What is a stationary solution to a SPDE?
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I'm reading Hairer's notes on SPDEs: http://www.hairer.org/notes/SPDEs.pdf
He says on page 6 that "the stationary solution to the stochastic heat equation is Gaussian free field". He never defines what "stationary solution" means and it's not obvious. Does it mean that the solution doesn't depend on time? Since the noise depends on time this can never happen.
What is the stationary solution to a SPDE?
stochastic-processes stochastic-calculus stochastic-analysis stochastic-pde
asked Dec 30 '18 at 2:33
RptoughsRptoughs
62
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add a comment |
$begingroup$
I'm reading Hairer's notes on SPDEs: http://www.hairer.org/notes/SPDEs.pdf
He says on page 6 that "the stationary solution to the stochastic heat equation is Gaussian free field". He never defines what "stationary solution" means and it's not obvious. Does it mean that the solution doesn't depend on time? Since the noise depends on time this can never happen.
What is the stationary solution to a SPDE?
stochastic-processes stochastic-calculus stochastic-analysis stochastic-pde
asked Dec 30 '18 at 2:33
RptoughsRptoughs
62
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1
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Loosely, it means that if you have some initial distribution and let it evolve according to the SDE, at any point in time the distribution will be the same.
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– tch
Dec 30 '18 at 2:40
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@TylerChen So the distribution is stationary. Do you have a source for this?
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– Rptoughs
Dec 30 '18 at 2:41
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Yes. Except in trivial cases, it only makes sense in the context of distributions (as you noted a single trajectory will generally depend on time because of the randomness). This book has a decent intro in 6.9.
$endgroup$
– tch
Dec 30 '18 at 5:36
add a comment |
$begingroup$
I'm reading Hairer's notes on SPDEs: http://www.hairer.org/notes/SPDEs.pdf
He says on page 6 that "the stationary solution to the stochastic heat equation is Gaussian free field". He never defines what "stationary solution" means and it's not obvious. Does it mean that the solution doesn't depend on time? Since the noise depends on time this can never happen.
What is the stationary solution to a SPDE?
stochastic-processes stochastic-calculus stochastic-analysis stochastic-pde
asked Dec 30 '18 at 2:33
RptoughsRptoughs
62
$endgroup$
I'm reading Hairer's notes on SPDEs: http://www.hairer.org/notes/SPDEs.pdf
He says on page 6 that "the stationary solution to the stochastic heat equation is Gaussian free field". He never defines what "stationary solution" means and it's not obvious. Does it mean that the solution doesn't depend on time? Since the noise depends on time this can never happen.
What is the stationary solution to a SPDE?
stochastic-processes stochastic-calculus stochastic-analysis stochastic-pde
stochastic-processes stochastic-calculus stochastic-analysis stochastic-pde
asked Dec 30 '18 at 2:33
RptoughsRptoughs
62
asked Dec 30 '18 at 2:33
RptoughsRptoughs
62
asked Dec 30 '18 at 2:33
RptoughsRptoughs
62
asked Dec 30 '18 at 2:33
RptoughsRptoughs
62
asked Dec 30 '18 at 2:33
RptoughsRptoughs
62
62
1
$begingroup$
Loosely, it means that if you have some initial distribution and let it evolve according to the SDE, at any point in time the distribution will be the same.
$endgroup$
– tch
Dec 30 '18 at 2:40
$begingroup$
@TylerChen So the distribution is stationary. Do you have a source for this?
$endgroup$
– Rptoughs
Dec 30 '18 at 2:41
$begingroup$
Yes. Except in trivial cases, it only makes sense in the context of distributions (as you noted a single trajectory will generally depend on time because of the randomness). This book has a decent intro in 6.9.
$endgroup$
– tch
Dec 30 '18 at 5:36
add a comment |
1
$begingroup$
Loosely, it means that if you have some initial distribution and let it evolve according to the SDE, at any point in time the distribution will be the same.
$endgroup$
– tch
Dec 30 '18 at 2:40
$begingroup$
@TylerChen So the distribution is stationary. Do you have a source for this?
$endgroup$
– Rptoughs
Dec 30 '18 at 2:41
$begingroup$
Yes. Except in trivial cases, it only makes sense in the context of distributions (as you noted a single trajectory will generally depend on time because of the randomness). This book has a decent intro in 6.9.
$endgroup$
– tch
Dec 30 '18 at 5:36
1
1
$begingroup$
Loosely, it means that if you have some initial distribution and let it evolve according to the SDE, at any point in time the distribution will be the same.
$endgroup$
– tch
Dec 30 '18 at 2:40
$begingroup$
Loosely, it means that if you have some initial distribution and let it evolve according to the SDE, at any point in time the distribution will be the same.
$endgroup$
– tch
Dec 30 '18 at 2:40
$begingroup$
@TylerChen So the distribution is stationary. Do you have a source for this?
$endgroup$
– Rptoughs
Dec 30 '18 at 2:41
$begingroup$
@TylerChen So the distribution is stationary. Do you have a source for this?
$endgroup$
– Rptoughs
Dec 30 '18 at 2:41
$begingroup$
Yes. Except in trivial cases, it only makes sense in the context of distributions (as you noted a single trajectory will generally depend on time because of the randomness). This book has a decent intro in 6.9.
$endgroup$
– tch
Dec 30 '18 at 5:36
$begingroup$
Yes. Except in trivial cases, it only makes sense in the context of distributions (as you noted a single trajectory will generally depend on time because of the randomness). This book has a decent intro in 6.9.
$endgroup$
– tch
Dec 30 '18 at 5:36
add a comment |
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$begingroup$
Loosely, it means that if you have some initial distribution and let it evolve according to the SDE, at any point in time the distribution will be the same.
$endgroup$
– tch
Dec 30 '18 at 2:40
$begingroup$
@TylerChen So the distribution is stationary. Do you have a source for this?
$endgroup$
– Rptoughs
Dec 30 '18 at 2:41
$begingroup$
Yes. Except in trivial cases, it only makes sense in the context of distributions (as you noted a single trajectory will generally depend on time because of the randomness). This book has a decent intro in 6.9.
$endgroup$
– tch
Dec 30 '18 at 5:36