Ito isometry and convergence in $L^2 (P)$?
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I am just wondering if my understanding of the importance of Ito isometry is correct.
The Ito isometry says that for $f$ satisfying certain conditions, we have
$$Eleft[left(int_S^T f(s,omega) dB_s (omega)right)^2right] = Eleft[int_S^T f(s,omega)^2 dsright]$$
Now if we have a function $f$ and elementary functions $f_n$ such that
$$Eleft[int_S^T (f(s,omega)-f_n(s,omega))^2 dsright] to 0$$
the Ito isometry implies
$$Eleft[left(int_S^T (f(s,omega)-f_n(s,omega)) dB_s (omega)right)^2right] to 0$$
Which I think implies that
$$int_S^T f_n(s,omega) dB_s (omega) to int_S^T f(s,omega) dB_s (omega)$$
in $L^2 (P)$.
Is this understanding corret, and if so, is the importance of the Ito isometry mainly due to this result? That Ito integrals of a function $f$ can be approximated by a sequence of Ito integrals of elementary functions which converge in $L^2(P)$ to it?
real-analysis probability
asked Dec 11 '18 at 0:57
XiaomiXiaomi
1,064115
$endgroup$
add a comment |
$begingroup$
I am just wondering if my understanding of the importance of Ito isometry is correct.
The Ito isometry says that for $f$ satisfying certain conditions, we have
$$Eleft[left(int_S^T f(s,omega) dB_s (omega)right)^2right] = Eleft[int_S^T f(s,omega)^2 dsright]$$
Now if we have a function $f$ and elementary functions $f_n$ such that
$$Eleft[int_S^T (f(s,omega)-f_n(s,omega))^2 dsright] to 0$$
the Ito isometry implies
$$Eleft[left(int_S^T (f(s,omega)-f_n(s,omega)) dB_s (omega)right)^2right] to 0$$
Which I think implies that
$$int_S^T f_n(s,omega) dB_s (omega) to int_S^T f(s,omega) dB_s (omega)$$
in $L^2 (P)$.
Is this understanding corret, and if so, is the importance of the Ito isometry mainly due to this result? That Ito integrals of a function $f$ can be approximated by a sequence of Ito integrals of elementary functions which converge in $L^2(P)$ to it?
real-analysis probability
asked Dec 11 '18 at 0:57
XiaomiXiaomi
1,064115
$endgroup$
add a comment |
$begingroup$
I am just wondering if my understanding of the importance of Ito isometry is correct.
The Ito isometry says that for $f$ satisfying certain conditions, we have
$$Eleft[left(int_S^T f(s,omega) dB_s (omega)right)^2right] = Eleft[int_S^T f(s,omega)^2 dsright]$$
Now if we have a function $f$ and elementary functions $f_n$ such that
$$Eleft[int_S^T (f(s,omega)-f_n(s,omega))^2 dsright] to 0$$
the Ito isometry implies
$$Eleft[left(int_S^T (f(s,omega)-f_n(s,omega)) dB_s (omega)right)^2right] to 0$$
Which I think implies that
$$int_S^T f_n(s,omega) dB_s (omega) to int_S^T f(s,omega) dB_s (omega)$$
in $L^2 (P)$.
Is this understanding corret, and if so, is the importance of the Ito isometry mainly due to this result? That Ito integrals of a function $f$ can be approximated by a sequence of Ito integrals of elementary functions which converge in $L^2(P)$ to it?
real-analysis probability
asked Dec 11 '18 at 0:57
XiaomiXiaomi
1,064115
$endgroup$
I am just wondering if my understanding of the importance of Ito isometry is correct.
The Ito isometry says that for $f$ satisfying certain conditions, we have
$$Eleft[left(int_S^T f(s,omega) dB_s (omega)right)^2right] = Eleft[int_S^T f(s,omega)^2 dsright]$$
Now if we have a function $f$ and elementary functions $f_n$ such that
$$Eleft[int_S^T (f(s,omega)-f_n(s,omega))^2 dsright] to 0$$
the Ito isometry implies
$$Eleft[left(int_S^T (f(s,omega)-f_n(s,omega)) dB_s (omega)right)^2right] to 0$$
Which I think implies that
$$int_S^T f_n(s,omega) dB_s (omega) to int_S^T f(s,omega) dB_s (omega)$$
in $L^2 (P)$.
Is this understanding corret, and if so, is the importance of the Ito isometry mainly due to this result? That Ito integrals of a function $f$ can be approximated by a sequence of Ito integrals of elementary functions which converge in $L^2(P)$ to it?
real-analysis probability
real-analysis probability
asked Dec 11 '18 at 0:57
XiaomiXiaomi
1,064115
asked Dec 11 '18 at 0:57
XiaomiXiaomi
1,064115
asked Dec 11 '18 at 0:57
XiaomiXiaomi
1,064115
asked Dec 11 '18 at 0:57
XiaomiXiaomi
1,064115
asked Dec 11 '18 at 0:57
XiaomiXiaomi
1,064115
1,064115
add a comment |
add a comment |
1 Answer
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As I know of, Ito isometry plays a key role in constructing Ito integral from simple processes (via Cauchy sequence of simple processes.) Thus, without Ito isometry, we may not be able to define properly and rigorously what an expression "$int_S^T f(s,omega) dB_s(omega)$" means for general $f$ (since $B_s$ is not of bounded variation.)
answered Dec 11 '18 at 2:10
SongSong
14.2k1634
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1 Answer
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1 Answer
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$begingroup$
As I know of, Ito isometry plays a key role in constructing Ito integral from simple processes (via Cauchy sequence of simple processes.) Thus, without Ito isometry, we may not be able to define properly and rigorously what an expression "$int_S^T f(s,omega) dB_s(omega)$" means for general $f$ (since $B_s$ is not of bounded variation.)
answered Dec 11 '18 at 2:10
SongSong
14.2k1634
$endgroup$
add a comment |
$begingroup$
As I know of, Ito isometry plays a key role in constructing Ito integral from simple processes (via Cauchy sequence of simple processes.) Thus, without Ito isometry, we may not be able to define properly and rigorously what an expression "$int_S^T f(s,omega) dB_s(omega)$" means for general $f$ (since $B_s$ is not of bounded variation.)
answered Dec 11 '18 at 2:10
SongSong
14.2k1634
$endgroup$
add a comment |
$begingroup$
As I know of, Ito isometry plays a key role in constructing Ito integral from simple processes (via Cauchy sequence of simple processes.) Thus, without Ito isometry, we may not be able to define properly and rigorously what an expression "$int_S^T f(s,omega) dB_s(omega)$" means for general $f$ (since $B_s$ is not of bounded variation.)
answered Dec 11 '18 at 2:10
SongSong
14.2k1634
$endgroup$
As I know of, Ito isometry plays a key role in constructing Ito integral from simple processes (via Cauchy sequence of simple processes.) Thus, without Ito isometry, we may not be able to define properly and rigorously what an expression "$int_S^T f(s,omega) dB_s(omega)$" means for general $f$ (since $B_s$ is not of bounded variation.)
answered Dec 11 '18 at 2:10
SongSong
14.2k1634
answered Dec 11 '18 at 2:10
SongSong
14.2k1634
answered Dec 11 '18 at 2:10
SongSong
14.2k1634
answered Dec 11 '18 at 2:10
SongSong
14.2k1634
14.2k1634
add a comment |
add a comment |
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