What is an Itô integral, what does it represent concretely?
1
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I'm really in truble to understand Itô integral. I can work with it without any problem, but I really don't understand what is it. And why is it an integral ? How can we interpret $$I=int_0^T f(s,B_s)dB_s ?$$
- Would it be : we consider the brownian motion on $[0,T]$ and a function $f:[0,infty )times mathbb Rto mathbb R$, then the area between $(B_t)_{tin [0,T]}$ and $(f(t,B_t))_{tin [0,T]}$ is a sort of Itô integral $I$ ? (in grey in the following picture)
- Why are we interested to this integral ? I also saw that if $H_s$ is the number of call we have then $int_0^T H_s dB_t$ is the profit we made after $[0,T]$. I so, why the price of a call is a Brownian motion ? This is very weird for me.
stochastic-calculus
asked Dec 29 '18 at 11:05
user623855user623855
1457
$endgroup$
|
show 4 more comments
1
$begingroup$
I'm really in truble to understand Itô integral. I can work with it without any problem, but I really don't understand what is it. And why is it an integral ? How can we interpret $$I=int_0^T f(s,B_s)dB_s ?$$
- Would it be : we consider the brownian motion on $[0,T]$ and a function $f:[0,infty )times mathbb Rto mathbb R$, then the area between $(B_t)_{tin [0,T]}$ and $(f(t,B_t))_{tin [0,T]}$ is a sort of Itô integral $I$ ? (in grey in the following picture)
- Why are we interested to this integral ? I also saw that if $H_s$ is the number of call we have then $int_0^T H_s dB_t$ is the profit we made after $[0,T]$. I so, why the price of a call is a Brownian motion ? This is very weird for me.
stochastic-calculus
asked Dec 29 '18 at 11:05
user623855user623855
1457
$endgroup$
$begingroup$
The stochastic integral doesn't represent an area as a Riemann integral. The stochastic integral is just a stochastic process that has the same properties than the integral, that's why we use the integral notation.
$endgroup$
– Surb
Dec 29 '18 at 11:09
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Your first suggestion is bizarre since usual integrals are never this area...
$endgroup$
– Did
Dec 29 '18 at 11:10
$begingroup$
@Did : I add a picture to be more clear. Maybe something in this spirit ? (probably not exactely that, but is to relate this to an area). I'm trying to make a comparaison with stiljes integral, but maybe such integral can't be related to an area, doest it ?
$endgroup$
– user623855
Dec 29 '18 at 11:18
$begingroup$
@Surb : And for the interpretation ?
$endgroup$
– user623855
Dec 29 '18 at 11:18
$begingroup$
But already for $f(B_t,t)=1$, this interpretation fails, as surely you can see...
$endgroup$
– Did
Dec 29 '18 at 11:19
|
show 4 more comments
1
1
1
$begingroup$
I'm really in truble to understand Itô integral. I can work with it without any problem, but I really don't understand what is it. And why is it an integral ? How can we interpret $$I=int_0^T f(s,B_s)dB_s ?$$
- Would it be : we consider the brownian motion on $[0,T]$ and a function $f:[0,infty )times mathbb Rto mathbb R$, then the area between $(B_t)_{tin [0,T]}$ and $(f(t,B_t))_{tin [0,T]}$ is a sort of Itô integral $I$ ? (in grey in the following picture)
- Why are we interested to this integral ? I also saw that if $H_s$ is the number of call we have then $int_0^T H_s dB_t$ is the profit we made after $[0,T]$. I so, why the price of a call is a Brownian motion ? This is very weird for me.
stochastic-calculus
asked Dec 29 '18 at 11:05
user623855user623855
1457
$endgroup$
I'm really in truble to understand Itô integral. I can work with it without any problem, but I really don't understand what is it. And why is it an integral ? How can we interpret $$I=int_0^T f(s,B_s)dB_s ?$$
- Would it be : we consider the brownian motion on $[0,T]$ and a function $f:[0,infty )times mathbb Rto mathbb R$, then the area between $(B_t)_{tin [0,T]}$ and $(f(t,B_t))_{tin [0,T]}$ is a sort of Itô integral $I$ ? (in grey in the following picture)
- Why are we interested to this integral ? I also saw that if $H_s$ is the number of call we have then $int_0^T H_s dB_t$ is the profit we made after $[0,T]$. I so, why the price of a call is a Brownian motion ? This is very weird for me.
stochastic-calculus
stochastic-calculus
asked Dec 29 '18 at 11:05
user623855user623855
1457
asked Dec 29 '18 at 11:05
user623855user623855
1457
edited Dec 29 '18 at 11:16
user623855
asked Dec 29 '18 at 11:05
user623855user623855
1457
asked Dec 29 '18 at 11:05
user623855user623855
1457
asked Dec 29 '18 at 11:05
user623855user623855
1457
1457
$begingroup$
The stochastic integral doesn't represent an area as a Riemann integral. The stochastic integral is just a stochastic process that has the same properties than the integral, that's why we use the integral notation.
$endgroup$
– Surb
Dec 29 '18 at 11:09
$begingroup$
Your first suggestion is bizarre since usual integrals are never this area...
$endgroup$
– Did
Dec 29 '18 at 11:10
$begingroup$
@Did : I add a picture to be more clear. Maybe something in this spirit ? (probably not exactely that, but is to relate this to an area). I'm trying to make a comparaison with stiljes integral, but maybe such integral can't be related to an area, doest it ?
$endgroup$
– user623855
Dec 29 '18 at 11:18
$begingroup$
@Surb : And for the interpretation ?
$endgroup$
– user623855
Dec 29 '18 at 11:18
$begingroup$
But already for $f(B_t,t)=1$, this interpretation fails, as surely you can see...
$endgroup$
– Did
Dec 29 '18 at 11:19
|
show 4 more comments
$begingroup$
The stochastic integral doesn't represent an area as a Riemann integral. The stochastic integral is just a stochastic process that has the same properties than the integral, that's why we use the integral notation.
$endgroup$
– Surb
Dec 29 '18 at 11:09
$begingroup$
Your first suggestion is bizarre since usual integrals are never this area...
$endgroup$
– Did
Dec 29 '18 at 11:10
$begingroup$
@Did : I add a picture to be more clear. Maybe something in this spirit ? (probably not exactely that, but is to relate this to an area). I'm trying to make a comparaison with stiljes integral, but maybe such integral can't be related to an area, doest it ?
$endgroup$
– user623855
Dec 29 '18 at 11:18
$begingroup$
@Surb : And for the interpretation ?
$endgroup$
– user623855
Dec 29 '18 at 11:18
$begingroup$
But already for $f(B_t,t)=1$, this interpretation fails, as surely you can see...
$endgroup$
– Did
Dec 29 '18 at 11:19
$begingroup$
The stochastic integral doesn't represent an area as a Riemann integral. The stochastic integral is just a stochastic process that has the same properties than the integral, that's why we use the integral notation.
$endgroup$
– Surb
Dec 29 '18 at 11:09
$begingroup$
The stochastic integral doesn't represent an area as a Riemann integral. The stochastic integral is just a stochastic process that has the same properties than the integral, that's why we use the integral notation.
$endgroup$
– Surb
Dec 29 '18 at 11:09
$begingroup$
Your first suggestion is bizarre since usual integrals are never this area...
$endgroup$
– Did
Dec 29 '18 at 11:10
$begingroup$
Your first suggestion is bizarre since usual integrals are never this area...
$endgroup$
– Did
Dec 29 '18 at 11:10
$begingroup$
@Did : I add a picture to be more clear. Maybe something in this spirit ? (probably not exactely that, but is to relate this to an area). I'm trying to make a comparaison with stiljes integral, but maybe such integral can't be related to an area, doest it ?
$endgroup$
– user623855
Dec 29 '18 at 11:18
$begingroup$
@Did : I add a picture to be more clear. Maybe something in this spirit ? (probably not exactely that, but is to relate this to an area). I'm trying to make a comparaison with stiljes integral, but maybe such integral can't be related to an area, doest it ?
$endgroup$
– user623855
Dec 29 '18 at 11:18
$begingroup$
@Surb : And for the interpretation ?
$endgroup$
– user623855
Dec 29 '18 at 11:18
$begingroup$
@Surb : And for the interpretation ?
$endgroup$
– user623855
Dec 29 '18 at 11:18
$begingroup$
But already for $f(B_t,t)=1$, this interpretation fails, as surely you can see...
$endgroup$
– Did
Dec 29 '18 at 11:19
$begingroup$
But already for $f(B_t,t)=1$, this interpretation fails, as surely you can see...
$endgroup$
– Did
Dec 29 '18 at 11:19
|
show 4 more comments
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$begingroup$
The stochastic integral doesn't represent an area as a Riemann integral. The stochastic integral is just a stochastic process that has the same properties than the integral, that's why we use the integral notation.
$endgroup$
– Surb
Dec 29 '18 at 11:09
$begingroup$
Your first suggestion is bizarre since usual integrals are never this area...
$endgroup$
– Did
Dec 29 '18 at 11:10
$begingroup$
@Did : I add a picture to be more clear. Maybe something in this spirit ? (probably not exactely that, but is to relate this to an area). I'm trying to make a comparaison with stiljes integral, but maybe such integral can't be related to an area, doest it ?
$endgroup$
– user623855
Dec 29 '18 at 11:18
$begingroup$
@Surb : And for the interpretation ?
$endgroup$
– user623855
Dec 29 '18 at 11:18
$begingroup$
But already for $f(B_t,t)=1$, this interpretation fails, as surely you can see...
$endgroup$
– Did
Dec 29 '18 at 11:19