
fn (x) =
n
2
2
2
i=1
E
n=1
Z X
k
fn (x) % f (x), thus by monotone convergence N(fn ) % N(f )
By 0 < e −f ≤ 1 and dominated convergence we have
i
h
h
i
lim ΨN (fn ) := lim E e −N(fn ) = E lim e −N(fn ) = ΨN (f )
1
n)
(1 − e −λn1(x∈A
)µ(dx) =
E n=1
Z
= exp[−
(1 − e −
Pk
n=1
Laplace functional
λn1(x∈A
1
n)
n→∞
)µ(dx)] =
n→∞
n→∞
E
Z
= exp[−
(1 − e −f (x) )µ(dx)]
E
Stochastic processes III, lecture 3
Stochastic processes III, lecture 3
Repetition
Laplace functional
Laplace functional
We still have to prove that
Z
−f (x)
)µ(dx) ,
ΨN (f ) = exp − (1 − e
The previous step yields
f ∈ B+
E
Z
−fn (x)
ΨN (fn ) = exp − (1 − e
)µ(dx)
E
By 1 − e −fn % 1 − e −f and monotone convergence, we have
Z
−f (x)
ΨN (f ) = lim ΨN (fn ) = exp − (1 − e
)µ(dx)
n→∞
Repetition
E
determines a Poisson process uniquely
We know that the Laplace transform uniquely determines a
point process (Theorem 4.7.2)
f (x) = λ1
1(x ∈ A) ⇒
Z
h
i
−λ1
1(x∈A)
ΨN (f ) = exp − (1 − e
)µ(dx) = exp −(1 − e −λ )µ(A) ,
E
which is the Laplace transform of a Poisson(µ(A)) distributed
random variable
Stochastic processes III, lecture 3
Stochastic processes III, lecture 3
Repetition
Laplace functional
Pk
∈ An ), with {An }kn=1 disjoint, then
Z
h Pk
i
P
− n=1 λn N(An )
1
− kn=1 λn1(x∈A
n)
E e
= exp − (1 − e
)µ(dx)
E
" Z k
#
X
−λn1(x∈A
1
)
n
= exp −
(1 − e
)µ(dx)
f (x) =
1
n=1 λn1(x
Repetition
Construction of general Poisson process
Theorem
Let S be a countable, possibly infinite set. Assume
{Ei , i ∈ S} are disjoint elements of E
E = ∪i∈S Ei
E n=1
=
=
k
Y
n=1
k
Y
Laplace functional
µ(Ei ) < ∞ for i ∈ S
h
i
exp −(1 − e −λn )µ(An )
h
i
E e −λn N(An )
n=1
So, the joint Laplace transform of (N(An ); 1 ≤ n ≤ k) factors into
a product of Laplace transforms, which shows independence
Consider an independent sequence of random variables (Yi ; i ∈ S),
where Yi ∼ Poisson(µ(Ei )).
Define a probability measure on Ei : Fi (dx) = µ(dx)/µ(Ei ).
Let {Xi,n ; 1 ≤ n ≤ Yi } be i.i.d. random variables on Ei with
distribution Fi . {Xi,n ; 1 ≤ n ≤ Yi , i ∈ S} are independent.
Yi
XX
Then, N =
Xi,n is PRM(µ).
i∈S n=1
Stochastic processes III, lecture 3
Stochastic processes III, lecture 3
Repetition
Laplace functional
Repetition
Laplace functional
Proof
For E = ∪i∈S Ei we have
Step 1: Check it for Ei
ΨN (f ) = E[e
−
P Yi
j=1
f (Xi,j )
] =
=
=
=
=
∞
X
h
E e
k=0
∞ X
−
Pk
j=1
f (Xi,j )
i (µ(E ))k
i
e −µ(Ei )
k!
ik (µ(E ))k
i
E e
e −µ(Ei )
k!
k=0
h
i
exp −µ(Ei )(1 − E[e −f (Xi,1 ) ])
Z
−f (x) µ(dx)
exp −µ(Ei ) 1 −
[e
]
µ(Ei )
Ei
Z
exp − (1 − e −f (x) )µ(dx)
h
ΨN (f ) =
Z
−f (x)
exp − (1 − e
)µ(dx)
Ei
i∈S
"
−f (Xi,1 )
= exp −
XZ
i∈S
#
(1 − e
−f (x)
)µ(dx)
Ei
Z
−f (x)
= exp − (1 − e
)µ(dx)
E
This completes the construction of a Poisson process on E
Ei
Stochastic processes III, lecture 3
Y
Stochastic processes III, lecture 3