Mixing
What represent ito integral ? What does means that $int_0^t B_sdB_s=frac{B_t^2}{2}-frac{t}{2}$
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I still have difficulties to understand Itô integral, and unfortunately, I don't really understand what represent the Itô integral.
Just to inform, I asked, but it has been erased without explanation.
I already ask this question previously here but it has been put on Hold without any explanation. Also, I found similar question on MSE here, here but I'm not really convinced by the answers. Let take an example $$int_0^t B_sdB_s=lim_{nto infty }sum_{i=0}^{n-1} B_{t_i}(B_{t_{i+1}}-B_{t_i}),$$
where ${t_0,...,t_{n+1}}$ is a subdivision of $[0,t]$ where $$max_{i}|t_{i+1}-t_i|to 0.$$
By calculation, we get $$int_0^t B_sdB_s=frac{B_t^2}{2}-frac{t}{2}.$$
But how can we interpret this result ? What does it mean exactly ? First I get crazy to understand what represent $int_0^t B_sdB_s$ (instead of the fact that it's continuous martingale). There is a picture of this integral on wikipedia, but I can't really understand what it represent.
For example, $int_0^af(x)dx$ is the area between $y=0$, $x=a$, $x=b$ and the curve $y=f(x)$.
If $g$ is a positive function, then $int_a^b fdg$ can be seen as the density, of the the surface delimited by $x=a$, $x=b$, $x=0$ and $y=f(x)$.
Now, what could represent $int_0^t B_sdB_s$ ?
stochastic-processes stochastic-integrals
asked Feb 3 at 12:20
DylanDylan
3218
$endgroup$
add a comment |
$begingroup$
I still have difficulties to understand Itô integral, and unfortunately, I don't really understand what represent the Itô integral.
Just to inform, I asked, but it has been erased without explanation.
I already ask this question previously here but it has been put on Hold without any explanation. Also, I found similar question on MSE here, here but I'm not really convinced by the answers. Let take an example $$int_0^t B_sdB_s=lim_{nto infty }sum_{i=0}^{n-1} B_{t_i}(B_{t_{i+1}}-B_{t_i}),$$
where ${t_0,...,t_{n+1}}$ is a subdivision of $[0,t]$ where $$max_{i}|t_{i+1}-t_i|to 0.$$
By calculation, we get $$int_0^t B_sdB_s=frac{B_t^2}{2}-frac{t}{2}.$$
But how can we interpret this result ? What does it mean exactly ? First I get crazy to understand what represent $int_0^t B_sdB_s$ (instead of the fact that it's continuous martingale). There is a picture of this integral on wikipedia, but I can't really understand what it represent.
For example, $int_0^af(x)dx$ is the area between $y=0$, $x=a$, $x=b$ and the curve $y=f(x)$.
If $g$ is a positive function, then $int_a^b fdg$ can be seen as the density, of the the surface delimited by $x=a$, $x=b$, $x=0$ and $y=f(x)$.
Now, what could represent $int_0^t B_sdB_s$ ?
stochastic-processes stochastic-integrals
asked Feb 3 at 12:20
DylanDylan
3218
$endgroup$
1
$begingroup$
Possible duplicate of Physical meaning of Ito integrals
$endgroup$
– Peter Foreman
Feb 3 at 12:25
$begingroup$
@PeterForeman: I agree that the question of the OP and the the question on your link is quite similar, but I don't really agree that the answers answer to the question that are a bit more "technical" than "intuitive". They just explain how Itô integral is, not what it represent. Nevertheless, seeing $B_{t_i+1}-B_{t_i}$ as a random length may be a good start.
$endgroup$
– Surb
Feb 3 at 15:44
add a comment |
$begingroup$
I still have difficulties to understand Itô integral, and unfortunately, I don't really understand what represent the Itô integral.
Just to inform, I asked, but it has been erased without explanation.
I already ask this question previously here but it has been put on Hold without any explanation. Also, I found similar question on MSE here, here but I'm not really convinced by the answers. Let take an example $$int_0^t B_sdB_s=lim_{nto infty }sum_{i=0}^{n-1} B_{t_i}(B_{t_{i+1}}-B_{t_i}),$$
where ${t_0,...,t_{n+1}}$ is a subdivision of $[0,t]$ where $$max_{i}|t_{i+1}-t_i|to 0.$$
By calculation, we get $$int_0^t B_sdB_s=frac{B_t^2}{2}-frac{t}{2}.$$
But how can we interpret this result ? What does it mean exactly ? First I get crazy to understand what represent $int_0^t B_sdB_s$ (instead of the fact that it's continuous martingale). There is a picture of this integral on wikipedia, but I can't really understand what it represent.
For example, $int_0^af(x)dx$ is the area between $y=0$, $x=a$, $x=b$ and the curve $y=f(x)$.
If $g$ is a positive function, then $int_a^b fdg$ can be seen as the density, of the the surface delimited by $x=a$, $x=b$, $x=0$ and $y=f(x)$.
Now, what could represent $int_0^t B_sdB_s$ ?
stochastic-processes stochastic-integrals
asked Feb 3 at 12:20
DylanDylan
3218
$endgroup$
I still have difficulties to understand Itô integral, and unfortunately, I don't really understand what represent the Itô integral.
Just to inform, I asked, but it has been erased without explanation.
I already ask this question previously here but it has been put on Hold without any explanation. Also, I found similar question on MSE here, here but I'm not really convinced by the answers. Let take an example $$int_0^t B_sdB_s=lim_{nto infty }sum_{i=0}^{n-1} B_{t_i}(B_{t_{i+1}}-B_{t_i}),$$
where ${t_0,...,t_{n+1}}$ is a subdivision of $[0,t]$ where $$max_{i}|t_{i+1}-t_i|to 0.$$
By calculation, we get $$int_0^t B_sdB_s=frac{B_t^2}{2}-frac{t}{2}.$$
But how can we interpret this result ? What does it mean exactly ? First I get crazy to understand what represent $int_0^t B_sdB_s$ (instead of the fact that it's continuous martingale). There is a picture of this integral on wikipedia, but I can't really understand what it represent.
For example, $int_0^af(x)dx$ is the area between $y=0$, $x=a$, $x=b$ and the curve $y=f(x)$.
If $g$ is a positive function, then $int_a^b fdg$ can be seen as the density, of the the surface delimited by $x=a$, $x=b$, $x=0$ and $y=f(x)$.
Now, what could represent $int_0^t B_sdB_s$ ?
stochastic-processes stochastic-integrals
stochastic-processes stochastic-integrals
asked Feb 3 at 12:20
DylanDylan
3218
asked Feb 3 at 12:20
DylanDylan
3218
asked Feb 3 at 12:20
DylanDylan
3218
asked Feb 3 at 12:20
DylanDylan
3218
asked Feb 3 at 12:20
DylanDylan
3218
3218
1
$begingroup$
Possible duplicate of Physical meaning of Ito integrals
$endgroup$
– Peter Foreman
Feb 3 at 12:25
$begingroup$
@PeterForeman: I agree that the question of the OP and the the question on your link is quite similar, but I don't really agree that the answers answer to the question that are a bit more "technical" than "intuitive". They just explain how Itô integral is, not what it represent. Nevertheless, seeing $B_{t_i+1}-B_{t_i}$ as a random length may be a good start.
$endgroup$
– Surb
Feb 3 at 15:44
add a comment |
1
$begingroup$
Possible duplicate of Physical meaning of Ito integrals
$endgroup$
– Peter Foreman
Feb 3 at 12:25
$begingroup$
@PeterForeman: I agree that the question of the OP and the the question on your link is quite similar, but I don't really agree that the answers answer to the question that are a bit more "technical" than "intuitive". They just explain how Itô integral is, not what it represent. Nevertheless, seeing $B_{t_i+1}-B_{t_i}$ as a random length may be a good start.
$endgroup$
– Surb
Feb 3 at 15:44
1
1
$begingroup$
Possible duplicate of Physical meaning of Ito integrals
$endgroup$
– Peter Foreman
Feb 3 at 12:25
$begingroup$
Possible duplicate of Physical meaning of Ito integrals
$endgroup$
– Peter Foreman
Feb 3 at 12:25
$begingroup$
@PeterForeman: I agree that the question of the OP and the the question on your link is quite similar, but I don't really agree that the answers answer to the question that are a bit more "technical" than "intuitive". They just explain how Itô integral is, not what it represent. Nevertheless, seeing $B_{t_i+1}-B_{t_i}$ as a random length may be a good start.
$endgroup$
– Surb
Feb 3 at 15:44
$begingroup$
@PeterForeman: I agree that the question of the OP and the the question on your link is quite similar, but I don't really agree that the answers answer to the question that are a bit more "technical" than "intuitive". They just explain how Itô integral is, not what it represent. Nevertheless, seeing $B_{t_i+1}-B_{t_i}$ as a random length may be a good start.
$endgroup$
– Surb
Feb 3 at 15:44
add a comment |
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1
$begingroup$
Possible duplicate of Physical meaning of Ito integrals
$endgroup$
– Peter Foreman
Feb 3 at 12:25
$begingroup$
@PeterForeman: I agree that the question of the OP and the the question on your link is quite similar, but I don't really agree that the answers answer to the question that are a bit more "technical" than "intuitive". They just explain how Itô integral is, not what it represent. Nevertheless, seeing $B_{t_i+1}-B_{t_i}$ as a random length may be a good start.
$endgroup$
– Surb
Feb 3 at 15:44