Why is an elementary Ito integral necessarily continuous?
$begingroup$
So I am working/reading through a proof that general Ito integrals have continuous versions. So that
$$I_f(t) = int_0^t f(s,omega)dB_s(omega)$$
Has a continuous version in $t$.
The proof I am reading does this by considering a sequence of elementary functions $phi_n$ that converge to $f$ in $L^2 (P)$. The author defines
$$I_n (t) = int_0^t phi_n (s,omega) dB_s (omega)$$
and states that $I_n (t)$ is continuous, but does not give the justification.
Here, $phi_n$ is assumed to take the form
$$phi_n = sum_j e_j^{(n)} (omega)times mathbb{1}_{left( t in[t_j,t_{j+1})right) }$$
My question is why must $I_n (t)$ be continuous in $t$ when we do not know this to be the case for the more general $f$ (though $f$ is also restricted to being $mathcal{F}_t$ adapted etc).
It is assumed throughout the book that the process $B_t$ is a continuous version of Brownian motion, so I don't find it difficult to believe, but I don't understand why it automatically holds for elementary functions and not the more general $f$?
edit
Following Saz's comment. So to show it's continuous for elementary functions, can we write:
$$|I_n (t) - I_n (s) | leq varepsilon$$
$$iff left| sum_{t_j in [s,t]} e_j^{(n)} Delta B_{t_j} right| leq varepsilon$$
And so we have
$$left| sum_{t_j in [s,t]} e_j^{(n)} Delta B_{t_j}right| leq |t-s| max_{t_j in [s,t]} |e_j^{(n)}| max_{t_j in [s,t]} |Delta B_j|$$
And so choosing
$$|t-s| leq varepsilon/ max_{t_j in [s,t]} |e_j^{(n)}| max_{t_j in [s,t]} |Delta B_j| = delta$$
It follows that if $|t-s|leq delta$ then $|I_n(t)-I_n(s)|leq varepsilon$
Is this correct?
probability continuity stochastic-calculus stochastic-integrals
asked Dec 12 '18 at 3:47
XiaomiXiaomi
1,057115
$endgroup$
add a comment |
$begingroup$
So I am working/reading through a proof that general Ito integrals have continuous versions. So that
$$I_f(t) = int_0^t f(s,omega)dB_s(omega)$$
Has a continuous version in $t$.
The proof I am reading does this by considering a sequence of elementary functions $phi_n$ that converge to $f$ in $L^2 (P)$. The author defines
$$I_n (t) = int_0^t phi_n (s,omega) dB_s (omega)$$
and states that $I_n (t)$ is continuous, but does not give the justification.
Here, $phi_n$ is assumed to take the form
$$phi_n = sum_j e_j^{(n)} (omega)times mathbb{1}_{left( t in[t_j,t_{j+1})right) }$$
My question is why must $I_n (t)$ be continuous in $t$ when we do not know this to be the case for the more general $f$ (though $f$ is also restricted to being $mathcal{F}_t$ adapted etc).
It is assumed throughout the book that the process $B_t$ is a continuous version of Brownian motion, so I don't find it difficult to believe, but I don't understand why it automatically holds for elementary functions and not the more general $f$?
edit
Following Saz's comment. So to show it's continuous for elementary functions, can we write:
$$|I_n (t) - I_n (s) | leq varepsilon$$
$$iff left| sum_{t_j in [s,t]} e_j^{(n)} Delta B_{t_j} right| leq varepsilon$$
And so we have
$$left| sum_{t_j in [s,t]} e_j^{(n)} Delta B_{t_j}right| leq |t-s| max_{t_j in [s,t]} |e_j^{(n)}| max_{t_j in [s,t]} |Delta B_j|$$
And so choosing
$$|t-s| leq varepsilon/ max_{t_j in [s,t]} |e_j^{(n)}| max_{t_j in [s,t]} |Delta B_j| = delta$$
It follows that if $|t-s|leq delta$ then $|I_n(t)-I_n(s)|leq varepsilon$
Is this correct?
probability continuity stochastic-calculus stochastic-integrals
asked Dec 12 '18 at 3:47
XiaomiXiaomi
1,057115
$endgroup$
$begingroup$
For elementary functions you can calculate the stochastic integral explicitly and check that it continuous in $t$. For general functions this approach doesn't work since it is not possible to calculate the integral explicitly...
$endgroup$
– saz
Dec 12 '18 at 6:29
$begingroup$
Thanks Saz, that makes sense. Sorry, I really struggle with this sort of stuff. I have added to the end of my post, is it correct? I worry that the $max Delta B_j$ is not properly defined in this instance.
$endgroup$
– Xiaomi
Dec 12 '18 at 7:11
add a comment |
$begingroup$
So I am working/reading through a proof that general Ito integrals have continuous versions. So that
$$I_f(t) = int_0^t f(s,omega)dB_s(omega)$$
Has a continuous version in $t$.
The proof I am reading does this by considering a sequence of elementary functions $phi_n$ that converge to $f$ in $L^2 (P)$. The author defines
$$I_n (t) = int_0^t phi_n (s,omega) dB_s (omega)$$
and states that $I_n (t)$ is continuous, but does not give the justification.
Here, $phi_n$ is assumed to take the form
$$phi_n = sum_j e_j^{(n)} (omega)times mathbb{1}_{left( t in[t_j,t_{j+1})right) }$$
My question is why must $I_n (t)$ be continuous in $t$ when we do not know this to be the case for the more general $f$ (though $f$ is also restricted to being $mathcal{F}_t$ adapted etc).
It is assumed throughout the book that the process $B_t$ is a continuous version of Brownian motion, so I don't find it difficult to believe, but I don't understand why it automatically holds for elementary functions and not the more general $f$?
edit
Following Saz's comment. So to show it's continuous for elementary functions, can we write:
$$|I_n (t) - I_n (s) | leq varepsilon$$
$$iff left| sum_{t_j in [s,t]} e_j^{(n)} Delta B_{t_j} right| leq varepsilon$$
And so we have
$$left| sum_{t_j in [s,t]} e_j^{(n)} Delta B_{t_j}right| leq |t-s| max_{t_j in [s,t]} |e_j^{(n)}| max_{t_j in [s,t]} |Delta B_j|$$
And so choosing
$$|t-s| leq varepsilon/ max_{t_j in [s,t]} |e_j^{(n)}| max_{t_j in [s,t]} |Delta B_j| = delta$$
It follows that if $|t-s|leq delta$ then $|I_n(t)-I_n(s)|leq varepsilon$
Is this correct?
probability continuity stochastic-calculus stochastic-integrals
asked Dec 12 '18 at 3:47
XiaomiXiaomi
1,057115
$endgroup$
So I am working/reading through a proof that general Ito integrals have continuous versions. So that
$$I_f(t) = int_0^t f(s,omega)dB_s(omega)$$
Has a continuous version in $t$.
The proof I am reading does this by considering a sequence of elementary functions $phi_n$ that converge to $f$ in $L^2 (P)$. The author defines
$$I_n (t) = int_0^t phi_n (s,omega) dB_s (omega)$$
and states that $I_n (t)$ is continuous, but does not give the justification.
Here, $phi_n$ is assumed to take the form
$$phi_n = sum_j e_j^{(n)} (omega)times mathbb{1}_{left( t in[t_j,t_{j+1})right) }$$
My question is why must $I_n (t)$ be continuous in $t$ when we do not know this to be the case for the more general $f$ (though $f$ is also restricted to being $mathcal{F}_t$ adapted etc).
It is assumed throughout the book that the process $B_t$ is a continuous version of Brownian motion, so I don't find it difficult to believe, but I don't understand why it automatically holds for elementary functions and not the more general $f$?
edit
Following Saz's comment. So to show it's continuous for elementary functions, can we write:
$$|I_n (t) - I_n (s) | leq varepsilon$$
$$iff left| sum_{t_j in [s,t]} e_j^{(n)} Delta B_{t_j} right| leq varepsilon$$
And so we have
$$left| sum_{t_j in [s,t]} e_j^{(n)} Delta B_{t_j}right| leq |t-s| max_{t_j in [s,t]} |e_j^{(n)}| max_{t_j in [s,t]} |Delta B_j|$$
And so choosing
$$|t-s| leq varepsilon/ max_{t_j in [s,t]} |e_j^{(n)}| max_{t_j in [s,t]} |Delta B_j| = delta$$
It follows that if $|t-s|leq delta$ then $|I_n(t)-I_n(s)|leq varepsilon$
Is this correct?
probability continuity stochastic-calculus stochastic-integrals
probability continuity stochastic-calculus stochastic-integrals
asked Dec 12 '18 at 3:47
XiaomiXiaomi
1,057115
asked Dec 12 '18 at 3:47
XiaomiXiaomi
1,057115
edited Dec 12 '18 at 9:27
Xiaomi
asked Dec 12 '18 at 3:47
XiaomiXiaomi
1,057115
asked Dec 12 '18 at 3:47
XiaomiXiaomi
1,057115
asked Dec 12 '18 at 3:47
XiaomiXiaomi
1,057115
1,057115
$begingroup$
For elementary functions you can calculate the stochastic integral explicitly and check that it continuous in $t$. For general functions this approach doesn't work since it is not possible to calculate the integral explicitly...
$endgroup$
– saz
Dec 12 '18 at 6:29
$begingroup$
Thanks Saz, that makes sense. Sorry, I really struggle with this sort of stuff. I have added to the end of my post, is it correct? I worry that the $max Delta B_j$ is not properly defined in this instance.
$endgroup$
– Xiaomi
Dec 12 '18 at 7:11
add a comment |
$begingroup$
For elementary functions you can calculate the stochastic integral explicitly and check that it continuous in $t$. For general functions this approach doesn't work since it is not possible to calculate the integral explicitly...
$endgroup$
– saz
Dec 12 '18 at 6:29
$begingroup$
Thanks Saz, that makes sense. Sorry, I really struggle with this sort of stuff. I have added to the end of my post, is it correct? I worry that the $max Delta B_j$ is not properly defined in this instance.
$endgroup$
– Xiaomi
Dec 12 '18 at 7:11
$begingroup$
For elementary functions you can calculate the stochastic integral explicitly and check that it continuous in $t$. For general functions this approach doesn't work since it is not possible to calculate the integral explicitly...
$endgroup$
– saz
Dec 12 '18 at 6:29
$begingroup$
For elementary functions you can calculate the stochastic integral explicitly and check that it continuous in $t$. For general functions this approach doesn't work since it is not possible to calculate the integral explicitly...
$endgroup$
– saz
Dec 12 '18 at 6:29
$begingroup$
Thanks Saz, that makes sense. Sorry, I really struggle with this sort of stuff. I have added to the end of my post, is it correct? I worry that the $max Delta B_j$ is not properly defined in this instance.
$endgroup$
– Xiaomi
Dec 12 '18 at 7:11
$begingroup$
Thanks Saz, that makes sense. Sorry, I really struggle with this sort of stuff. I have added to the end of my post, is it correct? I worry that the $max Delta B_j$ is not properly defined in this instance.
$endgroup$
– Xiaomi
Dec 12 '18 at 7:11
add a comment |
1 Answer
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$begingroup$
I don't see what you wrote as
$$
|I_n (t) - I_n (s) | leq varepsiloniff sum_{t_j in [s,t]} e_j^{(n)} Delta B_{t_j} < varepsilon.
$$ One can see that it is not true if one looks at how $I_n(t)$ is defined. In fact, each $I_n(t)$ is defined in the same manner as ordinary integral by
$$
I_n (t) = sum_{j=1}^n e_j(omega) (B_{twedge t_{j+1}}(omega) - B_{twedge t_{j}}(omega)).
$$ We can see immediately that $I_n(t)$ is continuous in $t$ since $tmapsto B_t(omega)$ is continuous for all $omegainOmega$.
answered Dec 12 '18 at 8:00
SongSong
11.7k628
$endgroup$
$begingroup$
Is the equivalence correct now that I've inserted absolute values and corrected $<$ to $leq$? I want to make sure I understand the error so I don't repeat it. Also, thanks for your answer, it is very clear and intuitive!
$endgroup$
– Xiaomi
Dec 12 '18 at 9:29
add a comment |
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$begingroup$
I don't see what you wrote as
$$
|I_n (t) - I_n (s) | leq varepsiloniff sum_{t_j in [s,t]} e_j^{(n)} Delta B_{t_j} < varepsilon.
$$ One can see that it is not true if one looks at how $I_n(t)$ is defined. In fact, each $I_n(t)$ is defined in the same manner as ordinary integral by
$$
I_n (t) = sum_{j=1}^n e_j(omega) (B_{twedge t_{j+1}}(omega) - B_{twedge t_{j}}(omega)).
$$ We can see immediately that $I_n(t)$ is continuous in $t$ since $tmapsto B_t(omega)$ is continuous for all $omegainOmega$.
answered Dec 12 '18 at 8:00
SongSong
11.7k628
$endgroup$
$begingroup$
Is the equivalence correct now that I've inserted absolute values and corrected $<$ to $leq$? I want to make sure I understand the error so I don't repeat it. Also, thanks for your answer, it is very clear and intuitive!
$endgroup$
– Xiaomi
Dec 12 '18 at 9:29
add a comment |
$begingroup$
I don't see what you wrote as
$$
|I_n (t) - I_n (s) | leq varepsiloniff sum_{t_j in [s,t]} e_j^{(n)} Delta B_{t_j} < varepsilon.
$$ One can see that it is not true if one looks at how $I_n(t)$ is defined. In fact, each $I_n(t)$ is defined in the same manner as ordinary integral by
$$
I_n (t) = sum_{j=1}^n e_j(omega) (B_{twedge t_{j+1}}(omega) - B_{twedge t_{j}}(omega)).
$$ We can see immediately that $I_n(t)$ is continuous in $t$ since $tmapsto B_t(omega)$ is continuous for all $omegainOmega$.
answered Dec 12 '18 at 8:00
SongSong
11.7k628
$endgroup$
$begingroup$
Is the equivalence correct now that I've inserted absolute values and corrected $<$ to $leq$? I want to make sure I understand the error so I don't repeat it. Also, thanks for your answer, it is very clear and intuitive!
$endgroup$
– Xiaomi
Dec 12 '18 at 9:29
add a comment |
$begingroup$
I don't see what you wrote as
$$
|I_n (t) - I_n (s) | leq varepsiloniff sum_{t_j in [s,t]} e_j^{(n)} Delta B_{t_j} < varepsilon.
$$ One can see that it is not true if one looks at how $I_n(t)$ is defined. In fact, each $I_n(t)$ is defined in the same manner as ordinary integral by
$$
I_n (t) = sum_{j=1}^n e_j(omega) (B_{twedge t_{j+1}}(omega) - B_{twedge t_{j}}(omega)).
$$ We can see immediately that $I_n(t)$ is continuous in $t$ since $tmapsto B_t(omega)$ is continuous for all $omegainOmega$.
answered Dec 12 '18 at 8:00
SongSong
11.7k628
$endgroup$
I don't see what you wrote as
$$
|I_n (t) - I_n (s) | leq varepsiloniff sum_{t_j in [s,t]} e_j^{(n)} Delta B_{t_j} < varepsilon.
$$ One can see that it is not true if one looks at how $I_n(t)$ is defined. In fact, each $I_n(t)$ is defined in the same manner as ordinary integral by
$$
I_n (t) = sum_{j=1}^n e_j(omega) (B_{twedge t_{j+1}}(omega) - B_{twedge t_{j}}(omega)).
$$ We can see immediately that $I_n(t)$ is continuous in $t$ since $tmapsto B_t(omega)$ is continuous for all $omegainOmega$.
answered Dec 12 '18 at 8:00
SongSong
11.7k628
answered Dec 12 '18 at 8:00
SongSong
11.7k628
answered Dec 12 '18 at 8:00
SongSong
11.7k628
answered Dec 12 '18 at 8:00
SongSong
11.7k628
11.7k628
$begingroup$
Is the equivalence correct now that I've inserted absolute values and corrected $<$ to $leq$? I want to make sure I understand the error so I don't repeat it. Also, thanks for your answer, it is very clear and intuitive!
$endgroup$
– Xiaomi
Dec 12 '18 at 9:29
add a comment |
$begingroup$
Is the equivalence correct now that I've inserted absolute values and corrected $<$ to $leq$? I want to make sure I understand the error so I don't repeat it. Also, thanks for your answer, it is very clear and intuitive!
$endgroup$
– Xiaomi
Dec 12 '18 at 9:29
$begingroup$
Is the equivalence correct now that I've inserted absolute values and corrected $<$ to $leq$? I want to make sure I understand the error so I don't repeat it. Also, thanks for your answer, it is very clear and intuitive!
$endgroup$
– Xiaomi
Dec 12 '18 at 9:29
$begingroup$
Is the equivalence correct now that I've inserted absolute values and corrected $<$ to $leq$? I want to make sure I understand the error so I don't repeat it. Also, thanks for your answer, it is very clear and intuitive!
$endgroup$
– Xiaomi
Dec 12 '18 at 9:29
add a comment |
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$begingroup$
For elementary functions you can calculate the stochastic integral explicitly and check that it continuous in $t$. For general functions this approach doesn't work since it is not possible to calculate the integral explicitly...
$endgroup$
– saz
Dec 12 '18 at 6:29
$begingroup$
Thanks Saz, that makes sense. Sorry, I really struggle with this sort of stuff. I have added to the end of my post, is it correct? I worry that the $max Delta B_j$ is not properly defined in this instance.
$endgroup$
– Xiaomi
Dec 12 '18 at 7:11