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:

0p−1T (t, s) dE[XT |FX

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

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 ) =

0g(t) dXt =

[∫ T

0gi (t)dXt

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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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 ) =

0g(t) dXt =

[∫ T

0gi (t)dXt

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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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

0gi (t) dXt = yi

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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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

0gi (t) dXt = yi

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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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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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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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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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

6 Open Questions

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Orthogonal Representation

Denote by 〈〈〈g〉〉〉 the matrix

〈〈〈g〉〉〉ij := 〈〈〈gi , gj〉〉〉 := Cov[∫ T

0gi (t) dXt ,

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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〈〈〈g〉〉〉ij := 〈〈〈gi , gj〉〉〉 := Cov[∫ T

0gi (t) dXt ,

0gj(t) dXt

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〈〈〈g〉〉〉ij := 〈〈〈gi , gj〉〉〉 := Cov[∫ T

0gi (t) dXt ,

0gj(t) dXt

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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

]= 〈〈〈1t , 1s〉〉〉 − 〈〈〈1t , g〉〉〉>〈〈〈g〉〉〉−1〈〈〈1s , g〉〉〉.

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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

]= 〈〈〈1t , 1s〉〉〉 − 〈〈〈1t , g〉〉〉>〈〈〈g〉〉〉−1〈〈〈1s , g〉〉〉.

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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

(m(t)− 〈〈〈1t , g〉〉〉>〈〈〈g〉〉〉−1

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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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

(m(t)− 〈〈〈1t , g〉〉〉>〈〈〈g〉〉〉−1

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

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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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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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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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

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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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

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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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) =

tgi (s)gj(s)d〈M〉s .

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`g(t, s) := −g>(t) 〈〈〈g〉〉〉−1(t) g(s). (3)

〈〈〈g〉〉〉ij(t) =

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`g(t, s) := −g>(t) 〈〈〈g〉〉〉−1(t) g(s). (3)

〈〈〈g〉〉〉ij(t) =

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`g(t, s) := −g>(t) 〈〈〈g〉〉〉−1(t) g(s). (3)

〈〈〈g〉〉〉ij(t) =

tgi (s)gj(s) d〈M〉s .

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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

0`g(s, u)dMudMs −

(∫ s

0`g(s, u) dMu

d〈M〉s .

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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

0`g(s, u)dMudMs −

(∫ s

0`g(s, u) dMu

d〈M〉s .

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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

(0) =p(

0; Gt(M), 〈〈〈g〉〉〉(t))

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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Proof.

αgt (dy) := P

[GT (M) ∈ dy

∣∣∣ FMt

(0) =p(

0; Gt(M), 〈〈〈g〉〉〉(t))

0; G0(M), 〈〈〈g〉〉〉(0)) .

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Proof.

αgt (dy) := P

[GT (M) ∈ dy

∣∣∣ FMt

(0) =p(

0; Gt(M), 〈〈〈g〉〉〉(t))

0; G0(M), 〈〈〈g〉〉〉(0)) .

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Corollary (Semimartingale Decomposition)

The canonical bridge representation Mg satisfies the stochasticdifferential equation

dMt = dMgt −

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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Corollary (Semimartingale Decomposition)

The canonical bridge representation Mg satisfies the stochasticdifferential equation

dMt = dMgt −

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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Remark (Equivalence and Singularity)

Note that ∫ T−

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`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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Remark (Equivalence and Singularity)

Note that ∫ T−

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`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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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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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 −

0`∗g(t, s) dMs d〈M〉t , (6)

i.e., (6) is the solution to (4).

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Proof.

Equation (6) is the solution to (4) if the kernel `∗g satisfies theresolvent equation

`g(t, s) + `∗g(t, s) =

s`g(t, u)`∗g(u, s)d〈M〉u. (7)

Indeed, suppose (6) is the solution to (4). This means that

(dMt −

0`∗g(t, s)dMs d〈M〉t

)−∫ t

0`g(t, s)

(dMs −

0`∗g(s, u)dMu d〈M〉s

)d〈M〉t ,

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Proof.

Equation (6) is the solution to (4) if the kernel `∗g satisfies theresolvent equation

`g(t, s) + `∗g(t, s) =

s`g(t, u)`∗g(u, s)d〈M〉u. (7)

Indeed, suppose (6) is the solution to (4). This means that

(dMt −

0`∗g(t, s)dMs d〈M〉t

)−∫ t

0`g(t, s)

(dMs −

0`∗g(s, u)dMu d〈M〉s

)d〈M〉t ,

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Proof.

In the integral form, by using the Fubini’s theorem, this means that

Mt = Mt −∫ t

s`∗g(u, s) d〈M〉udMs

−∫ t

s`g(u, s) d〈M〉udMs

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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Proof.

Mt = Mt −∫ t

−∫ t

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Proof.

Mt = Mt −∫ t

−∫ t

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Outline

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

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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Xt = E[XT

∣∣FXt

0p(t, s)dXs ,

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Xt = E[XT

∣∣FXt

0p(t, s)dXs ,

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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

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

0p−1(t, s) dXs .

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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

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

0p−1(t, s) dXs .

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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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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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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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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 =

0P−1[f ](t) dXt ,∫ T

0g(t) dXt =

0P[g ](t) dXt .

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0f (t) dXt =

0P−1[f ](t) dXt ,∫ T

0g(t) dXt =

0P[g ](t) dXt .

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0f (t) dXt =

0P−1[f ](t) dXt ,∫ T

0g(t) dXt =

0P[g ](t) dXt .

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Outline

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 =

0P−1[g](t)dX (t) = 0.

Theorem (Canonical Representation)

Let X be prediction-invertible Gaussian process. Then

X gt = Xt −

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〉〉〉X |(v).

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The next (our main) theorem follows basically by rewriting∫ T

0g(t)dXt =

0P−1[g](t)dX (t) = 0.

Theorem (Canonical Representation)

Let X be prediction-invertible Gaussian process. Then

X gt = Xt −

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〉〉〉X |(v).

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Proof.

In the prediction-martingale level we have

dX gs = dXs −

ˆ∗g(s, u)dXu d〈X 〉s .

Now, by operating with p−1 and P we get

X gt = Xt −

0p−1(t, s)

(∫ s

ˆ∗g(s, u) dXu

)d〈X 〉s

= Xt −∫ t

0p−1(t, s)

ˆ∗g(s, ·)

](u)dXu d〈X 〉s .

Finally, the claim follows by using the Fubini’s theorem.

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Proof.

In the prediction-martingale level we have

dX gs = dXs −

ˆ∗g(s, u)dXu d〈X 〉s .

Now, by operating with p−1 and P we get

X gt = Xt −

0p−1(t, s)

(∫ s

ˆ∗g(s, u) dXu

)d〈X 〉s

= Xt −∫ t

0p−1(t, s)

ˆ∗g(s, ·)

](u)dXu d〈X 〉s .

Finally, the claim follows by using the Fubini’s theorem.

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Outline

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

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Open Questions

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Open Questions

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Open Questions

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