Interchanging limit and expectation in Ito Isometry proof
$begingroup$
Let $phi_n$ be a sequence of elementary functions and $f$ a function satisfying:
$f(t,omega)$ is measurable wrt $mathcal{B}times mathcal{F}$
$f(t,omega)$ is $mathcal{F}_t$-adapted- $Eleft[int_S^T f(t,omega) dt right] < infty$
and suppose $E[int_S^T (f(t,omega) - phi_n(t,omega))^2 dt] to 0$.
The Ito integral of $phi_n$ is defined by
$$int_S^T phi_n (t,omega) dB_t (omega) = sum_j e_{n,j}(omega) Delta B(t_j)$$
and the Ito integral of $f$ is defined by
$$int_S^T f(t,omega) dB_t (omega) = lim_{n to infty} int_S^T phi_n (t,omega) dB_t (omega)$$
where the limit is with respect to $L^2 (P)$.
My question is about the proof of Ito's isometry, that
$$E left( int_S^T f(t,omega) dB_t (omega) right)^2 = E int_S^T f(t,omega)^2 dt$$
This clearly holds for elementary $phi_n$ due to linearity of expectation and the fact that $phi_n$ is $mathcal{F}_t$-adapted. But why does the above definition mean it holds for general $f$ satisfying $1,2,3$. Specifically, writing:
$$Eleft[left(int_S^T f(t,omega) dB_t (omega)right)^2 right]= Eleft[ lim_{nto infty} left(int_S^T phi_n(t,omega) dB_t (omega)right)^2 right]$$
What is the justification for interchanging the expectation with the $lim_{ntoinfty}$ in order to get the elementary Ito isometry to hold for the more general $f$?
real-analysis probability
asked Dec 11 '18 at 8:02
XiaomiXiaomi
1,057115
$endgroup$
add a comment |
$begingroup$
Let $phi_n$ be a sequence of elementary functions and $f$ a function satisfying:
$f(t,omega)$ is measurable wrt $mathcal{B}times mathcal{F}$
$f(t,omega)$ is $mathcal{F}_t$-adapted- $Eleft[int_S^T f(t,omega) dt right] < infty$
and suppose $E[int_S^T (f(t,omega) - phi_n(t,omega))^2 dt] to 0$.
The Ito integral of $phi_n$ is defined by
$$int_S^T phi_n (t,omega) dB_t (omega) = sum_j e_{n,j}(omega) Delta B(t_j)$$
and the Ito integral of $f$ is defined by
$$int_S^T f(t,omega) dB_t (omega) = lim_{n to infty} int_S^T phi_n (t,omega) dB_t (omega)$$
where the limit is with respect to $L^2 (P)$.
My question is about the proof of Ito's isometry, that
$$E left( int_S^T f(t,omega) dB_t (omega) right)^2 = E int_S^T f(t,omega)^2 dt$$
This clearly holds for elementary $phi_n$ due to linearity of expectation and the fact that $phi_n$ is $mathcal{F}_t$-adapted. But why does the above definition mean it holds for general $f$ satisfying $1,2,3$. Specifically, writing:
$$Eleft[left(int_S^T f(t,omega) dB_t (omega)right)^2 right]= Eleft[ lim_{nto infty} left(int_S^T phi_n(t,omega) dB_t (omega)right)^2 right]$$
What is the justification for interchanging the expectation with the $lim_{ntoinfty}$ in order to get the elementary Ito isometry to hold for the more general $f$?
real-analysis probability
asked Dec 11 '18 at 8:02
XiaomiXiaomi
1,057115
$endgroup$
1
$begingroup$
If $X_n to X$ in $L^2$ then $lim_{n to infty} mathbb{E}(X_n^2) = mathbb{E}(X^2)$. Use this for $$X := int_S^T phi_n(t) , dB_t$$ and $$X:= int_S^T f(t) , dB_t.$$
$endgroup$
– saz
Dec 11 '18 at 8:20
$begingroup$
Noted, thanks! I will make sure to remember that implication
$endgroup$
– Xiaomi
Dec 11 '18 at 10:08
add a comment |
$begingroup$
Let $phi_n$ be a sequence of elementary functions and $f$ a function satisfying:
$f(t,omega)$ is measurable wrt $mathcal{B}times mathcal{F}$
$f(t,omega)$ is $mathcal{F}_t$-adapted- $Eleft[int_S^T f(t,omega) dt right] < infty$
and suppose $E[int_S^T (f(t,omega) - phi_n(t,omega))^2 dt] to 0$.
The Ito integral of $phi_n$ is defined by
$$int_S^T phi_n (t,omega) dB_t (omega) = sum_j e_{n,j}(omega) Delta B(t_j)$$
and the Ito integral of $f$ is defined by
$$int_S^T f(t,omega) dB_t (omega) = lim_{n to infty} int_S^T phi_n (t,omega) dB_t (omega)$$
where the limit is with respect to $L^2 (P)$.
My question is about the proof of Ito's isometry, that
$$E left( int_S^T f(t,omega) dB_t (omega) right)^2 = E int_S^T f(t,omega)^2 dt$$
This clearly holds for elementary $phi_n$ due to linearity of expectation and the fact that $phi_n$ is $mathcal{F}_t$-adapted. But why does the above definition mean it holds for general $f$ satisfying $1,2,3$. Specifically, writing:
$$Eleft[left(int_S^T f(t,omega) dB_t (omega)right)^2 right]= Eleft[ lim_{nto infty} left(int_S^T phi_n(t,omega) dB_t (omega)right)^2 right]$$
What is the justification for interchanging the expectation with the $lim_{ntoinfty}$ in order to get the elementary Ito isometry to hold for the more general $f$?
real-analysis probability
asked Dec 11 '18 at 8:02
XiaomiXiaomi
1,057115
$endgroup$
Let $phi_n$ be a sequence of elementary functions and $f$ a function satisfying:
$f(t,omega)$ is measurable wrt $mathcal{B}times mathcal{F}$
$f(t,omega)$ is $mathcal{F}_t$-adapted- $Eleft[int_S^T f(t,omega) dt right] < infty$
and suppose $E[int_S^T (f(t,omega) - phi_n(t,omega))^2 dt] to 0$.
The Ito integral of $phi_n$ is defined by
$$int_S^T phi_n (t,omega) dB_t (omega) = sum_j e_{n,j}(omega) Delta B(t_j)$$
and the Ito integral of $f$ is defined by
$$int_S^T f(t,omega) dB_t (omega) = lim_{n to infty} int_S^T phi_n (t,omega) dB_t (omega)$$
where the limit is with respect to $L^2 (P)$.
My question is about the proof of Ito's isometry, that
$$E left( int_S^T f(t,omega) dB_t (omega) right)^2 = E int_S^T f(t,omega)^2 dt$$
This clearly holds for elementary $phi_n$ due to linearity of expectation and the fact that $phi_n$ is $mathcal{F}_t$-adapted. But why does the above definition mean it holds for general $f$ satisfying $1,2,3$. Specifically, writing:
$$Eleft[left(int_S^T f(t,omega) dB_t (omega)right)^2 right]= Eleft[ lim_{nto infty} left(int_S^T phi_n(t,omega) dB_t (omega)right)^2 right]$$
What is the justification for interchanging the expectation with the $lim_{ntoinfty}$ in order to get the elementary Ito isometry to hold for the more general $f$?
real-analysis probability
real-analysis probability
asked Dec 11 '18 at 8:02
XiaomiXiaomi
1,057115
asked Dec 11 '18 at 8:02
XiaomiXiaomi
1,057115
asked Dec 11 '18 at 8:02
XiaomiXiaomi
1,057115
asked Dec 11 '18 at 8:02
XiaomiXiaomi
1,057115
asked Dec 11 '18 at 8:02
XiaomiXiaomi
1,057115
1,057115
1
$begingroup$
If $X_n to X$ in $L^2$ then $lim_{n to infty} mathbb{E}(X_n^2) = mathbb{E}(X^2)$. Use this for $$X := int_S^T phi_n(t) , dB_t$$ and $$X:= int_S^T f(t) , dB_t.$$
$endgroup$
– saz
Dec 11 '18 at 8:20
$begingroup$
Noted, thanks! I will make sure to remember that implication
$endgroup$
– Xiaomi
Dec 11 '18 at 10:08
add a comment |
1
$begingroup$
If $X_n to X$ in $L^2$ then $lim_{n to infty} mathbb{E}(X_n^2) = mathbb{E}(X^2)$. Use this for $$X := int_S^T phi_n(t) , dB_t$$ and $$X:= int_S^T f(t) , dB_t.$$
$endgroup$
– saz
Dec 11 '18 at 8:20
$begingroup$
Noted, thanks! I will make sure to remember that implication
$endgroup$
– Xiaomi
Dec 11 '18 at 10:08
1
1
$begingroup$
If $X_n to X$ in $L^2$ then $lim_{n to infty} mathbb{E}(X_n^2) = mathbb{E}(X^2)$. Use this for $$X := int_S^T phi_n(t) , dB_t$$ and $$X:= int_S^T f(t) , dB_t.$$
$endgroup$
– saz
Dec 11 '18 at 8:20
$begingroup$
If $X_n to X$ in $L^2$ then $lim_{n to infty} mathbb{E}(X_n^2) = mathbb{E}(X^2)$. Use this for $$X := int_S^T phi_n(t) , dB_t$$ and $$X:= int_S^T f(t) , dB_t.$$
$endgroup$
– saz
Dec 11 '18 at 8:20
$begingroup$
Noted, thanks! I will make sure to remember that implication
$endgroup$
– Xiaomi
Dec 11 '18 at 10:08
$begingroup$
Noted, thanks! I will make sure to remember that implication
$endgroup$
– Xiaomi
Dec 11 '18 at 10:08
add a comment |
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$begingroup$
If $X_n to X$ in $L^2$ then $lim_{n to infty} mathbb{E}(X_n^2) = mathbb{E}(X^2)$. Use this for $$X := int_S^T phi_n(t) , dB_t$$ and $$X:= int_S^T f(t) , dB_t.$$
$endgroup$
– saz
Dec 11 '18 at 8:20
$begingroup$
Noted, thanks! I will make sure to remember that implication
$endgroup$
– Xiaomi
Dec 11 '18 at 10:08