Preliminaries
The coupling inequality
Convergence
Some applications
Introduction
Simulation
Skorokhod representation
Radon-Nikodym derivatives
Coupling and convergence
Saul Jacka, Warwick Statistics
OxWaSP, Warwick
7 November, 2014
Saul Jacka, Warwick Statistics
Coupling and convergence
Preliminaries
The coupling inequality
Convergence
Some applications
Introduction
Simulation
Skorokhod representation
Radon-Nikodym derivatives
Qn: What is coupling for?
A: To compare probabilities.
Three uses: ordering,
probabilities.
approximation
and
convergence
of
They all work by creating copies of random objects which
simultaneously live on the same probability space and have a
desirable relationship. OxWaSP students have already met the idea
with the Chen-Stein method.
Saul Jacka, Warwick Statistics
Coupling and convergence
Preliminaries
The coupling inequality
Convergence
Some applications
Introduction
Simulation
Skorokhod representation
Radon-Nikodym derivatives
Qn.: How do we simulate from an arbitrary distribution on R?
Answer: suppose the relevant distribution function is F . Take U , a
U[0, 1] random variable [under the probability measure P], and set
X = F −1 (U)
(where F −1 (t) = inf {x : F (x) ≥ t} for t ∈ [0, 1]).
def
Note: F is right-continuous and increasing so F (F −1 (t) ≥ t for all
t.
Check:
P(X ≤ a) = P(F −1 (U) ≤ a) = P(F (a) ≥ U) = F (a), since U is uniform,
so X has distribution function F as required.
Saul Jacka, Warwick Statistics
Coupling and convergence
Preliminaries
The coupling inequality
Convergence
Some applications
Introduction
Simulation
Skorokhod representation
Radon-Nikodym derivatives
Recall that the distribution under P of a random variable X
(dened on (Ω, F) and taking values in (E , E)) is PX given by
PX (A) = P(X ∈ A).
Theorem:
(Skorkohod, Dudley: Skorokhod representation)
Suppose that E is a separable metric space (e.g. Rn ) with Borel
σ -algebra E . Suppose that (Pn )n≤∞ are probability measures on
w
(E , E) with Pn ⇒ P∞ , then there exist random objects (Xn )n≤∞
and a probability space (Ω, F, P) such that
(i) PXn = Pn n ≤ ∞
and
a.s.
(ii) Xn −→ X∞ .
Saul Jacka, Warwick Statistics
Coupling and convergence
Preliminaries
The coupling inequality
Convergence
Some applications
Introduction
Simulation
Skorokhod representation
Radon-Nikodym derivatives
(in the real-valued case)
Proof:
Use the corresponding
distribution functions (Fn )n≤∞ . As before, construct a uniform
r.v. U and then set
Xn = Fn−1 (U).
−1 (t) for all but countably
It is (fairly) clear that Fn−1 (t) → F∞
a.s.
many t and so Xn −→ X∞ .
Saul Jacka, Warwick Statistics
Coupling and convergence
Preliminaries
The coupling inequality
Convergence
Some applications
Theorem:
Introduction
Simulation
Skorokhod representation
Radon-Nikodym derivatives
(Radon-Nikodym Theorem) Suppose that M and R are
σ -nite measures on (Ω, F) and whenever R(A) = 0, M(A) = 0,
then we write
M << R
and there exists an f ≥ 0, such that
Z
Z
XdM =
XfdR, for all M-integrable X .
We often write f =
M << N << R ⇒
write M ∼ R .
dM
dR and note that
dM
dM dN
dR = dN dR . If M
Exercise:[Ex1] Show that
M≤N
(i.e.
it satises the chain rule:
<< R and R << M we
d(M+N)
dR
M(A) ≤ N(A)
for
dM
= dM
dR + dR and deduce that
dM
dN
all A ∈ F ) then
dR ≤ dR .
Saul Jacka, Warwick Statistics
Coupling and convergence
if
Preliminaries
The coupling inequality
Convergence
Some applications
Denition: A family of measures
dominated
by a measure
µ
µ
{µθ ; θ ∈ Θ}
is said to be
if
µθ µ
and such a
Introduction
Simulation
Skorokhod representation
Radon-Nikodym derivatives
for all
θ ∈ Θ,
is said to be a dominating measure for the family.
Remark: Note that for any countable collection
measures on
that each
Pn
(Ω, F)
{Pn }
of probability
there is a dominating measure (call it
is absolutely continuous with respect to
R
R)
such
(and thus
has a density by the Radon-Nikodym theorem). To see this simply
set
1
2
R = (P∞ +
Note that, in fact,
R
∞
X
2−n Pn )
n=1
is a probability measure.
Saul Jacka, Warwick Statistics
Coupling and convergence
(1)
Preliminaries
The coupling inequality
Convergence
Some applications
Theorem:[Coupling]
(The fundamental inequality of coupling) If
P and Q are two probability measures on (Ω, F) then
(a) if X and Y are random objects: X , Y : (Ω0 , F 0 , P0 ) → (Ω, F)
with distributions P0X = P and P0Y = Q then
def
P0 (X 6= Y ) ≥ d(P, Q) = sup |P(A) − Q(A)| ;
A∈F
(b) there exists a probability space (Ω̃, F̃, P̃) and random objects
X , Y : (Ω̃, F̃, P̃) → (Ω, F) such that
(i)
P̃X = P
and
P̃Y = Q
and
(ii)
P̃(X 6= Y ) = d(P, Q).
Saul Jacka, Warwick Statistics
Coupling and convergence
Preliminaries
The coupling inequality
Convergence
Some applications
Proof: the proof of this result relies on the following observation:
suppose
R
P
(i.e. an
and
Q
are as above then, taking any dominating measure
dP
P, Q R), and, dening fP = dR
, fQ =
Z
Z
d(P, Q) = (fP − fQ )+ dR = (fP − fQ )− dR
R
s.t.
(since
Z
=
K
Z
(fP −fQ )dR =
where
Kc
R
dQ
dR ,
(fP − fQ )dR = 0)
(fQ −fP )dR = P(K )−Q(K ) = Q(K c )−P(K c )
K = {ω : fP (ω) > fQ (ω)}
Saul Jacka, Warwick Statistics
Coupling and convergence
Preliminaries
The coupling inequality
Convergence
Some applications
S by S(A) = P0 (X = Y ∈ A)
and observe that S ≤ P and S ≤ Q and so (by Exercise [Ex1]) S
has density dominated by fP ∧ fQ . Thus
Z
Z
1 − S(Ω) = P0 (X 6= Y ) ≥ 1 − fP ∧ fQ dR = (fP − fP ∧ fQ )dR
To prove (a), dene the measure
Ω
Z
=
Ω
Ω
(fP − fQ )+ dR = d(P, Q).
To prove (b) we construct independent random variables
def
X, Z
and
U on the measurable space (Ω̃, F̃) =
(Ω × Ω × [0, 1], F ⊗ F ⊗ B[0, 1]), with U being Uniform[0, 1], X
having distribution P and Z having density (wrt R )
def (fQ −fP )1K c
d(P,Q) .
fM =
Saul Jacka, Warwick Statistics
Coupling and convergence
Preliminaries
The coupling inequality
Convergence
Some applications
Thus the probability measure on (Ω̃, F̃) is P̃ = P ⊗ M ⊗ Λ, where Λ
is Lebesgue measure (on B[0, 1]) and, if ω̃ = (ω1 , ω2 , t), then
(X (ω̃), Z (ω̃), U(ω̃)) = (ω1 , ω2 , t), i.e. (X , Z , U) is the identity on
Ω̃.
Now dene Y (ω̃) = X 1
U≤
fP ∧fQ (ω1 )
fP (ω1 )
that Z takes values in K c , where
+ Z1
fP ∧fQ (ω1 )
fP (ω1 )
U>
fP ∧fQ (ω1 )
fP (ω1 )
. Notice
= 1.
It's clear that X has the right distribution and that
P̃(Y ∈ A) = P̃(Y = X ∈ A) + P̃(Y = Z ∈ A)
Z
=
A
fP ∧ fQ (ω1 )
dP(ω1 )
fP (ω1 )
Z
Z
+
1−
(ω1 ∈Ω)
(ω2 ∈A)
Saul Jacka, Warwick Statistics
fP ∧fQ (ω1 )
fP (ω1 )
dP(ω1 )dM(ω2 )
Coupling and convergence
Preliminaries
The coupling inequality
Convergence
Some applications
Z
Z
=
A
fP ∧ fQ (ω1 )dR +
(ω1 ∈Ω)
(fP − fQ ) dR(ω1 )
Z
=
A
A
(ω2 ∈A)
dM(ω2 )
Z
fP ∧ fQ (ω1 )dR + d(P, Q)
Z
=
Z
+
Z
fP ∧ fQ (ω1 )dR +
(ω2 ∈A)
+
(ω2 ∈A)
dM(ω2 )
Z
(fQ − fP ) dR(ω2 ) =
Y hasRdistribution Q, as required.
P̃(X = Y ) = fP ∧ fQ dR = 1 − d(P, Q),
so that
Saul Jacka, Warwick Statistics
A
fQ dR = Q(A),
Finally, it is clear that
Coupling and convergence
Preliminaries
The coupling inequality
Convergence
Some applications
Denitions
Skorokhod weak convergence
Skorokhod strong convergence
Counterexamples
Denition: Given a sequence of probability measures
the
Pn
converge
Skorokhod weakly to P∞ , written
(Pn )
we say
Sw
Pn ⇒ P∞ ,
if there is a dominating (probability) measure
prob(Q)
fn −→ f∞
as
Q
such that:
n → ∞,
dP
n
where fn is a version of
dQ .
Saul Jacka, Warwick Statistics
Coupling and convergence
Preliminaries
The coupling inequality
Convergence
Some applications
Denitions
Skorokhod weak convergence
Skorokhod strong convergence
Counterexamples
Given the (P ) , we say the P converge
Skorokhod strongly to P∞ , written
Denition:
n n≤∞
n
Ss
Pn ⇒ P∞ ,
if there exists a dominating probability measure Q such that:
f ∧ f −→ f as n → ∞ .
The reason for the nomenclature will become apparent
soon.
There is no need to restrict the choice of Q to
probability measuresany σ-nite measure will do.
∞
n
Qa.s.
∞
Remark:
Remark:
Saul Jacka, Warwick Statistics
Coupling and convergence
Preliminaries
The coupling inequality
Convergence
Some applications
Denitions
Skorokhod weak convergence
Skorokhod strong convergence
Counterexamples
Conversely, we gain nothing by allowing more general σ -nite
measures, since if R is a σ -nite dominating measure with Tn ↑ Ω
and R(Tn ) < ∞ for eachPn, then there exists a sequence
(an ) ⊂ (0, ∞) such that
an R(Tn \Tn−1 ) = 1 and, dening Q by
n
dQ X
=
an 1(Tn \Tn−1 ) ,
dR
n
we see that Q is a probability measure on (Ω, F) and Q ∼ R , so we
dPn dR
dPn
n
may substitute Q for R and dP
dQ (≡ dR dQ ) for dR .
Saul Jacka, Warwick Statistics
Coupling and convergence
Preliminaries
The coupling inequality
Convergence
Some applications
Theorem:[Sw] if
(Pn )
Denitions
Skorokhod weak convergence
Skorokhod strong convergence
Counterexamples
are probability measures on
(Ω, F)
then the
following are equivalent
Sw
(i) Pn ⇒ P∞
(ii) ∃ a probability
(Xn ) :
(a)
(Ω0 , F 0 , P0 ) and random
→ (Ω, F) such that
space
(Ω0 , F 0 , P0 )
objects
P0Xn = Pn
and
(b)
P0 (Xn 6= X∞ ) → 0
(iii) Pn → P∞
as
n→∞
with respect to the total variation metric
i.e.
d(Pn , P∞ ) → 0
(iv) ∃
as
a dominating probability measure
fn =
dPn
dQ satisfy
n→∞
Q
s.t. the densities
L1 (Q)
fn −→ f∞
(v) Pn (A) → P∞ (A)
uniformly in
Saul Jacka, Warwick Statistics
A ∈ F.
Coupling and convergence
Preliminaries
The coupling inequality
Convergence
Some applications
Denitions
Skorokhod weak convergence
Skorokhod strong convergence
Counterexamples
The equivalence of (i) and (iii) is Scheé's lemma (see
Billingsley (1968)).
Remark:
Proof: throughout the proof
(iv)
⇒
R
is a dominating measure.
(ii) This mimics part of the proof of the fundamental
inequality for coupling. Given
Ω0
Q
and the densities
(fn )n≤∞ ,
= Ω × Ω∞ × [0, 1],
F 0 = F ⊗ F ∗∞ ⊗ B([0, 1]),
Saul Jacka, Warwick Statistics
Coupling and convergence
dene
Preliminaries
The coupling inequality
Convergence
Some applications
and the probability measures
Denitions
Skorokhod weak convergence
Skorokhod strong convergence
Counterexamples
Mn
by
dMn
(fn − f∞ )+
=
.
dQ
d(Pn , P∞ )
Then dene
0
P = P∞ ⊗
∞
O
Mn ⊗ Λ
n=1
and dene, for each
ω 0 = (ω∞ , ω1 , . . . ; t) ∈ Ω0 ,
X∞ (ω 0 ) = ω∞ ,
Xn (ω 0 ) =
ω∞
1
+ ωn 1
t≤ f∞f ∧fn (ω1 )
∞
,
t> f∞f ∧fn (ω1 )
∞
and
Y (ω 0 ) =
t
.
Saul Jacka, Warwick Statistics
Coupling and convergence
Preliminaries
The coupling inequality
Convergence
Some applications
Denitions
Skorokhod weak convergence
Skorokhod strong convergence
Counterexamples
What we're doing is making all the coupling constructions
X∞ to have the right law under P0 ;
then, taking a single independent U[0, 1] r.v. (called U ), setting
n
Xn = X∞ if (and only if ) U ≤ f∞f∧f
(X∞ ) and otherwise giving Xn
n
simultaneously by constructing
for all n
a conditional distribution which gives it the right (unconditional)
distribution. It's not hard to check that
P0Xn = Pn
∧fn
(X∞ )
P0 (Xn 6= X∞ ) ≤ (=)P0 U > f∞f∞
R
+
∞)
= Ω (fn −f
dP∞
f∞
R
+
= Ω (fn − f∞ ) dQ,
and by (iv) the last term in (2) tends to 0.
Saul Jacka, Warwick Statistics
Coupling and convergence
, whilst
(2)
Preliminaries
The coupling inequality
Convergence
Some applications
(i)
L1
⇔
Denitions
Skorokhod weak convergence
Skorokhod strong convergence
Counterexamples
(iv) The reverse implication is obvious (since convergence in
is equivalent to
{convergence
in probability
and uniform
integrability}). The forward implication follows since (by virtue of
the fact that f∞ and fn are densities):
Z
|f∞ − fn |dQ = 2
Ω
Z
(f∞ − fn )+ dQ
(3)
Ω
and the integrand on the right of (3) is uniformly bounded by f∞
(which is, by denition, in
(ii)
⇒
L1 (Q)).
(iii) This follows immediately from the coupling inequality.
Saul Jacka, Warwick Statistics
Coupling and convergence
Preliminaries
The coupling inequality
Convergence
Some applications
(iii)
⇒
Denitions
Skorokhod weak convergence
Skorokhod strong convergence
Counterexamples
(iv) This follows on taking the dominating measure
R:
d(Pn , P∞ ) → 0
tells us that, letting densities with respect to
Z
R
R
be denoted by f· ,
R
(f∞
− fnR )+ dR → 0
Ω
and (as before)
R
Ω |f∞
R
− fnR |dR = 2
R
R
Ω (f∞
− fnR )+ dR
establishing
(iv).
(iii)
⇔
(v) This is obvious
Saul Jacka, Warwick Statistics
Coupling and convergence
Preliminaries
The coupling inequality
Convergence
Some applications
Theorem: [Ss] Suppose
(Pn )
Denitions
Skorokhod weak convergence
Skorokhod strong convergence
Counterexamples
are probability measures on
(Ω, F),
then the following are equivalent
Ss
(i) Pn ⇒ P∞
(ii) There exists
a probability space
(Ω0 , F 0 , P0 )
and random
objects
(Xn ) : (Ω0 , F 0 , P0 ) → (Ω, F)
such that
(a) P0Xn =
Pn
and
(b)
P0 (Xn 6= X∞ i.o.) = 0.
Saul Jacka, Warwick Statistics
Coupling and convergence
Preliminaries
The coupling inequality
Convergence
Some applications
Proof: (i)
⇒
Denitions
Skorokhod weak convergence
Skorokhod strong convergence
Counterexamples
(ii) Take the representation given in the proof of the
previous theorem, then
P0 (∃n ≥ N : Xn 6= X∞ ) = P0 (Y >
inf
n≥N
f∞ ∧fn
f∞ (X∞ ))
f∞ ∧ inf fn
R
n≥N
= Ω (1 −
(ω)) dP∞ (ω)
f∞
R
= Ω (f∞ (ω) −
fn (ω))+ dQ(ω)
inf
n≥N
and by monotone convergence this expression converges to
R
=
Ω (f∞
liminf f )
0 (by (i)).
−
Saul Jacka, Warwick Statistics
n
+ dP
∞
Coupling and convergence
Preliminaries
The coupling inequality
Convergence
Some applications
(ii)
⇒
(i) Given
P0
Denitions
Skorokhod weak convergence
Skorokhod strong convergence
Counterexamples
(Xn ) as in (ii), dene Q as in equation
m ≥ 1, the measure Sm on (Ω, F) by
and
and dene, for each
1,
Sm (A) = P0 (∃n ≥ m : Xn 6= X∞ , X∞ ∈ A),
(since
Sm (A) ≤ P0 (X∞ ∈ A) = P∞ (A)
by hypothesis)
Sm P∞ ,
whilst
Sm (Ω) = P0 (∃n ≥ m : Xn 6= X∞ ),
so that
lim S
m (Ω)
io
= P0 (Xn 6= X∞ . .).
Now
Sm (A) ≥ P0 (Xn 6= X∞ , X∞ ∈ A)
≥ P0 (Xn ∈ Ac , X∞ ∈ A)
≥ P0 (X∞ ∈ A) − P0 (Xn ∈ A)
= P∞ (A) − Pn (A)
Saul Jacka, Warwick Statistics
(for any
n ≥ m),
Coupling and convergence
Preliminaries
The coupling inequality
Convergence
Some applications
so that, for any
Denitions
Skorokhod weak convergence
Skorokhod strong convergence
Counterexamples
n ≥ m,
def
gm =
dSm
≥ f∞ − fn
dQ
(Q
a.s.),
so
gm ≥ (f∞ − fn )+
It follows that
gm ≥ (f∞ −
lim SR
≥ lim
0 =
inf
n≥m
(Q
a.s.) for any
n ≥ m.
fn )+ (Q a.s.) and hence
m (Ω)
Ω (f∞
lim R g dQ
− inf f ) dQ.
=
n≥m
Ω m
+
n
It follows (by monotone convergence) that lim inf fn
≥ f∞ (Q
a.s.)
from which we may easily deduce (using Fatou's lemma) that lim
inf fn
= f∞ (Q
a.s.) and hence
Qa.s.
Q
f∞
∧ fn −→ f∞
Saul Jacka, Warwick Statistics
Coupling and convergence
Preliminaries
The coupling inequality
Convergence
Some applications
Denitions
Skorokhod weak convergence
Skorokhod strong convergence
Counterexamples
Exercise:[Ex2] Using counting measure on the integers as a
reference measure show that if
integers with
w
Pn ⇒ P∞ ,
(Pn )n≤∞
are all measures on the
then
Ss
Pn ⇒ P∞ .
Saul Jacka, Warwick Statistics
Coupling and convergence
Preliminaries
The coupling inequality
Convergence
Some applications
Denitions
Skorokhod weak convergence
Skorokhod strong convergence
Counterexamples
In the examples (B ) are a sequence of
Bernoulli random variables:
Counterexamples
n
(Bn ) : (Ω, F) → ({0, 1}, 2{0,1} ),
and B is the random vector (B , B , . . .). Note that setting
Y = ·B B . . . [it being understood that a dyadic representation
is being given] it follows (from the fact that the Borel sets of
[0, 1] are generated by the intervals with dyadic rational
endpoints) that Y is a random variable:
1
2
1 2
Y : (Ω, F) → ([0, 1], B([0, 1])).
Saul Jacka, Warwick Statistics
Coupling and convergence
Preliminaries
The coupling inequality
Convergence
Some applications
Denitions
Skorokhod weak convergence
Skorokhod strong convergence
Counterexamples
Essentially we just want an example of a sequence
of densities which converge in probability, but not almost surely.
Given the (B ), dene P as follows: express k = 2 + r
(0 ≤ r ≤ 2 − 1), then
(i) under P , (B , . . . , B , B , . . .) are iid Bernoulli
(parameter );
(ii) if ·B . . . B is not the dyadic representation of
then
make B conditionally independent Bernoulli ( );
(iii) if ·B . . . B
the representation of , then set B = 1.
Example: Skorokhod weak does not imply Skorokhod strong
convergence.
k
n
n
n
k
1
1
1
2
n
n+2
r
2n
n
n+1
1
n is
Saul Jacka, Warwick Statistics
r
2n
Coupling and convergence
1
2
n+1
Preliminaries
The coupling inequality
Convergence
Some applications
It follows that fk , the density of
Denitions
Skorokhod weak convergence
Skorokhod strong convergence
Counterexamples
PkY
is given by


1: x ∈
/ [ 2rn , r 2+n1 )


r+ 1
fk (x) = 2 : x ∈ [ 2n2 , r 2+n1 )


0 : x ∈ [ r , r + 12 )
2n 2n
where
k
(as before) is
2n + r
( with
prob
0 ≤ r ≤ 2n − 1).
fn −→ f∞ (≡ 1), since fn diers from f∞
1
measure O(
), but equally clearly
log2 n
lim inf fn = 0
Clearly
only on a set of Lebesgue
Lebesgue a.e.
Saul Jacka, Warwick Statistics
Coupling and convergence
Preliminaries
The coupling inequality
Convergence
Some applications
Denitions
Skorokhod weak convergence
Skorokhod strong convergence
Counterexamples
. Here we just content ourselves with
Example: Skorokhod strong convergence does not imply a.s
giving f :
convergence of densities
k
fk (x) =
where, as usual, k = 2
n
but
1 − 2−n : x ∈
/ [ 2rn , r 2+n1 )
2 − 2−n : x ∈ [ 2rn , r2+n1 )
(
. Clearly,
+ r (0 ≤ r ≤ 2n − 1)
lim inf fn = 1,
lim sup fn = 2 (
Lebesgue a.e.)
Saul Jacka, Warwick Statistics
Coupling and convergence
Preliminaries
The coupling inequality
Convergence
Some applications
Poisson thinning and comparing skip-free random walks
Convergence of Markov chains
Conditioning MCs
BM and Girsanov's Theorem
Lipshitz continuity/Feller property for diusions
Suppose X and X are two continuous-time, skip-free Markov
chains on S = {0, 1, 2, ..., d} (generalised birth-and-death
processes), with birth rates λ and death rates µ and suppose
we know that X = x ≥ X = y and λ ≥ λ and µ ≤ µ for
each n.
Q: How do we show that X stochastically dominates X , i.e.
that
An example from nance (see [7])
1
2
1
0
i
n
2
0
1
n
2
i
n
n
1
1
2
n
n
2
t
t
P(Xt1 ≥ k) ≥ P(Xt2 ≥ k)
for each k and t ?
A: by a suitable coupling, using Poisson-thinning. Poisson
thinning is, at its simplest, the act of obtaining a
Poisson((1 − q)θ) process from Y , a Poisson(θ) process, by
removing jumps of Y independently with probability q.
Saul Jacka, Warwick Statistics
Coupling and convergence
Poisson thinning and comparing skip-free random walks
Convergence of Markov chains
Conditioning MCs
BM and Girsanov's Theorem
Lipshitz continuity/Feller property for diusions
Preliminaries
The coupling inequality
Convergence
Some applications
Let max(λ1n + µ2n ) = ρ. Construct a probability space with N , a
n∈S
Poisson process with rate 2ρ, and independent U[0, 1] r.v.s U1 , ....
Now start versions of X i at x and y respectively and construct
them as follows:
at Tk , the time of the k th jump of N , suppose XTi k − = xki , then
whenever xk2 < xk1 , X 1 jumps down by 1 if U <
if
1
2
>U>
1
2
−
λik
2ρ
µik
2ρ ,
jumps up by 1
otherwise X 1 doesn't move.
Similarly, X 2 jumps down by 1 if
λ2k
2ρ
1
2
<U<
1
2
+
µ2k
2ρ ,
jumps up by 1 if
otherwise X 1 doesn't move. Note that X 1 and X 2
cannot jump at the same time in this case.
U >1−
Saul Jacka, Warwick Statistics
Coupling and convergence
Preliminaries
The coupling inequality
Convergence
Some applications
Poisson thinning and comparing skip-free random walks
Convergence of Markov chains
Conditioning MCs
BM and Girsanov's Theorem
Lipshitz continuity/Feller property for diusions
However, if xk1 = xk2 then X i jumps down by 1 if U <
λik
µik
2ρ ,
jumps up
by 1 if U > 1 − 2ρ otherwise X i doesn't move. Note that in this
case, if X 1 jumps down, then so must X 2 , while if X 2 jumps up
then so must X 1 .
Since the resulting constructions cannot jump over one another, we
see that X 1 ≥ X 2 .
Exercise:[Ex3] Let
N
be a Poisson(θ ) process and construct
M
a
Poisson(pθ ) process by thinning
[0 , T ])
when
T =θ=2
and
N . Find P(N and M dier on
p = 0.95. Using your favourite
calculation package, compare this to the total variation distance
between the distributions of
N
and
Saul Jacka, Warwick Statistics
M
on [0,T].
Coupling and convergence
Preliminaries
The coupling inequality
Convergence
Some applications
Poisson thinning and comparing skip-free random walks
Convergence of Markov chains
Conditioning MCs
BM and Girsanov's Theorem
Lipshitz continuity/Feller property for diusions
[See [8]] We assume that (P ) are
a collection of probability measures on D([0, ∞); Z ): under P ,
X (given by X (ω) = ω ) is a time-inhomogeneous
Markov chain with initial distribution (p ). We assume the
existence of a dominating measure µ (nite on compact sets)
with respect to which each probability measure has transition
rates q (t) (t ≥ 0, i, j ∈ Z) and, as usual we write
q (t) = −q (t).
Now x T > 0 (temporarily) and denote the restriction of the
(P ) to the paths of X on [0, T ] by P |
.
n
Convergence of Markov chains
+
t
non-explosive
t
n
i
n
i,j
n
i
n
i,i
n
n
Saul Jacka, Warwick Statistics
n
[0,T ]
Coupling and convergence
Poisson thinning and comparing skip-free random walks
Convergence of Markov chains
Conditioning MCs
BM and Girsanov's Theorem
Lipshitz continuity/Feller property for diusions
Preliminaries
The coupling inequality
Convergence
Some applications
Theorem:[MC] (a)
If
as n → ∞ for each i;
as n → ∞ for each i;
(4)
pin → pi∞
L1 (µ)
qin −→ qi∞
n µa.e.
−→
qi,j
∞
qi,j
for each
i
and
j
in
and
Z+ ;
(5)
(6)
Ss
then
Pn |[0,T ] ⇒ P∞ |[0,T ] .
(b)
If (4) and (5) hold and
(µ)
∞
n
−→ qi,j
qi,j
then for, each
for each
i, j
in
Z+
T > 0,
Sw
Pn |[0,T ] ⇒ P∞ |[0,T ]
Saul Jacka, Warwick Statistics
Coupling and convergence
(7)
Preliminaries
The coupling inequality
Convergence
Some applications
Poisson thinning and comparing skip-free random walks
Convergence of Markov chains
Conditioning MCs
BM and Girsanov's Theorem
Lipshitz continuity/Feller property for diusions
We stress that we are assuming that, under P , the
chain is non-explosive.
∞
Remark:
Proof: We give rst a dominating (probability) measure
Q: it is
specied by having waiting time distribution exponential(µ) in
each state, i.e. qi (t) ≡ 1 for each i . Under Q, the jump chain
1
−(j+1)
forms a sequence of iid geometric( ) r.v.s so that qi,j (t) = 2
2
−(i+
1
)
and Q(X0 = i) = 2
. We assume that µ is continuous
wrt Q|
i.e. non-atomic. It is then fairly clear that the density of
Pk |[0,T ]
[0,T ] is
f k ≡ fTk
given by
Saul Jacka, Warwick Statistics
Coupling and convergence
Preliminaries
The coupling inequality
Convergence
Some applications
f k (ω) =
×
pωk 0
QN
n=1 2
QN
Poisson thinning and comparing skip-free random walks
Convergence of Markov chains
Conditioning MCs
BM and Girsanov's Theorem
Lipshitz continuity/Feller property for diusions
(ωTn +1) q k
ωTn−1 ,ωTn (Tn )
exp(R (1 − q (t))dµ(t))
R
× exp(
(1 − q (t))dµ(t)),
n=1
Tn
Tn−1
T
TN
k
ωTn−1
k
ω TN
(8)
N = N T (ω) = # {jumps of X on [0, T ]}, T0 = 0, and
Tn (1 ≤ n ≤ N) are the successive jump times of X (on [0, T ]).
Finally, since under Q the chain is non-explosive, notice that for
ε
any ε > 0, there is an n(ε) s.t. Q(N > n) ≤
2 and then ∃
m(n(ε), ε) s.t.
ε
Q(X
{0, . . . , m}
T) ≤ .
where
leaves
before
2
Denote the union of the two sets involved in these statements by
Aε .
We are now ready to prove (a).
Saul Jacka, Warwick Statistics
Coupling and convergence
Poisson thinning and comparing skip-free random walks
Convergence of Markov chains
Conditioning MCs
BM and Girsanov's Theorem
Lipshitz continuity/Feller property for diusions
Preliminaries
The coupling inequality
Convergence
Some applications
Under the assumption (5)
e−
for any
Rv
u
qik (t)dµ(t)
0 ≤ u ≤ v ≤ T.
Rv
→ e−
Hence, o
u
Aε ,
qi∞ (t)dµ(t)
,
there are only nitely many
terms in (8) and (by (4) and (6)) each converges
corresponding term in
since
ε
f ∞.
Thus
Q a.s. to the
Q(f k 6→ f ∞ ) ≤ Q(Aε ) ≤ ε and
is arbitrary we have established (a).
To prove (b) we need only to use the subsequence characterisation
of convergence in probability. Given a subsequence
sub-subsequence
(at least for
(nkj )
(nk )
take a
(by diagonalisation), along which (6) holds
t ∈ [0, T ])
then
f
nkj Qa.s.
−→ f ∞
n prob(Q)
subsequence is arbitrary so f
−→
Saul Jacka, Warwick Statistics
as
j →∞
f∞
Coupling and convergence
by (a). The
Preliminaries
The coupling inequality
Convergence
Some applications
Exercise:[Ex4] Check that
(characterising) family of
Pk (A) =
events A.
Poisson thinning and comparing skip-free random walks
Convergence of Markov chains
Conditioning MCs
BM and Girsanov's Theorem
Lipshitz continuity/Feller property for diusions
R
Af
k dQ for a suitable
Remark: The proof of Theorem [MC] only deals with the case
where
µ
is non-atomic; if
µ
exp(− R fdµ ) Q(1 − f ∆µ)
diculty: we simply need to replace
c
exp(− R fdµ)
has atoms there is no great additional
by
wherever such terms appear in (8).
This, in particular, allows us to deal with the discrete-time case.
Remark: It's easy to amend the proof to deal with semi-Markov
processes.
Saul Jacka, Warwick Statistics
Coupling and convergence
Preliminaries
The coupling inequality
Convergence
Some applications
Poisson thinning and comparing skip-free random walks
Convergence of Markov chains
Conditioning MCs
BM and Girsanov's Theorem
Lipshitz continuity/Feller property for diusions
Suppose that, under P , X is a
Markov chain started at x and τ is a hitting time for X i.e. the
rst time that X hits some set. Dene P to be the law of X
conditional on the event (τ > T ). Dene
Conditioning Markov chains
x
T
x
then for A ∈ F ,
h(x, t) = Px (τ > t),
t
PT
x (A) =
Ex [1A f (Xt , T − t)]
Px (A ∩ (τ > T ))
=
.
Px (τ > T )
f (x, T )
So, for any xed S < T , on F
dPT
x
dPx
S
= 1(τ >S)
f (XS , T − S)
f (x, T )
so if
→ ρ (y ) for each y where 1
wrt P then P ⇒ P .
f (y ,T −S)
f (x,T )
x
x,S
T Ss ∞
x
x
Saul Jacka, Warwick Statistics
(τ >S) ρ
Coupling and convergence
is a density
Preliminaries
The coupling inequality
Convergence
Some applications
Example: Suppose that, under
with
Q -matrix Q
and
τ
Poisson thinning and comparing skip-free random walks
Convergence of Markov chains
Conditioning MCs
BM and Girsanov's Theorem
Lipshitz continuity/Feller property for diusions
Px , X
is a MC on
{0, 1, 2, . . . , d}
is the hitting time of 0. The trick here is to
look at the asymptotic behaviour of
P̃(t),
the defective transition
probabilities for the process killed on hitting 0. So look at
restriction of
Q
to
{1, 2, . . . , d},
then
P̃t
Q̃ ,
the
is
P̃ = exp(t Q̃).
Q̃ = E ΛD then, assuming we have
−λt e d
written the largest eigenvalue of Q̃ as λ = Λ1,1 , P̃t ∼ e
i,1 1,j
P −λt
and it follows that f (i, t) ∼
e
e
d
and
thus
i,1 1,j
j
ρi,S (j) = e λS h(j)/h(i), where h = e·,1 is the right eigenvector of Q̃
corresponding to the principal eigenvalue. Since h is an eigenvector
T Ss ∞ on [0, S], where P∞
it follows that ρ is a density and so P ⇒ P
is the law of a MC on with Q -matrix Q̄ given by
Q̄i,j = λδi,j + hj Q̃i,j /h(i).
Now if we can diagonalise
Q̃
as
Saul Jacka, Warwick Statistics
Coupling and convergence
Preliminaries
The coupling inequality
Convergence
Some applications
Poisson thinning and comparing skip-free random walks
Convergence of Markov chains
Conditioning MCs
BM and Girsanov's Theorem
Lipshitz continuity/Feller property for diusions
Theorem:[Girsanov] Suppose that, under
Brownian Motion,
we dene
Q
µ
with drift
is a standard
by
dQ
= exp(
dP
def
T
Z
Q, Zt = Bt −
rate µ.
then under
P, B
is a bounded continuous adapted process and
0
Rt
0
1
µs dBs −
2
µs ds
Z
0
T
µ2s ds),
is a standard BM, or
B
is a BM
This allows us to copy what we did for convergence of Markov
chains in the Ito diusion setting.
Saul Jacka, Warwick Statistics
Coupling and convergence
Preliminaries
The coupling inequality
Convergence
Some applications
Poisson thinning and comparing skip-free random walks
Convergence of Markov chains
Conditioning MCs
BM and Girsanov's Theorem
Lipshitz continuity/Feller property for diusions
The idea is to couple two copies X and X of a
diusion starting at dierent positions by coupling the driving
BMs using a mirror coupling. This gives a lower bound on the
probability of coupling by time T (note, they don't actually
have to have the same SDE but...). So
x
Feller property
Xtz
Z
=z+
0
t
σ(Xsz )dBs
Z
+
0
t
µ(Xsz )ds,
y
(9)
where is a standard BM and σ is bounded below by η > 0 and
. Take a copy of X , call it Y , where the SDE (9) is
driven by −B (also a BM). This will have the same law. Now
take another copy, call it Z , driven by −B until the stopping
time τ and then driven by B , where τ is the coupling time:
B
|µ| ≤ M
y
τ = inf {t : Xt = Yt }.
Saul Jacka, Warwick Statistics
Coupling and convergence
Preliminaries
The coupling inequality
Convergence
Some applications
Poisson thinning and comparing skip-free random walks
Convergence of Markov chains
Conditioning MCs
BM and Girsanov's Theorem
Lipshitz continuity/Feller property for diusions
Now suppose that g is a bounded measurable function,then
|E(g (XT ) − E(g (ZT )| ≤ 2MP(τ > T )
where M = sup |g |. Wlog x > y , so
P(τ > T ) =
P(x−y + inf
0≤t≤T
Z
0
t
Z t
(σ(Xs )+σ(Ys ))dBs + (µ(Xs )−µ(Ys ))ds > 0)
0
2
≤ Px−y (τB > 4η T ),
where B is a one-dimensional BM with drift 2Mη2 and τB is the rst
time that B hits the origin. Now Pz (τB > T ) is O( √zT ) (as z → 0)
and hence we obtain the required result (see ([6]) and ([10])).
Saul Jacka, Warwick Statistics
Coupling and convergence
Preliminaries
The coupling inequality
Convergence
Some applications
Poisson thinning and comparing skip-free random walks
Convergence of Markov chains
Conditioning MCs
BM and Girsanov's Theorem
Lipshitz continuity/Feller property for diusions
[1] R. Atar and K. Burdzy, Mirror Couplings and Neumann Eigenfunctions.
American Mathematical Society 17(2), 243265, 2004.
Journal of the
[2] M. T. Barlow and S. D. Jacka, Tracking a Diusion, and an Application to Weak
Convergence. Advances in Applied Probability 18, 1525, 1986.
[3] K. Burdzy and W. S. Kendall, Ecient Markovian Couplings: Examples and Counterexamples.
The Annals of Probability 10(2), 362409, 2000.
[4] E. P. Hsu and K.T. Sturm, Maximal Coupling of Euclidean Brownian Motions. SFB Preprint
85, University of Bonn, 2003.
[5] Saul Jacka and Aleksandar Mijatovi¢, Coupling and Tracking of Regime-Switching
Martingales. ArXiv:1209.0180, 2012.
[6] Saul Jacka, Aleksandar Mijatovic and Dejan Siraj, Mirror and synchronous couplings of
Geometric Brownian motion, Stochastic Processes and their Applications, 124, 1055-1069 (2014)
[7] Saul Jacka and Adriana Ocejo, American-type options with parameter uncertainty,
arXiv:1309.1404 , 2013.
[8] Saul Jacka and Gareth Roberts, 'On strong forms of weak convergence' Stoch. Proc. & Appl.
67, 41-53 (1997)
[9] T. Lindvall,
. Dover Publications, New York, 2002.
Lectures on the Coupling Method
[10] T. Lindvall and L. C. G. Rogers, Coupling of Multidimensional Diusions by Reection.
Annals of Probability 14(3), 860872, 1986.
Saul Jacka, Warwick Statistics
Coupling and convergence
The
Preliminaries
The coupling inequality
Convergence
Some applications
Poisson thinning and comparing skip-free random walks
Convergence of Markov chains
Conditioning MCs
BM and Girsanov's Theorem
Lipshitz continuity/Feller property for diusions
[11] M. N. Pascu, Mirror Coupling of Reecting Brownian motion and an Application to Chavel's
Conjecture. Electronic Journal of Probability 16, 505530, 2011.
[12] H. Thorisson,
. Springer-Verlag New York, 2000.
Coupling, Stationarity, and Regeneration
Saul Jacka, Warwick Statistics
Coupling and convergence