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

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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










0












$begingroup$


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




  1. $(t,omega) mapsto f(t,omega)$ is $mathcalB otimes mathcalF$ - measurable, where $mathcalB$ is the Borel $sigma$-algebra on $[0,infty)$,


  2. $f$ is $mathcalF_t$-adapted,


  3. $E[ int_S^T f(t,omega)^2 dt ] < infty$.

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
$$
int_S^T (h - g_n)^2 dt = 0.~as~nrightarrow infty, for~each~omega
$$










share|cite|improve this question











$endgroup$











  • $begingroup$
    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















0












$begingroup$


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




  1. $(t,omega) mapsto f(t,omega)$ is $mathcalB otimes mathcalF$ - measurable, where $mathcalB$ is the Borel $sigma$-algebra on $[0,infty)$,


  2. $f$ is $mathcalF_t$-adapted,


  3. $E[ int_S^T f(t,omega)^2 dt ] < infty$.

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
$$
int_S^T (h - g_n)^2 dt = 0.~as~nrightarrow infty, for~each~omega
$$










share|cite|improve this question











$endgroup$











  • $begingroup$
    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













0












0








0


0



$begingroup$


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




  1. $(t,omega) mapsto f(t,omega)$ is $mathcalB otimes mathcalF$ - measurable, where $mathcalB$ is the Borel $sigma$-algebra on $[0,infty)$,


  2. $f$ is $mathcalF_t$-adapted,


  3. $E[ int_S^T f(t,omega)^2 dt ] < infty$.

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
$$
int_S^T (h - g_n)^2 dt = 0.~as~nrightarrow infty, for~each~omega
$$










share|cite|improve this question











$endgroup$




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




  1. $(t,omega) mapsto f(t,omega)$ is $mathcalB otimes mathcalF$ - measurable, where $mathcalB$ is the Borel $sigma$-algebra on $[0,infty)$,


  2. $f$ is $mathcalF_t$-adapted,


  3. $E[ int_S^T f(t,omega)^2 dt ] < infty$.

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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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
















  • $begingroup$
    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










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