Adaptedness of Solution to SPDE
$begingroup$
I'm trying to understand existence of solutions to SDPEs of the form
$$
dX = AXdt + F(t,X)dt + B(t,X)dW
$$
from the Hilbert space point of view, following Da Prato & Zabcyzk. They rely on the standard fixe point method, showing that if the interval $[0,T]$, is sufficiently small, the mapping
$$
mathcal{K}(Y)(t)= S(t)xi + int_0^t S(t-s)F(t,Y)dt + S(t-s)B(t,Y)dW
$$
has a fixed point in the space
$$
C([0,T];L^2(Omega, mathbb{P};H))
$$
equipped with the norm
$$
|X|_T^2= sup_{0leq tleq T}mathbb{E}|X(t)|_H^2.
$$
The existence argument is standard, using the sequence $X_0 =xi$, $X_{n+1} = mathcal{K}(X_n)$.
What I am curious/unsure about is the adaptedness, both of the sequence and the limiting solution. Assuming $xi$ is $mathcal{F}_0$, how can I see that:
- Each $X_n$ is adapted
- The limit, $X$, is also adapted
probability measure-theory stochastic-processes
asked Dec 21 '18 at 15:38
user2379888user2379888
24019
$endgroup$
add a comment |
$begingroup$
I'm trying to understand existence of solutions to SDPEs of the form
$$
dX = AXdt + F(t,X)dt + B(t,X)dW
$$
from the Hilbert space point of view, following Da Prato & Zabcyzk. They rely on the standard fixe point method, showing that if the interval $[0,T]$, is sufficiently small, the mapping
$$
mathcal{K}(Y)(t)= S(t)xi + int_0^t S(t-s)F(t,Y)dt + S(t-s)B(t,Y)dW
$$
has a fixed point in the space
$$
C([0,T];L^2(Omega, mathbb{P};H))
$$
equipped with the norm
$$
|X|_T^2= sup_{0leq tleq T}mathbb{E}|X(t)|_H^2.
$$
The existence argument is standard, using the sequence $X_0 =xi$, $X_{n+1} = mathcal{K}(X_n)$.
What I am curious/unsure about is the adaptedness, both of the sequence and the limiting solution. Assuming $xi$ is $mathcal{F}_0$, how can I see that:
- Each $X_n$ is adapted
- The limit, $X$, is also adapted
probability measure-theory stochastic-processes
asked Dec 21 '18 at 15:38
user2379888user2379888
24019
$endgroup$
$begingroup$
Is it possible that you forgot an integral in the definition of $mathcal{K}$?
$endgroup$
– eddie
Dec 22 '18 at 9:53
$begingroup$
do you mean against the stochastic term? I had meant that single integral to be against both the $dt$ and the $dW$. I will edit if you think that would be helpful.
$endgroup$
– user2379888
Dec 22 '18 at 17:31
1
$begingroup$
I think it follows inductively. $X_0$ is measurable as a constant. The stochastic integral is a martingale and hence $X_n$ is adapted.
$endgroup$
– eddie
Dec 22 '18 at 23:57
$begingroup$
I can see the merit to an inductive argument, but I could use some tips on filling in the details.
$endgroup$
– user2379888
Dec 28 '18 at 14:31
add a comment |
$begingroup$
I'm trying to understand existence of solutions to SDPEs of the form
$$
dX = AXdt + F(t,X)dt + B(t,X)dW
$$
from the Hilbert space point of view, following Da Prato & Zabcyzk. They rely on the standard fixe point method, showing that if the interval $[0,T]$, is sufficiently small, the mapping
$$
mathcal{K}(Y)(t)= S(t)xi + int_0^t S(t-s)F(t,Y)dt + S(t-s)B(t,Y)dW
$$
has a fixed point in the space
$$
C([0,T];L^2(Omega, mathbb{P};H))
$$
equipped with the norm
$$
|X|_T^2= sup_{0leq tleq T}mathbb{E}|X(t)|_H^2.
$$
The existence argument is standard, using the sequence $X_0 =xi$, $X_{n+1} = mathcal{K}(X_n)$.
What I am curious/unsure about is the adaptedness, both of the sequence and the limiting solution. Assuming $xi$ is $mathcal{F}_0$, how can I see that:
- Each $X_n$ is adapted
- The limit, $X$, is also adapted
probability measure-theory stochastic-processes
asked Dec 21 '18 at 15:38
user2379888user2379888
24019
$endgroup$
I'm trying to understand existence of solutions to SDPEs of the form
$$
dX = AXdt + F(t,X)dt + B(t,X)dW
$$
from the Hilbert space point of view, following Da Prato & Zabcyzk. They rely on the standard fixe point method, showing that if the interval $[0,T]$, is sufficiently small, the mapping
$$
mathcal{K}(Y)(t)= S(t)xi + int_0^t S(t-s)F(t,Y)dt + S(t-s)B(t,Y)dW
$$
has a fixed point in the space
$$
C([0,T];L^2(Omega, mathbb{P};H))
$$
equipped with the norm
$$
|X|_T^2= sup_{0leq tleq T}mathbb{E}|X(t)|_H^2.
$$
The existence argument is standard, using the sequence $X_0 =xi$, $X_{n+1} = mathcal{K}(X_n)$.
What I am curious/unsure about is the adaptedness, both of the sequence and the limiting solution. Assuming $xi$ is $mathcal{F}_0$, how can I see that:
- Each $X_n$ is adapted
- The limit, $X$, is also adapted
probability measure-theory stochastic-processes
probability measure-theory stochastic-processes
asked Dec 21 '18 at 15:38
user2379888user2379888
24019
asked Dec 21 '18 at 15:38
user2379888user2379888
24019
asked Dec 21 '18 at 15:38
user2379888user2379888
24019
asked Dec 21 '18 at 15:38
user2379888user2379888
24019
asked Dec 21 '18 at 15:38
user2379888user2379888
24019
24019
$begingroup$
Is it possible that you forgot an integral in the definition of $mathcal{K}$?
$endgroup$
– eddie
Dec 22 '18 at 9:53
$begingroup$
do you mean against the stochastic term? I had meant that single integral to be against both the $dt$ and the $dW$. I will edit if you think that would be helpful.
$endgroup$
– user2379888
Dec 22 '18 at 17:31
1
$begingroup$
I think it follows inductively. $X_0$ is measurable as a constant. The stochastic integral is a martingale and hence $X_n$ is adapted.
$endgroup$
– eddie
Dec 22 '18 at 23:57
$begingroup$
I can see the merit to an inductive argument, but I could use some tips on filling in the details.
$endgroup$
– user2379888
Dec 28 '18 at 14:31
add a comment |
$begingroup$
Is it possible that you forgot an integral in the definition of $mathcal{K}$?
$endgroup$
– eddie
Dec 22 '18 at 9:53
$begingroup$
do you mean against the stochastic term? I had meant that single integral to be against both the $dt$ and the $dW$. I will edit if you think that would be helpful.
$endgroup$
– user2379888
Dec 22 '18 at 17:31
1
$begingroup$
I think it follows inductively. $X_0$ is measurable as a constant. The stochastic integral is a martingale and hence $X_n$ is adapted.
$endgroup$
– eddie
Dec 22 '18 at 23:57
$begingroup$
I can see the merit to an inductive argument, but I could use some tips on filling in the details.
$endgroup$
– user2379888
Dec 28 '18 at 14:31
$begingroup$
Is it possible that you forgot an integral in the definition of $mathcal{K}$?
$endgroup$
– eddie
Dec 22 '18 at 9:53
$begingroup$
Is it possible that you forgot an integral in the definition of $mathcal{K}$?
$endgroup$
– eddie
Dec 22 '18 at 9:53
$begingroup$
do you mean against the stochastic term? I had meant that single integral to be against both the $dt$ and the $dW$. I will edit if you think that would be helpful.
$endgroup$
– user2379888
Dec 22 '18 at 17:31
$begingroup$
do you mean against the stochastic term? I had meant that single integral to be against both the $dt$ and the $dW$. I will edit if you think that would be helpful.
$endgroup$
– user2379888
Dec 22 '18 at 17:31
1
1
$begingroup$
I think it follows inductively. $X_0$ is measurable as a constant. The stochastic integral is a martingale and hence $X_n$ is adapted.
$endgroup$
– eddie
Dec 22 '18 at 23:57
$begingroup$
I think it follows inductively. $X_0$ is measurable as a constant. The stochastic integral is a martingale and hence $X_n$ is adapted.
$endgroup$
– eddie
Dec 22 '18 at 23:57
$begingroup$
I can see the merit to an inductive argument, but I could use some tips on filling in the details.
$endgroup$
– user2379888
Dec 28 '18 at 14:31
$begingroup$
I can see the merit to an inductive argument, but I could use some tips on filling in the details.
$endgroup$
– user2379888
Dec 28 '18 at 14:31
add a comment |
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$begingroup$
Is it possible that you forgot an integral in the definition of $mathcal{K}$?
$endgroup$
– eddie
Dec 22 '18 at 9:53
$begingroup$
do you mean against the stochastic term? I had meant that single integral to be against both the $dt$ and the $dW$. I will edit if you think that would be helpful.
$endgroup$
– user2379888
Dec 22 '18 at 17:31
1
$begingroup$
I think it follows inductively. $X_0$ is measurable as a constant. The stochastic integral is a martingale and hence $X_n$ is adapted.
$endgroup$
– eddie
Dec 22 '18 at 23:57
$begingroup$
I can see the merit to an inductive argument, but I could use some tips on filling in the details.
$endgroup$
– user2379888
Dec 28 '18 at 14:31