Generalized Gaussian Bridges of Prediction-Invertible...
Transcript of Generalized Gaussian Bridges of Prediction-Invertible...
Generalized Gaussian Bridges ofPrediction-Invertible Processes
Tommi Sottinen1 and Adil Yazigi
University of Vaasa, Finland
Modern Stochastics: Theory and Applications III
September 10, 2012, Kyiv, Ukraine
1http://www.uwasa.fi/∼tsottine/1 / 37
Abstract
A generalized bridge is the law of a stochastic process that isconditioned on multiple linear functionals of its path. We considercanonical representation of the bridges. In the canonicalrepresentation the linear spaces Lt(X ) = span Xs ; s ≤ t coinsidefor all t < T for both the original process and its bridgerepresentation.
A Gaussian process X = (Xt)t∈[0,T ] is prediction-invertible ifit can be recovered (in law, at least) from its prediction martingale:
Xt =
∫ t
0p−1T (t, s) dE[XT |FX
s ].
In discrete time all non-degenerate Gaussian processes areprediction-invertible. In continuous time this is most probably nottrue.
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References
This work, S. and Yazigi (2012), combines and extends the resultsof Alili (2002) and Gasbarra S. and Valkeila (2007).
Alili. Canonical decompositions of certain generalizedBrownian bridges. Electron. Comm. Probab., 7, 2002, 27–36.
Gasbarra, S. and Valkeila. Gaussian bridges. Stochasticanalysis and applications, Abel Symp. 2, 2007, 361–382.
S. and Yazigi Generalized Gaussian Bridges. PreprintarXiv:1205.3405v1, May 2012.
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Outline
1 Generalized (linearly conditioned) Bridges
2 Orthogonal Representation
3 Canonical Representation for Martingales
4 Prediction-Invertible Processes
5 Canonical Representation forPrediction-Invertible Processes
6 Open Questions
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Outline
1 Generalized (linearly conditioned) Bridges
2 Orthogonal Representation
3 Canonical Representation for Martingales
4 Prediction-Invertible Processes
5 Canonical Representation forPrediction-Invertible Processes
6 Open Questions
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Generalized (linearly conditioned) Bridges
Let X = (Xt)t∈[0,T ] be a continuous Gaussian process with positivedefinite covariance function R, mean function m of boundedvariation, and X0 = m(0). We consider the conditioning, orbridging, of X on N linear functionals GT = [G i
T ]Ni=1 of its paths:
GT (X ) =
∫ T
0g(t) dXt =
[∫ T
0gi (t)dXt
]Ni=1
.
Remark (On Linear Independence of Conditionings)
We assume, without any loss of generality, that the functions giare linearly independent. Indeed, if this is not the case then thelinearly dependent, or redundant, components of g can simply beremoved from the conditioning (1) below without changing it.
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Generalized (linearly conditioned) Bridges
Let X = (Xt)t∈[0,T ] be a continuous Gaussian process with positivedefinite covariance function R, mean function m of boundedvariation, and X0 = m(0). We consider the conditioning, orbridging, of X on N linear functionals GT = [G i
T ]Ni=1 of its paths:
GT (X ) =
∫ T
0g(t) dXt =
[∫ T
0gi (t)dXt
]Ni=1
.
Remark (On Linear Independence of Conditionings)
We assume, without any loss of generality, that the functions giare linearly independent. Indeed, if this is not the case then thelinearly dependent, or redundant, components of g can simply beremoved from the conditioning (1) below without changing it.
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Generalized (linearly conditioned) Bridges
Informally, the generalized (Gaussian) bridge X g;y is (the law of)the (Gaussian) process X conditioned on the set∫ T
0g(t)dXt = y
=
N⋂i=1
∫ T
0gi (t) dXt = yi
. (1)
The rigorous definition is given in the next slide.
Remark (Canonical Space Framework)
For the sake of convenience, we will work on the canonicalfiltered probability space (Ω,F ,F,P), where Ω = C [0,T ],P corresponds to the Gaussian coordinate processXt(ω) = ω(t): P = P[X ∈ · ]. The filtration F = (Ft)t∈[0,T ] is theintrinsic filtration of the coordinate process X that is augmentedwith the null-sets and made right-continuous, and F = FT .
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Generalized (linearly conditioned) Bridges
Informally, the generalized (Gaussian) bridge X g;y is (the law of)the (Gaussian) process X conditioned on the set∫ T
0g(t)dXt = y
=
N⋂i=1
∫ T
0gi (t) dXt = yi
. (1)
The rigorous definition is given in the next slide.
Remark (Canonical Space Framework)
For the sake of convenience, we will work on the canonicalfiltered probability space (Ω,F ,F,P), where Ω = C [0,T ],P corresponds to the Gaussian coordinate processXt(ω) = ω(t): P = P[X ∈ · ]. The filtration F = (Ft)t∈[0,T ] is theintrinsic filtration of the coordinate process X that is augmentedwith the null-sets and made right-continuous, and F = FT .
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Generalized (linearly conditioned) Bridges
Definition (Generalized Gaussian Bridge)
The generalized bridge measure Pg;y is the regularconditional law
Pg;y = Pg;y [X ∈ · ] = P[
X ∈ ·∣∣∣∣ ∫ T
0g(t) dXt = y
].
A representation of the generalized Gaussian bridgeis any process X g;y satisfying
P [X g;y ∈ · ] = Pg;y [X ∈ · ] = P[
X ∈ ·∣∣∣∣ ∫ T
0g(t)dXt = y
].
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Generalized (linearly conditioned) Bridges
Definition (Generalized Gaussian Bridge)
The generalized bridge measure Pg;y is the regularconditional law
Pg;y = Pg;y [X ∈ · ] = P[
X ∈ ·∣∣∣∣ ∫ T
0g(t) dXt = y
].
A representation of the generalized Gaussian bridgeis any process X g;y satisfying
P [X g;y ∈ · ] = Pg;y [X ∈ · ] = P[
X ∈ ·∣∣∣∣ ∫ T
0g(t)dXt = y
].
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Generalized (linearly conditioned) Bridges
Remark (Technical Observations)
1 Note that the conditioning on the P-null-set is not a problem,since the canonical space of continuous processes is smallenough to admit regular conditional laws.
2 As a measure Pg;y the generalized Gaussian bridge is unique,but it has several different representations X g;y. Indeed,for any representation of the bridge one can combine it withany P-measure-preserving transformation to get a newrepresentation.
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Generalized (linearly conditioned) Bridges
Remark (Technical Observations)
1 Note that the conditioning on the P-null-set is not a problem,since the canonical space of continuous processes is smallenough to admit regular conditional laws.
2 As a measure Pg;y the generalized Gaussian bridge is unique,but it has several different representations X g;y. Indeed,for any representation of the bridge one can combine it withany P-measure-preserving transformation to get a newrepresentation.
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Outline
1 Generalized (linearly conditioned) Bridges
2 Orthogonal Representation
3 Canonical Representation for Martingales
4 Prediction-Invertible Processes
5 Canonical Representation forPrediction-Invertible Processes
6 Open Questions
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Orthogonal Representation
Denote by 〈〈〈g〉〉〉 the matrix
〈〈〈g〉〉〉ij := 〈〈〈gi , gj〉〉〉 := Cov[∫ T
0gi (t) dXt ,
∫ T
0gj(t) dXt
].
Remark (〈〈〈·〉〉〉 is invertible and depend on T and R only)
1 Note that 〈〈〈g〉〉〉 does not depend on the mean of X nor on theconditioned values y: 〈〈〈g〉〉〉 depends only on the conditioningfunctions g = [gi ]
Ni=1 and the covariance R.
2 Since gi ’s are linearly independent and R is positive definite,the matrix 〈〈〈g〉〉〉 is invertible.
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Orthogonal Representation
Denote by 〈〈〈g〉〉〉 the matrix
〈〈〈g〉〉〉ij := 〈〈〈gi , gj〉〉〉 := Cov[∫ T
0gi (t) dXt ,
∫ T
0gj(t) dXt
].
Remark (〈〈〈·〉〉〉 is invertible and depend on T and R only)
1 Note that 〈〈〈g〉〉〉 does not depend on the mean of X nor on theconditioned values y: 〈〈〈g〉〉〉 depends only on the conditioningfunctions g = [gi ]
Ni=1 and the covariance R.
2 Since gi ’s are linearly independent and R is positive definite,the matrix 〈〈〈g〉〉〉 is invertible.
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Orthogonal Representation
Denote by 〈〈〈g〉〉〉 the matrix
〈〈〈g〉〉〉ij := 〈〈〈gi , gj〉〉〉 := Cov[∫ T
0gi (t) dXt ,
∫ T
0gj(t) dXt
].
Remark (〈〈〈·〉〉〉 is invertible and depend on T and R only)
1 Note that 〈〈〈g〉〉〉 does not depend on the mean of X nor on theconditioned values y: 〈〈〈g〉〉〉 depends only on the conditioningfunctions g = [gi ]
Ni=1 and the covariance R.
2 Since gi ’s are linearly independent and R is positive definite,the matrix 〈〈〈g〉〉〉 is invertible.
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Orthogonal Representation
Theorem (Orthogonal Representation)
The generalized Gaussian bridge X g;y can be represented as
X g;yt = Xt − 〈〈〈1t , g〉〉〉>〈〈〈g〉〉〉−1
(∫ T
0g(u) dXu − y
). (2)
Moreover, any generalized Gaussian bridge X g;y is a Gaussianprocess with
E[X g;yt
]= m(t)−
〈〈〈1t , g〉〉〉>〈〈〈g〉〉〉−1(∫ T
0g(u)dm(u)− y
),
Cov[X g;yt ,X g;y
s
]= 〈〈〈1t , 1s〉〉〉 − 〈〈〈1t , g〉〉〉>〈〈〈g〉〉〉−1〈〈〈1s , g〉〉〉.
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Orthogonal Representation
Theorem (Orthogonal Representation)
The generalized Gaussian bridge X g;y can be represented as
X g;yt = Xt − 〈〈〈1t , g〉〉〉>〈〈〈g〉〉〉−1
(∫ T
0g(u) dXu − y
). (2)
Moreover, any generalized Gaussian bridge X g;y is a Gaussianprocess with
E[X g;yt
]= m(t)−
〈〈〈1t , g〉〉〉>〈〈〈g〉〉〉−1(∫ T
0g(u)dm(u)− y
),
Cov[X g;yt ,X g;y
s
]= 〈〈〈1t , 1s〉〉〉 − 〈〈〈1t , g〉〉〉>〈〈〈g〉〉〉−1〈〈〈1s , g〉〉〉.
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Orthogonal Representation
Corollary (Mean-Conditioning Invariance)
Let X be a centered Gaussian process with X0 = 0 and let m be afunction of bounded variation. Let X g := X g;0 be a bridge wherethe conditional functionals are conditioned to zero. Then
(X + m)g;yt = X g
t +
(m(t)− 〈〈〈1t , g〉〉〉>〈〈〈g〉〉〉−1
∫ T
0g(u) dm(u)
)+ 〈〈〈1t , g〉〉〉>〈〈〈g〉〉〉−1y.
Remark (Normalization)
Because of the corollary above we can, and will, assume in whatfollows that m = 0 and y = 0.
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Orthogonal Representation
Corollary (Mean-Conditioning Invariance)
Let X be a centered Gaussian process with X0 = 0 and let m be afunction of bounded variation. Let X g := X g;0 be a bridge wherethe conditional functionals are conditioned to zero. Then
(X + m)g;yt = X g
t +
(m(t)− 〈〈〈1t , g〉〉〉>〈〈〈g〉〉〉−1
∫ T
0g(u) dm(u)
)+ 〈〈〈1t , g〉〉〉>〈〈〈g〉〉〉−1y.
Remark (Normalization)
Because of the corollary above we can, and will, assume in whatfollows that m = 0 and y = 0.
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Outline
1 Generalized (linearly conditioned) Bridges
2 Orthogonal Representation
3 Canonical Representation for Martingales
4 Prediction-Invertible Processes
5 Canonical Representation forPrediction-Invertible Processes
6 Open Questions
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Canonical Representation for Martingales
The problem with the orthogonal bridge representation (2) of X g;y
is that in order to construct it at any point t ∈ [0,T ) one needsthe whole path of the underlying process X up to time T . In thissection and the following sections we construct a bridgerepresentation that is canonical in the following sense:
Definition (Canonical Representation)
The bridge X g;y is of canonical representation if, for allt ∈ [0,T ), X g;y
t ∈ Lt(X ) and Xt ∈ Lt(X g;y).
Here Lt(Y ) is the closed linear subspace of L2(Ω,F ,P) generatedby the random variables Ys , s ≤ t.
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Canonical Representation for Martingales
The problem with the orthogonal bridge representation (2) of X g;y
is that in order to construct it at any point t ∈ [0,T ) one needsthe whole path of the underlying process X up to time T . In thissection and the following sections we construct a bridgerepresentation that is canonical in the following sense:
Definition (Canonical Representation)
The bridge X g;y is of canonical representation if, for allt ∈ [0,T ), X g;y
t ∈ Lt(X ) and Xt ∈ Lt(X g;y).
Here Lt(Y ) is the closed linear subspace of L2(Ω,F ,P) generatedby the random variables Ys , s ≤ t.
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Canonical Representation for Martingales
The problem with the orthogonal bridge representation (2) of X g;y
is that in order to construct it at any point t ∈ [0,T ) one needsthe whole path of the underlying process X up to time T . In thissection and the following sections we construct a bridgerepresentation that is canonical in the following sense:
Definition (Canonical Representation)
The bridge X g;y is of canonical representation if, for allt ∈ [0,T ), X g;y
t ∈ Lt(X ) and Xt ∈ Lt(X g;y).
Here Lt(Y ) is the closed linear subspace of L2(Ω,F ,P) generatedby the random variables Ys , s ≤ t.
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Canonical Representation for Martingales
Remark (Gaussian Specialities)
1 Since the conditional laws of Gaussian processes are Gaussianand Gaussian spaces are linear, the assumptions X g;y
t ∈ Lt(X )and Xt ∈ Lt(X g;y) are the same as assuming that X g;y
t isFXt -measurable and Xt is FX g;y
t -measurable (and,consequently, FX
t = FX g;y
t ). This fact is very special toGaussian processes. Indeed, in general conditioned processessuch as generalized bridges are not linear transformations ofthe underlying process.
2 Another “Gaussian fact” here is that the bridges are Gaussian.In general conditioned processes do not belong to the samefamily of distributions as the original process.
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Canonical Representation for Martingales
Remark (Gaussian Specialities)
1 Since the conditional laws of Gaussian processes are Gaussianand Gaussian spaces are linear, the assumptions X g;y
t ∈ Lt(X )and Xt ∈ Lt(X g;y) are the same as assuming that X g;y
t isFXt -measurable and Xt is FX g;y
t -measurable (and,consequently, FX
t = FX g;y
t ). This fact is very special toGaussian processes. Indeed, in general conditioned processessuch as generalized bridges are not linear transformations ofthe underlying process.
2 Another “Gaussian fact” here is that the bridges are Gaussian.In general conditioned processes do not belong to the samefamily of distributions as the original process.
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Canonical Representation for Martingales
We shall require that the restricted measures Pg,yt := Pg;y|Ft and
Pt := P|Ft are equivalent for all t < T (they are obviously singularfor t = T ). To this end we assume that the matrix
〈〈〈g〉〉〉ij(t) := E[(
G iT (X )− G i
t (X ))(
G jT (X )− G j
t (X ))]
= E[∫ T
tgi (s) dXs
∫ T
tgj(s) dXs
]is invertible for all t < T .
Remark (On Notation)
In the previous section we considered the matrix 〈〈〈g〉〉〉, but fromnow on we consider the function 〈〈〈g〉〉〉(·). Their connection is ofcourse 〈〈〈g〉〉〉 = 〈〈〈g〉〉〉(0). We hope that is overloading of notationdoes not cause confusion to the reader.
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Canonical Representation for Martingales
We shall require that the restricted measures Pg,yt := Pg;y|Ft and
Pt := P|Ft are equivalent for all t < T (they are obviously singularfor t = T ). To this end we assume that the matrix
〈〈〈g〉〉〉ij(t) := E[(
G iT (X )− G i
t (X ))(
G jT (X )− G j
t (X ))]
= E[∫ T
tgi (s) dXs
∫ T
tgj(s) dXs
]is invertible for all t < T .
Remark (On Notation)
In the previous section we considered the matrix 〈〈〈g〉〉〉, but fromnow on we consider the function 〈〈〈g〉〉〉(·). Their connection is ofcourse 〈〈〈g〉〉〉 = 〈〈〈g〉〉〉(0). We hope that is overloading of notationdoes not cause confusion to the reader.
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Canonical Representation for Martingales
Let now M be a Gaussian martingale with strictly increasingbracket 〈M〉 and M0 = 0.
Remark (Bracket and Non-Degeneracy)
Note that the bracket is strictly increasing if and only if thecovariance R is positive definite. Indeed, for Gaussian martingaleswe have R(t, s) = Var(Mt∧s) = 〈M〉t∧s .
Define a Volterra kernel
`g(t, s) := −g>(t) 〈〈〈g〉〉〉−1(t) g(s). (3)
Note that the kernel `g depends on the process M through itscovariance 〈〈〈·, ·〉〉〉, and in the Gaussian martingale case we have
〈〈〈g〉〉〉ij(t) =
∫ T
tgi (s)gj(s)d〈M〉s .
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Canonical Representation for Martingales
Let now M be a Gaussian martingale with strictly increasingbracket 〈M〉 and M0 = 0.
Remark (Bracket and Non-Degeneracy)
Note that the bracket is strictly increasing if and only if thecovariance R is positive definite. Indeed, for Gaussian martingaleswe have R(t, s) = Var(Mt∧s) = 〈M〉t∧s .
Define a Volterra kernel
`g(t, s) := −g>(t) 〈〈〈g〉〉〉−1(t) g(s). (3)
Note that the kernel `g depends on the process M through itscovariance 〈〈〈·, ·〉〉〉, and in the Gaussian martingale case we have
〈〈〈g〉〉〉ij(t) =
∫ T
tgi (s)gj(s)d〈M〉s .
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Canonical Representation for Martingales
Let now M be a Gaussian martingale with strictly increasingbracket 〈M〉 and M0 = 0.
Remark (Bracket and Non-Degeneracy)
Note that the bracket is strictly increasing if and only if thecovariance R is positive definite. Indeed, for Gaussian martingaleswe have R(t, s) = Var(Mt∧s) = 〈M〉t∧s .
Define a Volterra kernel
`g(t, s) := −g>(t) 〈〈〈g〉〉〉−1(t) g(s). (3)
Note that the kernel `g depends on the process M through itscovariance 〈〈〈·, ·〉〉〉, and in the Gaussian martingale case we have
〈〈〈g〉〉〉ij(t) =
∫ T
tgi (s)gj(s)d〈M〉s .
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Canonical Representation for Martingales
Let now M be a Gaussian martingale with strictly increasingbracket 〈M〉 and M0 = 0.
Remark (Bracket and Non-Degeneracy)
Note that the bracket is strictly increasing if and only if thecovariance R is positive definite. Indeed, for Gaussian martingaleswe have R(t, s) = Var(Mt∧s) = 〈M〉t∧s .
Define a Volterra kernel
`g(t, s) := −g>(t) 〈〈〈g〉〉〉−1(t) g(s). (3)
Note that the kernel `g depends on the process M through itscovariance 〈〈〈·, ·〉〉〉, and in the Gaussian martingale case we have
〈〈〈g〉〉〉ij(t) =
∫ T
tgi (s)gj(s) d〈M〉s .
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Canonical Representation for Martingales
The following lemma is the key observation in finding the canonicalgeneralized bridge representation. Actually, it is a multivariateversion of Proposition 6 of Gasbarra, S. and Valkeila (2007).
Lemma (Radon-Nikodym for Bridges)
Let `g be given by (3) and let M be a continuous Gaussianmartingale with strictly increasing bracket 〈M〉 and M0 = 0. Then
logdPg
t
dPt
=
∫ t
0
∫ s
0`g(s, u)dMudMs −
1
2
∫ t
0
(∫ s
0`g(s, u) dMu
)2
d〈M〉s .
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Canonical Representation for Martingales
The following lemma is the key observation in finding the canonicalgeneralized bridge representation. Actually, it is a multivariateversion of Proposition 6 of Gasbarra, S. and Valkeila (2007).
Lemma (Radon-Nikodym for Bridges)
Let `g be given by (3) and let M be a continuous Gaussianmartingale with strictly increasing bracket 〈M〉 and M0 = 0. Then
logdPg
t
dPt
=
∫ t
0
∫ s
0`g(s, u)dMudMs −
1
2
∫ t
0
(∫ s
0`g(s, u) dMu
)2
d〈M〉s .
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Canonical Representation for Martingales
Proof.
Let p(·;µ,Σ) be the Gaussian density on RN and let
αgt (dy) := P
[GT (M) ∈ dy
∣∣∣ FMt
].
By the Bayes’ formula and the martingale property
dPgt
dPt=
dαgt
dαg0
(0) =p(
0; Gt(M), 〈〈〈g〉〉〉(t))
p(
0; G0(M), 〈〈〈g〉〉〉(0)) .
Denote F (t,Mt) := −12G>t 〈〈〈g〉〉〉−1(0)Gt and the claim follows by
using the Ito formula.
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Canonical Representation for Martingales
Proof.
Let p(·;µ,Σ) be the Gaussian density on RN and let
αgt (dy) := P
[GT (M) ∈ dy
∣∣∣ FMt
].
By the Bayes’ formula and the martingale property
dPgt
dPt=
dαgt
dαg0
(0) =p(
0; Gt(M), 〈〈〈g〉〉〉(t))
p(
0; G0(M), 〈〈〈g〉〉〉(0)) .
Denote F (t,Mt) := −12G>t 〈〈〈g〉〉〉−1(0)Gt and the claim follows by
using the Ito formula.
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Canonical Representation for Martingales
Proof.
Let p(·;µ,Σ) be the Gaussian density on RN and let
αgt (dy) := P
[GT (M) ∈ dy
∣∣∣ FMt
].
By the Bayes’ formula and the martingale property
dPgt
dPt=
dαgt
dαg0
(0) =p(
0; Gt(M), 〈〈〈g〉〉〉(t))
p(
0; G0(M), 〈〈〈g〉〉〉(0)) .
Denote F (t,Mt) := −12G>t 〈〈〈g〉〉〉−1(0)Gt and the claim follows by
using the Ito formula.
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Canonical Representation for Martingales
Corollary (Semimartingale Decomposition)
The canonical bridge representation Mg satisfies the stochasticdifferential equation
dMt = dMgt −
∫ t
0`g(t, s) dMg
s d〈M〉t , (4)
where `g is given by (3). Moreover 〈M〉 = 〈Mg〉.
Proof.
The claim follows by using the Girsanov’s theorem.
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Canonical Representation for Martingales
Corollary (Semimartingale Decomposition)
The canonical bridge representation Mg satisfies the stochasticdifferential equation
dMt = dMgt −
∫ t
0`g(t, s) dMg
s d〈M〉t , (4)
where `g is given by (3). Moreover 〈M〉 = 〈Mg〉.
Proof.
The claim follows by using the Girsanov’s theorem.
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Canonical Representation for Martingales
Remark (Equivalence and Singularity)
Note that ∫ T−
0
∫ s
0`g(s, u)2 du ds <∞.
In view of (4) this means that M and Mg are equivalent on [0,T ).Indeed, equation (4) can be viewed as the Hitsudarepresentation between two equivalent Gaussian processes.
Also note that ∫ T
0
∫ s
0`g(s, u)2 du ds =∞
meaning, by the Hitsuda representation theorem, that M and Mg
are singular on [0,T ].
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Canonical Representation for Martingales
Remark (Equivalence and Singularity)
Note that ∫ T−
0
∫ s
0`g(s, u)2 du ds <∞.
In view of (4) this means that M and Mg are equivalent on [0,T ).Indeed, equation (4) can be viewed as the Hitsudarepresentation between two equivalent Gaussian processes.
Also note that ∫ T
0
∫ s
0`g(s, u)2 du ds =∞
meaning, by the Hitsuda representation theorem, that M and Mg
are singular on [0,T ].
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Canonical Representation for Martingales
Next we solve the stochastic differential equation (4) of Corollary6. In general, solving a Volterra–Stieltjes equation like (4) ina closed form is difficult.
Of course, the general theory of Volterra equations suggests thatthe solution will be of the form (6) of next theorem, where `∗g isthe resolvent kernel of `g determined by the resolventequation (7) given below.
Also, the general theory suggests that the resolvent kernel can becalculated implicitly by using the Neumann series. In our casethe kernel `g is a quadratic form that factorizes in its argument.This allows us to calculate the resolvent `∗g explicitly as (5) below.
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Canonical Representation for Martingales
Next we solve the stochastic differential equation (4) of Corollary6. In general, solving a Volterra–Stieltjes equation like (4) ina closed form is difficult.
Of course, the general theory of Volterra equations suggests thatthe solution will be of the form (6) of next theorem, where `∗g isthe resolvent kernel of `g determined by the resolventequation (7) given below.
Also, the general theory suggests that the resolvent kernel can becalculated implicitly by using the Neumann series. In our casethe kernel `g is a quadratic form that factorizes in its argument.This allows us to calculate the resolvent `∗g explicitly as (5) below.
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Canonical Representation for Martingales
Next we solve the stochastic differential equation (4) of Corollary6. In general, solving a Volterra–Stieltjes equation like (4) ina closed form is difficult.
Of course, the general theory of Volterra equations suggests thatthe solution will be of the form (6) of next theorem, where `∗g isthe resolvent kernel of `g determined by the resolventequation (7) given below.
Also, the general theory suggests that the resolvent kernel can becalculated implicitly by using the Neumann series. In our casethe kernel `g is a quadratic form that factorizes in its argument.This allows us to calculate the resolvent `∗g explicitly as (5) below.
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Canonical Representation for Martingales
Theorem (Solution to an SDE)
Let s ≤ t ∈ [0,T ]. Define the Volterra kernel
`∗g(t, s) := −`g(t, s)|〈〈〈g〉〉〉|(t)
|〈〈〈g〉〉〉|(s)(5)
= |〈〈〈g〉〉〉|(t)g>(t)〈〈〈g〉〉〉−1(t)g(s)
|〈〈〈g〉〉〉|(s).
Then the bridge Mg has the canonical representation
dMgt = dMt −
∫ t
0`∗g(t, s) dMs d〈M〉t , (6)
i.e., (6) is the solution to (4).
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Canonical Representation for Martingales
Proof.
Equation (6) is the solution to (4) if the kernel `∗g satisfies theresolvent equation
`g(t, s) + `∗g(t, s) =
∫ t
s`g(t, u)`∗g(u, s)d〈M〉u. (7)
Indeed, suppose (6) is the solution to (4). This means that
dMt =
(dMt −
∫ t
0`∗g(t, s)dMs d〈M〉t
)−∫ t
0`g(t, s)
(dMs −
∫ s
0`∗g(s, u)dMu d〈M〉s
)d〈M〉t ,
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Canonical Representation for Martingales
Proof.
Equation (6) is the solution to (4) if the kernel `∗g satisfies theresolvent equation
`g(t, s) + `∗g(t, s) =
∫ t
s`g(t, u)`∗g(u, s)d〈M〉u. (7)
Indeed, suppose (6) is the solution to (4). This means that
dMt =
(dMt −
∫ t
0`∗g(t, s)dMs d〈M〉t
)−∫ t
0`g(t, s)
(dMs −
∫ s
0`∗g(s, u)dMu d〈M〉s
)d〈M〉t ,
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Canonical Representation for Martingales
Proof.
In the integral form, by using the Fubini’s theorem, this means that
Mt = Mt −∫ t
0
∫ t
s`∗g(u, s) d〈M〉udMs
−∫ t
0
∫ t
s`g(u, s) d〈M〉udMs
+
∫ t
0
∫ t
s
∫ s
u`g(s, v)`∗g(v , u)d〈M〉v d〈M〉udMs .
The resolvent criterion (7) follows by identifying the integrands inthe d〈M〉udMs -integrals above.
Now it is straightforward, but tedious, to check that the resolventequation holds. We omit the details.
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Canonical Representation for Martingales
Proof.
In the integral form, by using the Fubini’s theorem, this means that
Mt = Mt −∫ t
0
∫ t
s`∗g(u, s) d〈M〉udMs
−∫ t
0
∫ t
s`g(u, s) d〈M〉udMs
+
∫ t
0
∫ t
s
∫ s
u`g(s, v)`∗g(v , u)d〈M〉v d〈M〉udMs .
The resolvent criterion (7) follows by identifying the integrands inthe d〈M〉udMs -integrals above.
Now it is straightforward, but tedious, to check that the resolventequation holds. We omit the details.
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Canonical Representation for Martingales
Proof.
In the integral form, by using the Fubini’s theorem, this means that
Mt = Mt −∫ t
0
∫ t
s`∗g(u, s) d〈M〉udMs
−∫ t
0
∫ t
s`g(u, s) d〈M〉udMs
+
∫ t
0
∫ t
s
∫ s
u`g(s, v)`∗g(v , u)d〈M〉v d〈M〉udMs .
The resolvent criterion (7) follows by identifying the integrands inthe d〈M〉udMs -integrals above.
Now it is straightforward, but tedious, to check that the resolventequation holds. We omit the details.
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Outline
1 Generalized (linearly conditioned) Bridges
2 Orthogonal Representation
3 Canonical Representation for Martingales
4 Prediction-Invertible Processes
5 Canonical Representation forPrediction-Invertible Processes
6 Open Questions
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Prediction-Invertible Processes
Let us now consider a Gaussian process X that is not a martingale.For a (Gaussian) process X its prediction martingale is theprocess X defined as
Xt = E[XT
∣∣FXt
].
Since for Gaussian processes Xt ∈ Lt(X ), we may write, at leastformally, that
Xt =
∫ t
0p(t, s)dXs ,
where the abstract kernel p depends also on T (since X dependson T ).
In the next definition we assume, among other things, that thekernel p exists as a real, and not only formal, function. We alsoassume that the kernel p is invertible.
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Prediction-Invertible Processes
Let us now consider a Gaussian process X that is not a martingale.For a (Gaussian) process X its prediction martingale is theprocess X defined as
Xt = E[XT
∣∣FXt
].
Since for Gaussian processes Xt ∈ Lt(X ), we may write, at leastformally, that
Xt =
∫ t
0p(t, s)dXs ,
where the abstract kernel p depends also on T (since X dependson T ).
In the next definition we assume, among other things, that thekernel p exists as a real, and not only formal, function. We alsoassume that the kernel p is invertible.
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Prediction-Invertible Processes
Let us now consider a Gaussian process X that is not a martingale.For a (Gaussian) process X its prediction martingale is theprocess X defined as
Xt = E[XT
∣∣FXt
].
Since for Gaussian processes Xt ∈ Lt(X ), we may write, at leastformally, that
Xt =
∫ t
0p(t, s)dXs ,
where the abstract kernel p depends also on T (since X dependson T ).
In the next definition we assume, among other things, that thekernel p exists as a real, and not only formal, function. We alsoassume that the kernel p is invertible.
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Prediction-Invertible Processes
Definition (Prediction-Invertibility)
A Gaussian process X is prediction-invertible if there exists akernel p such that its prediction martingale X is continuous, canbe represented as
Xt =
∫ t
0p(t, s)dXs ,
and there exists an inverse kernel p−1 such that, for all t ∈ [0,T ],p−1(t, ·) ∈ L2([0,T ],d〈X 〉) and X can be recovered from X by
Xt =
∫ t
0p−1(t, s) dXs .
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Prediction-Invertible Processes
Definition (Prediction-Invertibility)
A Gaussian process X is prediction-invertible if there exists akernel p such that its prediction martingale X is continuous, canbe represented as
Xt =
∫ t
0p(t, s)dXs ,
and there exists an inverse kernel p−1 such that, for all t ∈ [0,T ],p−1(t, ·) ∈ L2([0,T ], d〈X 〉) and X can be recovered from X by
Xt =
∫ t
0p−1(t, s) dXs .
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Prediction-Invertible Processes
Remark
In general it seems to be a difficult problem to determine whethera Gaussian process is prediction-invertible or not. In the discretetime non-degenerate case all Gaussian processes areprediction-invertible. In continuous time the situation is moredifficult, as the next example illustrates.
Nevertheless, we can immediately see that we must have
R(t, s) =
∫ t∧s
0p−1(t, u) p−1(s, u) d〈X 〉u,
where〈X 〉u = Var (E [XT |Fu]) .
However, this criterion does not seem to be very helpful in practice.
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Prediction-Invertible Processes
Remark
In general it seems to be a difficult problem to determine whethera Gaussian process is prediction-invertible or not. In the discretetime non-degenerate case all Gaussian processes areprediction-invertible. In continuous time the situation is moredifficult, as the next example illustrates.
Nevertheless, we can immediately see that we must have
R(t, s) =
∫ t∧s
0p−1(t, u) p−1(s, u) d〈X 〉u,
where〈X 〉u = Var (E [XT |Fu]) .
However, this criterion does not seem to be very helpful in practice.
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Prediction-Invertible Processes
Example (Fine Points of Prediction-Invertibility)
Consider the Gaussian slope Xt = tξ, t ∈ [0,T ], where ξ is astandard normal random variable. Now, if we consider the “rawfiltration” GXt = σ(Xs ; s ≤ t), then X is not prediction invertible.Indeed, then X0 = 0 but Xt = XT , if t ∈ (0,T ]. So, X is notcontinuous. On the other hand, the augmented filtration is simplyFXt = σ(ξ) for all t ∈ [0,T ]. So, X = XT . Note, however, that in
both cases the slope X can be recovered from the predictionmartingale: Xt = t
T Xt .
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Prediction-Invertible Processes
We want to represent integrals w.r.t. X w.r.t. X .
Definition (Linear Extensions of Kernels)
Let X be prediction-invertible. Let P and P−1 extend the relationsP[1t ] = p(t, ·) and P−1[1t ] = p−1(t, ·) linearly.
Lemma (Abstract vs. Concrete Wiener Integrals)
For f and g nice enough∫ T
0f (t) dXt =
∫ T
0P−1[f ](t) dXt ,∫ T
0g(t) dXt =
∫ T
0P[g ](t) dXt .
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Prediction-Invertible Processes
We want to represent integrals w.r.t. X w.r.t. X .
Definition (Linear Extensions of Kernels)
Let X be prediction-invertible. Let P and P−1 extend the relationsP[1t ] = p(t, ·) and P−1[1t ] = p−1(t, ·) linearly.
Lemma (Abstract vs. Concrete Wiener Integrals)
For f and g nice enough∫ T
0f (t) dXt =
∫ T
0P−1[f ](t) dXt ,∫ T
0g(t) dXt =
∫ T
0P[g ](t) dXt .
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Prediction-Invertible Processes
We want to represent integrals w.r.t. X w.r.t. X .
Definition (Linear Extensions of Kernels)
Let X be prediction-invertible. Let P and P−1 extend the relationsP[1t ] = p(t, ·) and P−1[1t ] = p−1(t, ·) linearly.
Lemma (Abstract vs. Concrete Wiener Integrals)
For f and g nice enough∫ T
0f (t) dXt =
∫ T
0P−1[f ](t) dXt ,∫ T
0g(t) dXt =
∫ T
0P[g ](t) dXt .
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Outline
1 Generalized (linearly conditioned) Bridges
2 Orthogonal Representation
3 Canonical Representation for Martingales
4 Prediction-Invertible Processes
5 Canonical Representation forPrediction-Invertible Processes
6 Open Questions
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Canonical Representation forPrediction-Invertible Processes
The next (our main) theorem follows basically by rewriting∫ T
0g(t)dXt =
∫ T
0P−1[g](t)dX (t) = 0.
Theorem (Canonical Representation)
Let X be prediction-invertible Gaussian process. Then
X gt = Xt −
∫ t
0
∫ t
sp−1(t, u)P
[ˆ∗g(u, ·)
](s)d〈X 〉u dXs ,
where g = P−1[g ] and
ˆ∗g(u, v) = |〈〈〈g〉〉〉X |(u)g>(u)(〈〈〈g〉〉〉X )
−1(u)
g(v)
|〈〈〈g〉〉〉X |(v).
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Canonical Representation forPrediction-Invertible Processes
The next (our main) theorem follows basically by rewriting∫ T
0g(t)dXt =
∫ T
0P−1[g](t)dX (t) = 0.
Theorem (Canonical Representation)
Let X be prediction-invertible Gaussian process. Then
X gt = Xt −
∫ t
0
∫ t
sp−1(t, u)P
[ˆ∗g(u, ·)
](s) d〈X 〉u dXs ,
where g = P−1[g ] and
ˆ∗g(u, v) = |〈〈〈g〉〉〉X |(u)g>(u)(〈〈〈g〉〉〉X )
−1(u)
g(v)
|〈〈〈g〉〉〉X |(v).
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Canonical Representation forPrediction-Invertible Processes
Proof.
In the prediction-martingale level we have
dX gs = dXs −
∫ s
0
ˆ∗g(s, u)dXu d〈X 〉s .
Now, by operating with p−1 and P we get
X gt = Xt −
∫ t
0p−1(t, s)
(∫ s
0
ˆ∗g(s, u) dXu
)d〈X 〉s
= Xt −∫ t
0p−1(t, s)
∫ s
0P[
ˆ∗g(s, ·)
](u)dXu d〈X 〉s .
Finally, the claim follows by using the Fubini’s theorem.
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Canonical Representation forPrediction-Invertible Processes
Proof.
In the prediction-martingale level we have
dX gs = dXs −
∫ s
0
ˆ∗g(s, u)dXu d〈X 〉s .
Now, by operating with p−1 and P we get
X gt = Xt −
∫ t
0p−1(t, s)
(∫ s
0
ˆ∗g(s, u) dXu
)d〈X 〉s
= Xt −∫ t
0p−1(t, s)
∫ s
0P[
ˆ∗g(s, ·)
](u)dXu d〈X 〉s .
Finally, the claim follows by using the Fubini’s theorem.
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Outline
1 Generalized (linearly conditioned) Bridges
2 Orthogonal Representation
3 Canonical Representation for Martingales
4 Prediction-Invertible Processes
5 Canonical Representation forPrediction-Invertible Processes
6 Open Questions
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Open Questions
1 What kind of condition (for the covariance) would implyprediction-invertibility?
2 How to calculate the prediction-inversion kernel p−1(t, s)?
3 What would be an example of continuous Gaussian processthat is not prediction-invertible?
4 What would ensure that Ls(X ) ( Lt(X ) for s < t andLs(X ) = ∩t>sLt(X ) and Lt(X ) = Lt(X g)? Would this beenough for prediction-invertiblity?
– The End –Thank you for your attention!
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Open Questions
1 What kind of condition (for the covariance) would implyprediction-invertibility?
2 How to calculate the prediction-inversion kernel p−1(t, s)?
3 What would be an example of continuous Gaussian processthat is not prediction-invertible?
4 What would ensure that Ls(X ) ( Lt(X ) for s < t andLs(X ) = ∩t>sLt(X ) and Lt(X ) = Lt(X g)? Would this beenough for prediction-invertiblity?
– The End –Thank you for your attention!
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Open Questions
1 What kind of condition (for the covariance) would implyprediction-invertibility?
2 How to calculate the prediction-inversion kernel p−1(t, s)?
3 What would be an example of continuous Gaussian processthat is not prediction-invertible?
4 What would ensure that Ls(X ) ( Lt(X ) for s < t andLs(X ) = ∩t>sLt(X ) and Lt(X ) = Lt(X g)? Would this beenough for prediction-invertiblity?
– The End –Thank you for your attention!
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Open Questions
1 What kind of condition (for the covariance) would implyprediction-invertibility?
2 How to calculate the prediction-inversion kernel p−1(t, s)?
3 What would be an example of continuous Gaussian processthat is not prediction-invertible?
4 What would ensure that Ls(X ) ( Lt(X ) for s < t andLs(X ) = ∩t>sLt(X ) and Lt(X ) = Lt(X g)? Would this beenough for prediction-invertiblity?
– The End –Thank you for your attention!
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Open Questions
1 What kind of condition (for the covariance) would implyprediction-invertibility?
2 How to calculate the prediction-inversion kernel p−1(t, s)?
3 What would be an example of continuous Gaussian processthat is not prediction-invertible?
4 What would ensure that Ls(X ) ( Lt(X ) for s < t andLs(X ) = ∩t>sLt(X ) and Lt(X ) = Lt(X g)? Would this beenough for prediction-invertiblity?
– The End –Thank you for your attention!
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