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Ito Integral and brownian motion question [duplicate]
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Prove directly from the definition of the Ito's integral
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Can you prove directly from the definition of Ito integral (By that I mean limit of simple functions)
$$int_0^tB(s)^2 dB(s) = frac13B(t)^3-int_0^t B(s) ds $$
So I started by writing $B(t)^3 - B(0)^3= sum_0^n-1B(s_i+1)^3-B(s_i)^3$ Where $0=s_0<s_1<...<s_n=t $ and $s_i+1-s_i$ tend to 0 as n tends to infinity. Where to go from here? Maybe bring in $(B(s_i+1)-B(s_i))^3$.
Thanks!
stochastic-processes stochastic-calculus brownian-motion
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Monty
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This question already has an answer here:
Prove directly from the definition of the Ito's integral
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Can you prove directly from the definition of Ito integral (By that I mean limit of simple functions)
$$int_0^tB(s)^2 dB(s) = frac13B(t)^3-int_0^t B(s) ds $$
So I started by writing $B(t)^3 - B(0)^3= sum_0^n-1B(s_i+1)^3-B(s_i)^3$ Where $0=s_0<s_1<...<s_n=t $ and $s_i+1-s_i$ tend to 0 as n tends to infinity. Where to go from here? Maybe bring in $(B(s_i+1)-B(s_i))^3$.
Thanks!
stochastic-processes stochastic-calculus brownian-motion
asked yesterday
Monty
15212
marked as duplicate by saz stochastic-processes
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This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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This question already has an answer here:
Prove directly from the definition of the Ito's integral
1 answer
Can you prove directly from the definition of Ito integral (By that I mean limit of simple functions)
$$int_0^tB(s)^2 dB(s) = frac13B(t)^3-int_0^t B(s) ds $$
So I started by writing $B(t)^3 - B(0)^3= sum_0^n-1B(s_i+1)^3-B(s_i)^3$ Where $0=s_0<s_1<...<s_n=t $ and $s_i+1-s_i$ tend to 0 as n tends to infinity. Where to go from here? Maybe bring in $(B(s_i+1)-B(s_i))^3$.
Thanks!
stochastic-processes stochastic-calculus brownian-motion
asked yesterday
Monty
15212
This question already has an answer here:
Prove directly from the definition of the Ito's integral
1 answer
Can you prove directly from the definition of Ito integral (By that I mean limit of simple functions)
$$int_0^tB(s)^2 dB(s) = frac13B(t)^3-int_0^t B(s) ds $$
So I started by writing $B(t)^3 - B(0)^3= sum_0^n-1B(s_i+1)^3-B(s_i)^3$ Where $0=s_0<s_1<...<s_n=t $ and $s_i+1-s_i$ tend to 0 as n tends to infinity. Where to go from here? Maybe bring in $(B(s_i+1)-B(s_i))^3$.
Thanks!
This question already has an answer here:
Prove directly from the definition of the Ito's integral
1 answer
stochastic-processes stochastic-calculus brownian-motion
asked yesterday
Monty
15212
asked yesterday
Monty
15212
asked yesterday
Monty
15212
asked yesterday
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For simplicity I suppose $t=1$, even if all arguments should be adabtable for a generic $t>0$.
We introduce the mesh $a_i=fraci2^n,i = 0,..,2^n$ and note that:
$B(1)^3=sum_k=0^2^n-1B^3(a_k+1)-B^3(a_k)$
Now we can exploit that $a^3-b^3=(a-b)(a^2+ab+b^2)$. Therefore:
beginalign
B(1)^3=&sum_k=0^2^n-1(B(a_k+1)-B(a_k))(B^2(a_k+1)+B^2(a_k)+B(a_k)B(a_k+1))=\
&=3sum_k=0^2^n-1 (B(a_k+1)-B(a_k))B^2(a_k)+
sum_k=0^2^n-1 (B(a_k+1)-B(a_k))^2(B(a_k+1)+2B(a_k))
endalign
, where only algebraic manipulations have been used.
At this level by definition:
$sum_k=0^2^n-1 (B(a_k+1)-B(a_k))B^2(a_k) rightarrow int_0^1 B^2(s)dB(s)$
, where convergence in $L^2$, under the limit of fine meshes (large n) is implied.
The final result follows by observing that:
beginalign
&F(n)equivsum_k=0^2^n-1 (B(a_k+1)-B(a_k))^2 B(a_k+1) rightarrow int_0^1 B(s) ds\
&G(n)equivsum_k=0^2^n-1 (B(a_k+1)-B(a_k))^2 B(a_k) rightarrow int_0^1 B(s) ds
endalign
This is expected to hold by the naive argument $dB^2(s) sim ds $ and, with a bit of patience, can be formally derived. I sketch a proof, leaving out the details.
First we show that $F(n)-G(n) rightarrow 0$:
$E[|F(n)-G(n)|^2]=sum_k,k'E[(B(a_k+1)-B(a_k))^3(B(a_k'+1)-B(a_k'))^3]=sum_kE[(B(a_k+1)-B(a_k))^6] rightarrow 0$
, because $E[(B(a_k+1)-B(a_k))^6] sim (a_k+1-a_k)^3$ and the independence of the increments has been used.
Then we prove that $G(n) rightarrow int_0^1 B(s) ds$. Since $sum_k B(a_k)(a_k+1-a_k) rightarrow int_0^1 B(s)ds$ it is sufficient to show that:
$E[left[sum_k=0^2^n-1 [(B(a_k+1)-B(a_k))^2-(a_k+1-a_k)] B(a_k) right]^2]le
sum_k=0^2^n-1 Eleft[ [(B(a_k+1)-B(a_k))^2-(a_k+1-a_k)]^2 B^2(a_k)right]=
sum_k=0^2^n-1 E[ [(B(a_k+1)-B(a_k))^2-(a_k+1-a_k)]^2]a_k sim \
sum_k=0^2^n-1 (a_k+1-a_k)^2 a_k rightarrow 0$
, where the moments of the gaussian distributions have been used and constants neglected in the last estimation. The independence of $B(a_k)$ from $B(a_k+1)-B(a_k)$ and the value $B^2(a_k)=a_k$ have also been used.
answered 17 hours ago
Thomas
938
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1 Answer
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1 Answer
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active
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active
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For simplicity I suppose $t=1$, even if all arguments should be adabtable for a generic $t>0$.
We introduce the mesh $a_i=fraci2^n,i = 0,..,2^n$ and note that:
$B(1)^3=sum_k=0^2^n-1B^3(a_k+1)-B^3(a_k)$
Now we can exploit that $a^3-b^3=(a-b)(a^2+ab+b^2)$. Therefore:
beginalign
B(1)^3=&sum_k=0^2^n-1(B(a_k+1)-B(a_k))(B^2(a_k+1)+B^2(a_k)+B(a_k)B(a_k+1))=\
&=3sum_k=0^2^n-1 (B(a_k+1)-B(a_k))B^2(a_k)+
sum_k=0^2^n-1 (B(a_k+1)-B(a_k))^2(B(a_k+1)+2B(a_k))
endalign
, where only algebraic manipulations have been used.
At this level by definition:
$sum_k=0^2^n-1 (B(a_k+1)-B(a_k))B^2(a_k) rightarrow int_0^1 B^2(s)dB(s)$
, where convergence in $L^2$, under the limit of fine meshes (large n) is implied.
The final result follows by observing that:
beginalign
&F(n)equivsum_k=0^2^n-1 (B(a_k+1)-B(a_k))^2 B(a_k+1) rightarrow int_0^1 B(s) ds\
&G(n)equivsum_k=0^2^n-1 (B(a_k+1)-B(a_k))^2 B(a_k) rightarrow int_0^1 B(s) ds
endalign
This is expected to hold by the naive argument $dB^2(s) sim ds $ and, with a bit of patience, can be formally derived. I sketch a proof, leaving out the details.
First we show that $F(n)-G(n) rightarrow 0$:
$E[|F(n)-G(n)|^2]=sum_k,k'E[(B(a_k+1)-B(a_k))^3(B(a_k'+1)-B(a_k'))^3]=sum_kE[(B(a_k+1)-B(a_k))^6] rightarrow 0$
, because $E[(B(a_k+1)-B(a_k))^6] sim (a_k+1-a_k)^3$ and the independence of the increments has been used.
Then we prove that $G(n) rightarrow int_0^1 B(s) ds$. Since $sum_k B(a_k)(a_k+1-a_k) rightarrow int_0^1 B(s)ds$ it is sufficient to show that:
$E[left[sum_k=0^2^n-1 [(B(a_k+1)-B(a_k))^2-(a_k+1-a_k)] B(a_k) right]^2]le
sum_k=0^2^n-1 Eleft[ [(B(a_k+1)-B(a_k))^2-(a_k+1-a_k)]^2 B^2(a_k)right]=
sum_k=0^2^n-1 E[ [(B(a_k+1)-B(a_k))^2-(a_k+1-a_k)]^2]a_k sim \
sum_k=0^2^n-1 (a_k+1-a_k)^2 a_k rightarrow 0$
, where the moments of the gaussian distributions have been used and constants neglected in the last estimation. The independence of $B(a_k)$ from $B(a_k+1)-B(a_k)$ and the value $B^2(a_k)=a_k$ have also been used.
answered 17 hours ago
Thomas
938
add a comment |Â
up vote
0
down vote
For simplicity I suppose $t=1$, even if all arguments should be adabtable for a generic $t>0$.
We introduce the mesh $a_i=fraci2^n,i = 0,..,2^n$ and note that:
$B(1)^3=sum_k=0^2^n-1B^3(a_k+1)-B^3(a_k)$
Now we can exploit that $a^3-b^3=(a-b)(a^2+ab+b^2)$. Therefore:
beginalign
B(1)^3=&sum_k=0^2^n-1(B(a_k+1)-B(a_k))(B^2(a_k+1)+B^2(a_k)+B(a_k)B(a_k+1))=\
&=3sum_k=0^2^n-1 (B(a_k+1)-B(a_k))B^2(a_k)+
sum_k=0^2^n-1 (B(a_k+1)-B(a_k))^2(B(a_k+1)+2B(a_k))
endalign
, where only algebraic manipulations have been used.
At this level by definition:
$sum_k=0^2^n-1 (B(a_k+1)-B(a_k))B^2(a_k) rightarrow int_0^1 B^2(s)dB(s)$
, where convergence in $L^2$, under the limit of fine meshes (large n) is implied.
The final result follows by observing that:
beginalign
&F(n)equivsum_k=0^2^n-1 (B(a_k+1)-B(a_k))^2 B(a_k+1) rightarrow int_0^1 B(s) ds\
&G(n)equivsum_k=0^2^n-1 (B(a_k+1)-B(a_k))^2 B(a_k) rightarrow int_0^1 B(s) ds
endalign
This is expected to hold by the naive argument $dB^2(s) sim ds $ and, with a bit of patience, can be formally derived. I sketch a proof, leaving out the details.
First we show that $F(n)-G(n) rightarrow 0$:
$E[|F(n)-G(n)|^2]=sum_k,k'E[(B(a_k+1)-B(a_k))^3(B(a_k'+1)-B(a_k'))^3]=sum_kE[(B(a_k+1)-B(a_k))^6] rightarrow 0$
, because $E[(B(a_k+1)-B(a_k))^6] sim (a_k+1-a_k)^3$ and the independence of the increments has been used.
Then we prove that $G(n) rightarrow int_0^1 B(s) ds$. Since $sum_k B(a_k)(a_k+1-a_k) rightarrow int_0^1 B(s)ds$ it is sufficient to show that:
$E[left[sum_k=0^2^n-1 [(B(a_k+1)-B(a_k))^2-(a_k+1-a_k)] B(a_k) right]^2]le
sum_k=0^2^n-1 Eleft[ [(B(a_k+1)-B(a_k))^2-(a_k+1-a_k)]^2 B^2(a_k)right]=
sum_k=0^2^n-1 E[ [(B(a_k+1)-B(a_k))^2-(a_k+1-a_k)]^2]a_k sim \
sum_k=0^2^n-1 (a_k+1-a_k)^2 a_k rightarrow 0$
, where the moments of the gaussian distributions have been used and constants neglected in the last estimation. The independence of $B(a_k)$ from $B(a_k+1)-B(a_k)$ and the value $B^2(a_k)=a_k$ have also been used.
answered 17 hours ago
Thomas
938
add a comment |Â
up vote
0
down vote
up vote
0
down vote
For simplicity I suppose $t=1$, even if all arguments should be adabtable for a generic $t>0$.
We introduce the mesh $a_i=fraci2^n,i = 0,..,2^n$ and note that:
$B(1)^3=sum_k=0^2^n-1B^3(a_k+1)-B^3(a_k)$
Now we can exploit that $a^3-b^3=(a-b)(a^2+ab+b^2)$. Therefore:
beginalign
B(1)^3=&sum_k=0^2^n-1(B(a_k+1)-B(a_k))(B^2(a_k+1)+B^2(a_k)+B(a_k)B(a_k+1))=\
&=3sum_k=0^2^n-1 (B(a_k+1)-B(a_k))B^2(a_k)+
sum_k=0^2^n-1 (B(a_k+1)-B(a_k))^2(B(a_k+1)+2B(a_k))
endalign
, where only algebraic manipulations have been used.
At this level by definition:
$sum_k=0^2^n-1 (B(a_k+1)-B(a_k))B^2(a_k) rightarrow int_0^1 B^2(s)dB(s)$
, where convergence in $L^2$, under the limit of fine meshes (large n) is implied.
The final result follows by observing that:
beginalign
&F(n)equivsum_k=0^2^n-1 (B(a_k+1)-B(a_k))^2 B(a_k+1) rightarrow int_0^1 B(s) ds\
&G(n)equivsum_k=0^2^n-1 (B(a_k+1)-B(a_k))^2 B(a_k) rightarrow int_0^1 B(s) ds
endalign
This is expected to hold by the naive argument $dB^2(s) sim ds $ and, with a bit of patience, can be formally derived. I sketch a proof, leaving out the details.
First we show that $F(n)-G(n) rightarrow 0$:
$E[|F(n)-G(n)|^2]=sum_k,k'E[(B(a_k+1)-B(a_k))^3(B(a_k'+1)-B(a_k'))^3]=sum_kE[(B(a_k+1)-B(a_k))^6] rightarrow 0$
, because $E[(B(a_k+1)-B(a_k))^6] sim (a_k+1-a_k)^3$ and the independence of the increments has been used.
Then we prove that $G(n) rightarrow int_0^1 B(s) ds$. Since $sum_k B(a_k)(a_k+1-a_k) rightarrow int_0^1 B(s)ds$ it is sufficient to show that:
$E[left[sum_k=0^2^n-1 [(B(a_k+1)-B(a_k))^2-(a_k+1-a_k)] B(a_k) right]^2]le
sum_k=0^2^n-1 Eleft[ [(B(a_k+1)-B(a_k))^2-(a_k+1-a_k)]^2 B^2(a_k)right]=
sum_k=0^2^n-1 E[ [(B(a_k+1)-B(a_k))^2-(a_k+1-a_k)]^2]a_k sim \
sum_k=0^2^n-1 (a_k+1-a_k)^2 a_k rightarrow 0$
, where the moments of the gaussian distributions have been used and constants neglected in the last estimation. The independence of $B(a_k)$ from $B(a_k+1)-B(a_k)$ and the value $B^2(a_k)=a_k$ have also been used.
answered 17 hours ago
Thomas
938
For simplicity I suppose $t=1$, even if all arguments should be adabtable for a generic $t>0$.
We introduce the mesh $a_i=fraci2^n,i = 0,..,2^n$ and note that:
$B(1)^3=sum_k=0^2^n-1B^3(a_k+1)-B^3(a_k)$
Now we can exploit that $a^3-b^3=(a-b)(a^2+ab+b^2)$. Therefore:
beginalign
B(1)^3=&sum_k=0^2^n-1(B(a_k+1)-B(a_k))(B^2(a_k+1)+B^2(a_k)+B(a_k)B(a_k+1))=\
&=3sum_k=0^2^n-1 (B(a_k+1)-B(a_k))B^2(a_k)+
sum_k=0^2^n-1 (B(a_k+1)-B(a_k))^2(B(a_k+1)+2B(a_k))
endalign
, where only algebraic manipulations have been used.
At this level by definition:
$sum_k=0^2^n-1 (B(a_k+1)-B(a_k))B^2(a_k) rightarrow int_0^1 B^2(s)dB(s)$
, where convergence in $L^2$, under the limit of fine meshes (large n) is implied.
The final result follows by observing that:
beginalign
&F(n)equivsum_k=0^2^n-1 (B(a_k+1)-B(a_k))^2 B(a_k+1) rightarrow int_0^1 B(s) ds\
&G(n)equivsum_k=0^2^n-1 (B(a_k+1)-B(a_k))^2 B(a_k) rightarrow int_0^1 B(s) ds
endalign
This is expected to hold by the naive argument $dB^2(s) sim ds $ and, with a bit of patience, can be formally derived. I sketch a proof, leaving out the details.
First we show that $F(n)-G(n) rightarrow 0$:
$E[|F(n)-G(n)|^2]=sum_k,k'E[(B(a_k+1)-B(a_k))^3(B(a_k'+1)-B(a_k'))^3]=sum_kE[(B(a_k+1)-B(a_k))^6] rightarrow 0$
, because $E[(B(a_k+1)-B(a_k))^6] sim (a_k+1-a_k)^3$ and the independence of the increments has been used.
Then we prove that $G(n) rightarrow int_0^1 B(s) ds$. Since $sum_k B(a_k)(a_k+1-a_k) rightarrow int_0^1 B(s)ds$ it is sufficient to show that:
$E[left[sum_k=0^2^n-1 [(B(a_k+1)-B(a_k))^2-(a_k+1-a_k)] B(a_k) right]^2]le
sum_k=0^2^n-1 Eleft[ [(B(a_k+1)-B(a_k))^2-(a_k+1-a_k)]^2 B^2(a_k)right]=
sum_k=0^2^n-1 E[ [(B(a_k+1)-B(a_k))^2-(a_k+1-a_k)]^2]a_k sim \
sum_k=0^2^n-1 (a_k+1-a_k)^2 a_k rightarrow 0$
, where the moments of the gaussian distributions have been used and constants neglected in the last estimation. The independence of $B(a_k)$ from $B(a_k+1)-B(a_k)$ and the value $B^2(a_k)=a_k$ have also been used.
answered 17 hours ago
Thomas
938
edited 17 hours ago
answered 17 hours ago
Thomas
938
answered 17 hours ago
Thomas
938
answered 17 hours ago
Thomas
938
938
add a comment |Â
add a comment |Â