Why is the rate function given by Schilder's theorem infinite outside of CM space? Can we understand Schilder's theorem through CM theorem?

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Schilder's theorem from large deviations theory tells us that scaled Brownian motion $sqrtvarepsilon W_t$ on Wiener space $C_0([0,T],Bbb R^d)$ satisfies a large deviation principle with good rate function:

$$I(omega)=begincasesfrac12int_0^T |dotomega(t)|^2~dt &text if omegain mathcal H\
infty &text if omeganotinmathcal Hendcases$$

Where $mathcal H$ is the Cameron Martin space of Brownian motion (see also this).

This seems to be too coincidental. Why do we only care about the paths that are in the Cameron Martin space? Cameron martin theorem tells us that the law of Brownian motion is quasi invariant under translations by Cameron Martin directions. This theorem seems to tell us that paths outside the CM space decay very quickly in probability.

What is the connection? Is there anyway of understanding Schilder's theorem by CM theorem?

probability-theory stochastic-processes large-deviation-theory

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edited Jul 31 at 17:52

asked Jul 31 at 17:21

user223391

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up vote
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Schilder's theorem from large deviations theory tells us that scaled Brownian motion $sqrtvarepsilon W_t$ on Wiener space $C_0([0,T],Bbb R^d)$ satisfies a large deviation principle with good rate function:

$$I(omega)=begincasesfrac12int_0^T |dotomega(t)|^2~dt &text if omegain mathcal H\
infty &text if omeganotinmathcal Hendcases$$

Where $mathcal H$ is the Cameron Martin space of Brownian motion (see also this).

This seems to be too coincidental. Why do we only care about the paths that are in the Cameron Martin space? Cameron martin theorem tells us that the law of Brownian motion is quasi invariant under translations by Cameron Martin directions. This theorem seems to tell us that paths outside the CM space decay very quickly in probability.

What is the connection? Is there anyway of understanding Schilder's theorem by CM theorem?

probability-theory stochastic-processes large-deviation-theory

share|cite|improve this question

edited Jul 31 at 17:52

asked Jul 31 at 17:21

user223391

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

Schilder's theorem from large deviations theory tells us that scaled Brownian motion $sqrtvarepsilon W_t$ on Wiener space $C_0([0,T],Bbb R^d)$ satisfies a large deviation principle with good rate function:

$$I(omega)=begincasesfrac12int_0^T |dotomega(t)|^2~dt &text if omegain mathcal H\
infty &text if omeganotinmathcal Hendcases$$

Where $mathcal H$ is the Cameron Martin space of Brownian motion (see also this).

This seems to be too coincidental. Why do we only care about the paths that are in the Cameron Martin space? Cameron martin theorem tells us that the law of Brownian motion is quasi invariant under translations by Cameron Martin directions. This theorem seems to tell us that paths outside the CM space decay very quickly in probability.

What is the connection? Is there anyway of understanding Schilder's theorem by CM theorem?

probability-theory stochastic-processes large-deviation-theory

share|cite|improve this question

edited Jul 31 at 17:52

asked Jul 31 at 17:21

user223391

Schilder's theorem from large deviations theory tells us that scaled Brownian motion $sqrtvarepsilon W_t$ on Wiener space $C_0([0,T],Bbb R^d)$ satisfies a large deviation principle with good rate function:

$$I(omega)=begincasesfrac12int_0^T |dotomega(t)|^2~dt &text if omegain mathcal H\
infty &text if omeganotinmathcal Hendcases$$

Where $mathcal H$ is the Cameron Martin space of Brownian motion (see also this).

This seems to be too coincidental. Why do we only care about the paths that are in the Cameron Martin space? Cameron martin theorem tells us that the law of Brownian motion is quasi invariant under translations by Cameron Martin directions. This theorem seems to tell us that paths outside the CM space decay very quickly in probability.

What is the connection? Is there anyway of understanding Schilder's theorem by CM theorem?

probability-theory stochastic-processes large-deviation-theory

share|cite|improve this question

edited Jul 31 at 17:52

asked Jul 31 at 17:21

user223391

share|cite|improve this question

share|cite|improve this question

edited Jul 31 at 17:52

edited Jul 31 at 17:52

edited Jul 31 at 17:52

asked Jul 31 at 17:21

user223391

asked Jul 31 at 17:21

user223391

asked Jul 31 at 17:21

user223391

add a comment | 

add a comment | 

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Figured it out.

Let $(mathcal B, mu)$ be a separable Banach space $mathcal B$ with Gaussian measure $mu$.

Theorem (Cameron-Martin): for $hin mathcal B$, define the map $T_h:mathcal Bto mathcal B$ by $T_h(x)=x+h$. Then the measure $T_h^astmu$ is absolutely continuous with respect to $mu$ if and only if $hin mathcal H_mu$, the Cameron Martin space.

We have the Radon Nikodym derivative:

$$fracdT_h^ast mudmu=expleft( h^ast(x)-frac12|h|_mathcal H_mu^2right)$$

In computing the good rate function we have to apply a Girsanov-type change of measure. However this transformation only makes sense if $h in mathcal H_mu$. The expession in the $exp$ is exactly what you take the $sup$ of to get the Legendre transform (which via Gaertner-Ellis/Cramer is the good rate function).

So (without full details checked, but I believe this should be correct):

Theorem: Given a separable Banach space $mathcal B$ with centered Gaussian measure $mu$ and covariance $C_mu$, then the measures $mu_epsilon$ given by the covariance $C_mu^epsilon=epsilon C_mu$ satisfy a LDP with good rate function:

$$I(omega)=begincasesfrac12|omega|_mathcal H_mu^2&text for omega in mathcal H_mu\infty &text for omeganotin mathcal H_mu endcases$$

share|cite|improve this answer

answered Aug 1 at 18:25

user223391

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1 Answer
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active

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votes

1 Answer
1

active

oldest

votes

active

oldest

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active

oldest

votes

up vote
1
down vote

Figured it out.

Let $(mathcal B, mu)$ be a separable Banach space $mathcal B$ with Gaussian measure $mu$.

Theorem (Cameron-Martin): for $hin mathcal B$, define the map $T_h:mathcal Bto mathcal B$ by $T_h(x)=x+h$. Then the measure $T_h^astmu$ is absolutely continuous with respect to $mu$ if and only if $hin mathcal H_mu$, the Cameron Martin space.

We have the Radon Nikodym derivative:

$$fracdT_h^ast mudmu=expleft( h^ast(x)-frac12|h|_mathcal H_mu^2right)$$

In computing the good rate function we have to apply a Girsanov-type change of measure. However this transformation only makes sense if $h in mathcal H_mu$. The expession in the $exp$ is exactly what you take the $sup$ of to get the Legendre transform (which via Gaertner-Ellis/Cramer is the good rate function).

So (without full details checked, but I believe this should be correct):

Theorem: Given a separable Banach space $mathcal B$ with centered Gaussian measure $mu$ and covariance $C_mu$, then the measures $mu_epsilon$ given by the covariance $C_mu^epsilon=epsilon C_mu$ satisfy a LDP with good rate function:

$$I(omega)=begincasesfrac12|omega|_mathcal H_mu^2&text for omega in mathcal H_mu\infty &text for omeganotin mathcal H_mu endcases$$

share|cite|improve this answer

answered Aug 1 at 18:25

user223391

add a comment | 

up vote
1
down vote

Figured it out.

Let $(mathcal B, mu)$ be a separable Banach space $mathcal B$ with Gaussian measure $mu$.

Theorem (Cameron-Martin): for $hin mathcal B$, define the map $T_h:mathcal Bto mathcal B$ by $T_h(x)=x+h$. Then the measure $T_h^astmu$ is absolutely continuous with respect to $mu$ if and only if $hin mathcal H_mu$, the Cameron Martin space.

We have the Radon Nikodym derivative:

$$fracdT_h^ast mudmu=expleft( h^ast(x)-frac12|h|_mathcal H_mu^2right)$$

In computing the good rate function we have to apply a Girsanov-type change of measure. However this transformation only makes sense if $h in mathcal H_mu$. The expession in the $exp$ is exactly what you take the $sup$ of to get the Legendre transform (which via Gaertner-Ellis/Cramer is the good rate function).

So (without full details checked, but I believe this should be correct):

Theorem: Given a separable Banach space $mathcal B$ with centered Gaussian measure $mu$ and covariance $C_mu$, then the measures $mu_epsilon$ given by the covariance $C_mu^epsilon=epsilon C_mu$ satisfy a LDP with good rate function:

$$I(omega)=begincasesfrac12|omega|_mathcal H_mu^2&text for omega in mathcal H_mu\infty &text for omeganotin mathcal H_mu endcases$$

share|cite|improve this answer

answered Aug 1 at 18:25

user223391

add a comment | 

up vote
1
down vote

up vote
1
down vote

Figured it out.

Let $(mathcal B, mu)$ be a separable Banach space $mathcal B$ with Gaussian measure $mu$.

Theorem (Cameron-Martin): for $hin mathcal B$, define the map $T_h:mathcal Bto mathcal B$ by $T_h(x)=x+h$. Then the measure $T_h^astmu$ is absolutely continuous with respect to $mu$ if and only if $hin mathcal H_mu$, the Cameron Martin space.

We have the Radon Nikodym derivative:

$$fracdT_h^ast mudmu=expleft( h^ast(x)-frac12|h|_mathcal H_mu^2right)$$

In computing the good rate function we have to apply a Girsanov-type change of measure. However this transformation only makes sense if $h in mathcal H_mu$. The expession in the $exp$ is exactly what you take the $sup$ of to get the Legendre transform (which via Gaertner-Ellis/Cramer is the good rate function).

So (without full details checked, but I believe this should be correct):

Theorem: Given a separable Banach space $mathcal B$ with centered Gaussian measure $mu$ and covariance $C_mu$, then the measures $mu_epsilon$ given by the covariance $C_mu^epsilon=epsilon C_mu$ satisfy a LDP with good rate function:

$$I(omega)=begincasesfrac12|omega|_mathcal H_mu^2&text for omega in mathcal H_mu\infty &text for omeganotin mathcal H_mu endcases$$

share|cite|improve this answer

answered Aug 1 at 18:25

user223391

Figured it out.

Let $(mathcal B, mu)$ be a separable Banach space $mathcal B$ with Gaussian measure $mu$.

Theorem (Cameron-Martin): for $hin mathcal B$, define the map $T_h:mathcal Bto mathcal B$ by $T_h(x)=x+h$. Then the measure $T_h^astmu$ is absolutely continuous with respect to $mu$ if and only if $hin mathcal H_mu$, the Cameron Martin space.

We have the Radon Nikodym derivative:

$$fracdT_h^ast mudmu=expleft( h^ast(x)-frac12|h|_mathcal H_mu^2right)$$

In computing the good rate function we have to apply a Girsanov-type change of measure. However this transformation only makes sense if $h in mathcal H_mu$. The expession in the $exp$ is exactly what you take the $sup$ of to get the Legendre transform (which via Gaertner-Ellis/Cramer is the good rate function).

So (without full details checked, but I believe this should be correct):

Theorem: Given a separable Banach space $mathcal B$ with centered Gaussian measure $mu$ and covariance $C_mu$, then the measures $mu_epsilon$ given by the covariance $C_mu^epsilon=epsilon C_mu$ satisfy a LDP with good rate function:

$$I(omega)=begincasesfrac12|omega|_mathcal H_mu^2&text for omega in mathcal H_mu\infty &text for omeganotin mathcal H_mu endcases$$

share|cite|improve this answer

answered Aug 1 at 18:25

user223391

share|cite|improve this answer

share|cite|improve this answer

answered Aug 1 at 18:25

user223391

answered Aug 1 at 18:25

user223391

answered Aug 1 at 18:25

user223391

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add a comment | 

 

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