the continuity in the proof of Ito integral The 2019 Stack Overflow Developer Survey Results Are In Unicorn Meta Zoo #1: Why another podcast? Announcing the arrival of Valued Associate #679: Cesar ManaraContinuity of the Ito integralIto Integral representation for bounded claimsDefinition of Ito IntegralConstruction of Ito integralExtending a property of Ito integrals of elementary functionsPull out measurable function out of a stochastic integralA confusion about adapted processes and filtrationsExpected Value, Brownian Motion, Ito IntegralHow to bound increments of Ito integral?Justification of interchanging Expectation and Limit in Ito Integral Approximation
Working through the single responsibility principle (SRP) in Python when calls are expensive
Can each chord in a progression create its own key?
Can the DM override racial traits?
Using dividends to reduce short term capital gains?
60's-70's movie: home appliances revolting against the owners
What does "spokes" mean in this context?
Was credit for the black hole image misappropriated?
Homework question about an engine pulling a train
Is 'stolen' appropriate word?
Simulating Exploding Dice
"... to apply for a visa" or "... and applied for a visa"?
The following signatures were invalid: EXPKEYSIG 1397BC53640DB551
How to politely respond to generic emails requesting a PhD/job in my lab? Without wasting too much time
How to determine omitted units in a publication
Why doesn't shell automatically fix "useless use of cat"?
Could an empire control the whole planet with today's comunication methods?
What other Star Trek series did the main TNG cast show up in?
Do warforged have souls?
How do you keep chess fun when your opponent constantly beats you?
Single author papers against my advisor's will?
Is there a way to generate uniformly distributed points on a sphere from a fixed amount of random real numbers per point?
How did the crowd guess the pentatonic scale in Bobby McFerrin's presentation?
How to read αἱμύλιος or when to aspirate
Sub-subscripts in strings cause different spacings than subscripts
the continuity in the proof of Ito integral
The 2019 Stack Overflow Developer Survey Results Are In
Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar ManaraContinuity of the Ito integralIto Integral representation for bounded claimsDefinition of Ito IntegralConstruction of Ito integralExtending a property of Ito integrals of elementary functionsPull out measurable function out of a stochastic integralA confusion about adapted processes and filtrationsExpected Value, Brownian Motion, Ito IntegralHow to bound increments of Ito integral?Justification of interchanging Expectation and Limit in Ito Integral Approximation
$begingroup$
This is in regards to constructing the Ito integral, specifically the second step of approximating bounded functions by bounded and continuous functions.
Let $(Omega, mathcalF, P)$ be a probability space and let $V = V(S,T)$ be the class of functions $f: [0,infty) times Omega to mathbbR$ such that
$(t,omega) mapsto f(t,omega)$ is $mathcalB otimes mathcalF$ - measurable, where $mathcalB$ is the Borel $sigma$-algebra on $[0,infty)$,
$f$ is $mathcalF_t$-adapted,
$E[ int_S^T f(t,omega)^2 dt ] < infty$.
Oksendal's 6th ed. of "Stochastic Differential Equations," on page 27, states:
enter image description here
Let $h in V$ be bounded. Then there exist bounded functions $g_nin V$ such that $g(cdot,omega)$ continuous for each $omega$ and $n$,and
$$
lim_n to infty Eleft[ int_S^T (h - g_n)^2 dtright] = 0.
$$
$Proof$:Let $|h(t,omega)|leq M$ for all $(t,omega)$.For each $n$ let $psi_n$ be a non-negative,continuous function on R such that
(i)let $psi_n=0$ for $xleq -1/n$ and $xgeq 0$
(ii)$int_-infty^inftypsi_n(x)dx=1$
Define
$$
g_n(t,omega)=int_0^t psi_n(s-t)h(s,omega)ds.
$$
Then $g_n(.,omega)$ is continuous for each $omega$ and $|g_n(t,omega)|leq M$. Since $hin mathcalV$ we can show that $g_n(t,.)$ is $mathcalF_t$-measurable for all $t$.Moreover,
$$
int_S^T (h - g_n)^2 dt = 0.~as~nrightarrow infty,~for~ each~ omega
$$
In the proof, the author constructs a convolution to make the bounded function h continuous and bounded. I'm not sure why $g_n(.,w)$ is continuous for each $omega$,and the most important,why
$$
int_S^T (h - g_n)^2 dt = 0.~as~nrightarrow infty, for~each~omega
$$
stochastic-calculus brownian-motion
$endgroup$
add a comment |
$begingroup$
This is in regards to constructing the Ito integral, specifically the second step of approximating bounded functions by bounded and continuous functions.
Let $(Omega, mathcalF, P)$ be a probability space and let $V = V(S,T)$ be the class of functions $f: [0,infty) times Omega to mathbbR$ such that
$(t,omega) mapsto f(t,omega)$ is $mathcalB otimes mathcalF$ - measurable, where $mathcalB$ is the Borel $sigma$-algebra on $[0,infty)$,
$f$ is $mathcalF_t$-adapted,
$E[ int_S^T f(t,omega)^2 dt ] < infty$.
Oksendal's 6th ed. of "Stochastic Differential Equations," on page 27, states:
enter image description here
Let $h in V$ be bounded. Then there exist bounded functions $g_nin V$ such that $g(cdot,omega)$ continuous for each $omega$ and $n$,and
$$
lim_n to infty Eleft[ int_S^T (h - g_n)^2 dtright] = 0.
$$
$Proof$:Let $|h(t,omega)|leq M$ for all $(t,omega)$.For each $n$ let $psi_n$ be a non-negative,continuous function on R such that
(i)let $psi_n=0$ for $xleq -1/n$ and $xgeq 0$
(ii)$int_-infty^inftypsi_n(x)dx=1$
Define
$$
g_n(t,omega)=int_0^t psi_n(s-t)h(s,omega)ds.
$$
Then $g_n(.,omega)$ is continuous for each $omega$ and $|g_n(t,omega)|leq M$. Since $hin mathcalV$ we can show that $g_n(t,.)$ is $mathcalF_t$-measurable for all $t$.Moreover,
$$
int_S^T (h - g_n)^2 dt = 0.~as~nrightarrow infty,~for~ each~ omega
$$
In the proof, the author constructs a convolution to make the bounded function h continuous and bounded. I'm not sure why $g_n(.,w)$ is continuous for each $omega$,and the most important,why
$$
int_S^T (h - g_n)^2 dt = 0.~as~nrightarrow infty, for~each~omega
$$
stochastic-calculus brownian-motion
$endgroup$
$begingroup$
1) $g_n$ is a convolution which means that it inherits regularity properties from $psi_n$. Look up results on continuity of convolutions. 2) The convergence holds for a suitable subsequence $g_n_k$ of $g_n$ (recall that $L^2(mathbbP)$ convergence implies a.s. convergence of a subsequence).
$endgroup$
– saz
Apr 8 at 9:32
$begingroup$
@saz I'm sorry but for 2), do you mean the $(h-g_n)^2rightarrow 0$ in $L^2(P)$? Could you be a little more detailed?
$endgroup$
– user8281063
Apr 8 at 13:22
$begingroup$
Ah, sorry, I got something wrong about 2), forget what I wrote.
$endgroup$
– saz
Apr 8 at 13:34
add a comment |
$begingroup$
This is in regards to constructing the Ito integral, specifically the second step of approximating bounded functions by bounded and continuous functions.
Let $(Omega, mathcalF, P)$ be a probability space and let $V = V(S,T)$ be the class of functions $f: [0,infty) times Omega to mathbbR$ such that
$(t,omega) mapsto f(t,omega)$ is $mathcalB otimes mathcalF$ - measurable, where $mathcalB$ is the Borel $sigma$-algebra on $[0,infty)$,
$f$ is $mathcalF_t$-adapted,
$E[ int_S^T f(t,omega)^2 dt ] < infty$.
Oksendal's 6th ed. of "Stochastic Differential Equations," on page 27, states:
enter image description here
Let $h in V$ be bounded. Then there exist bounded functions $g_nin V$ such that $g(cdot,omega)$ continuous for each $omega$ and $n$,and
$$
lim_n to infty Eleft[ int_S^T (h - g_n)^2 dtright] = 0.
$$
$Proof$:Let $|h(t,omega)|leq M$ for all $(t,omega)$.For each $n$ let $psi_n$ be a non-negative,continuous function on R such that
(i)let $psi_n=0$ for $xleq -1/n$ and $xgeq 0$
(ii)$int_-infty^inftypsi_n(x)dx=1$
Define
$$
g_n(t,omega)=int_0^t psi_n(s-t)h(s,omega)ds.
$$
Then $g_n(.,omega)$ is continuous for each $omega$ and $|g_n(t,omega)|leq M$. Since $hin mathcalV$ we can show that $g_n(t,.)$ is $mathcalF_t$-measurable for all $t$.Moreover,
$$
int_S^T (h - g_n)^2 dt = 0.~as~nrightarrow infty,~for~ each~ omega
$$
In the proof, the author constructs a convolution to make the bounded function h continuous and bounded. I'm not sure why $g_n(.,w)$ is continuous for each $omega$,and the most important,why
$$
int_S^T (h - g_n)^2 dt = 0.~as~nrightarrow infty, for~each~omega
$$
stochastic-calculus brownian-motion
$endgroup$
This is in regards to constructing the Ito integral, specifically the second step of approximating bounded functions by bounded and continuous functions.
Let $(Omega, mathcalF, P)$ be a probability space and let $V = V(S,T)$ be the class of functions $f: [0,infty) times Omega to mathbbR$ such that
$(t,omega) mapsto f(t,omega)$ is $mathcalB otimes mathcalF$ - measurable, where $mathcalB$ is the Borel $sigma$-algebra on $[0,infty)$,
$f$ is $mathcalF_t$-adapted,
$E[ int_S^T f(t,omega)^2 dt ] < infty$.
Oksendal's 6th ed. of "Stochastic Differential Equations," on page 27, states:
enter image description here
Let $h in V$ be bounded. Then there exist bounded functions $g_nin V$ such that $g(cdot,omega)$ continuous for each $omega$ and $n$,and
$$
lim_n to infty Eleft[ int_S^T (h - g_n)^2 dtright] = 0.
$$
$Proof$:Let $|h(t,omega)|leq M$ for all $(t,omega)$.For each $n$ let $psi_n$ be a non-negative,continuous function on R such that
(i)let $psi_n=0$ for $xleq -1/n$ and $xgeq 0$
(ii)$int_-infty^inftypsi_n(x)dx=1$
Define
$$
g_n(t,omega)=int_0^t psi_n(s-t)h(s,omega)ds.
$$
Then $g_n(.,omega)$ is continuous for each $omega$ and $|g_n(t,omega)|leq M$. Since $hin mathcalV$ we can show that $g_n(t,.)$ is $mathcalF_t$-measurable for all $t$.Moreover,
$$
int_S^T (h - g_n)^2 dt = 0.~as~nrightarrow infty,~for~ each~ omega
$$
In the proof, the author constructs a convolution to make the bounded function h continuous and bounded. I'm not sure why $g_n(.,w)$ is continuous for each $omega$,and the most important,why
$$
int_S^T (h - g_n)^2 dt = 0.~as~nrightarrow infty, for~each~omega
$$
stochastic-calculus brownian-motion
stochastic-calculus brownian-motion
edited Apr 8 at 8:02
Bernard
124k741117
124k741117
asked Apr 8 at 6:44
user8281063user8281063
33
33
$begingroup$
1) $g_n$ is a convolution which means that it inherits regularity properties from $psi_n$. Look up results on continuity of convolutions. 2) The convergence holds for a suitable subsequence $g_n_k$ of $g_n$ (recall that $L^2(mathbbP)$ convergence implies a.s. convergence of a subsequence).
$endgroup$
– saz
Apr 8 at 9:32
$begingroup$
@saz I'm sorry but for 2), do you mean the $(h-g_n)^2rightarrow 0$ in $L^2(P)$? Could you be a little more detailed?
$endgroup$
– user8281063
Apr 8 at 13:22
$begingroup$
Ah, sorry, I got something wrong about 2), forget what I wrote.
$endgroup$
– saz
Apr 8 at 13:34
add a comment |
$begingroup$
1) $g_n$ is a convolution which means that it inherits regularity properties from $psi_n$. Look up results on continuity of convolutions. 2) The convergence holds for a suitable subsequence $g_n_k$ of $g_n$ (recall that $L^2(mathbbP)$ convergence implies a.s. convergence of a subsequence).
$endgroup$
– saz
Apr 8 at 9:32
$begingroup$
@saz I'm sorry but for 2), do you mean the $(h-g_n)^2rightarrow 0$ in $L^2(P)$? Could you be a little more detailed?
$endgroup$
– user8281063
Apr 8 at 13:22
$begingroup$
Ah, sorry, I got something wrong about 2), forget what I wrote.
$endgroup$
– saz
Apr 8 at 13:34
$begingroup$
1) $g_n$ is a convolution which means that it inherits regularity properties from $psi_n$. Look up results on continuity of convolutions. 2) The convergence holds for a suitable subsequence $g_n_k$ of $g_n$ (recall that $L^2(mathbbP)$ convergence implies a.s. convergence of a subsequence).
$endgroup$
– saz
Apr 8 at 9:32
$begingroup$
1) $g_n$ is a convolution which means that it inherits regularity properties from $psi_n$. Look up results on continuity of convolutions. 2) The convergence holds for a suitable subsequence $g_n_k$ of $g_n$ (recall that $L^2(mathbbP)$ convergence implies a.s. convergence of a subsequence).
$endgroup$
– saz
Apr 8 at 9:32
$begingroup$
@saz I'm sorry but for 2), do you mean the $(h-g_n)^2rightarrow 0$ in $L^2(P)$? Could you be a little more detailed?
$endgroup$
– user8281063
Apr 8 at 13:22
$begingroup$
@saz I'm sorry but for 2), do you mean the $(h-g_n)^2rightarrow 0$ in $L^2(P)$? Could you be a little more detailed?
$endgroup$
– user8281063
Apr 8 at 13:22
$begingroup$
Ah, sorry, I got something wrong about 2), forget what I wrote.
$endgroup$
– saz
Apr 8 at 13:34
$begingroup$
Ah, sorry, I got something wrong about 2), forget what I wrote.
$endgroup$
– saz
Apr 8 at 13:34
add a comment |
0
active
oldest
votes
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3179265%2fthe-continuity-in-the-proof-of-ito-integral%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3179265%2fthe-continuity-in-the-proof-of-ito-integral%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
1) $g_n$ is a convolution which means that it inherits regularity properties from $psi_n$. Look up results on continuity of convolutions. 2) The convergence holds for a suitable subsequence $g_n_k$ of $g_n$ (recall that $L^2(mathbbP)$ convergence implies a.s. convergence of a subsequence).
$endgroup$
– saz
Apr 8 at 9:32
$begingroup$
@saz I'm sorry but for 2), do you mean the $(h-g_n)^2rightarrow 0$ in $L^2(P)$? Could you be a little more detailed?
$endgroup$
– user8281063
Apr 8 at 13:22
$begingroup$
Ah, sorry, I got something wrong about 2), forget what I wrote.
$endgroup$
– saz
Apr 8 at 13:34