Stochastic integral involving Wiener process
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Suppose $W_t$ is the standard Wiener process. I am wondering whether I can use the independent increment property of the process as well as the fact that $W_t$ is zero-mean Gaussian to evaluate expectation of the following stochastic integral as I have:
$$ mathbb{E}Big[ intlimits_{0}^t W^3_s dW_s Big] = intlimits_{0}^t mathbb{E}[W^3_s] ; mathbb{E}[dW_s] \ = intlimits_{0}^t underbrace{mathbb{E}[W^3_s]}_{=0} ; mathbb{E}[dW_s] \ = 0.$$
I am wondering whether I have used the techniques soundly.
stochastic-processes stochastic-calculus stochastic-integrals stochastic-analysis
asked Nov 20 at 1:26
user2348674
1558
|
show 1 more comment
up vote
1
down vote
favorite
Suppose $W_t$ is the standard Wiener process. I am wondering whether I can use the independent increment property of the process as well as the fact that $W_t$ is zero-mean Gaussian to evaluate expectation of the following stochastic integral as I have:
$$ mathbb{E}Big[ intlimits_{0}^t W^3_s dW_s Big] = intlimits_{0}^t mathbb{E}[W^3_s] ; mathbb{E}[dW_s] \ = intlimits_{0}^t underbrace{mathbb{E}[W^3_s]}_{=0} ; mathbb{E}[dW_s] \ = 0.$$
I am wondering whether I have used the techniques soundly.
stochastic-processes stochastic-calculus stochastic-integrals stochastic-analysis
asked Nov 20 at 1:26
user2348674
1558
$mathbb{E}(dW_s)$ doesn't make any sense...
– saz
Nov 20 at 7:21
@saz $mathbb{E}[dW_s]$ is expectation of the infinitesimal increment in the Wiener process which is equal to zero. What would be wrong with that?
– user2348674
Nov 21 at 0:20
If it equals zero, then you are already done after the first "=". To make your argumentation sound, you will need to prove rigorously the first "=".
– saz
Nov 21 at 6:54
@saz can't it be done via just writing the integral as an infinite sum of product of increment and the cubed brownian motion? These turn out to be independent, so the expectation can be split
– Makina
Nov 24 at 4:01
1
Well, yes, I didn't say that it is impossible. There are also general statements which say that $M_t := int_0^t f(s) , dW_s$ is a martingale under certain assumptions on $f$, this implies in particular $mathbb{E}(M_t)=mathbb{E}(M_0)=0$.The best approach really depends on your background.
– saz
Nov 24 at 7:03
|
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Suppose $W_t$ is the standard Wiener process. I am wondering whether I can use the independent increment property of the process as well as the fact that $W_t$ is zero-mean Gaussian to evaluate expectation of the following stochastic integral as I have:
$$ mathbb{E}Big[ intlimits_{0}^t W^3_s dW_s Big] = intlimits_{0}^t mathbb{E}[W^3_s] ; mathbb{E}[dW_s] \ = intlimits_{0}^t underbrace{mathbb{E}[W^3_s]}_{=0} ; mathbb{E}[dW_s] \ = 0.$$
I am wondering whether I have used the techniques soundly.
stochastic-processes stochastic-calculus stochastic-integrals stochastic-analysis
asked Nov 20 at 1:26
user2348674
1558
Suppose $W_t$ is the standard Wiener process. I am wondering whether I can use the independent increment property of the process as well as the fact that $W_t$ is zero-mean Gaussian to evaluate expectation of the following stochastic integral as I have:
$$ mathbb{E}Big[ intlimits_{0}^t W^3_s dW_s Big] = intlimits_{0}^t mathbb{E}[W^3_s] ; mathbb{E}[dW_s] \ = intlimits_{0}^t underbrace{mathbb{E}[W^3_s]}_{=0} ; mathbb{E}[dW_s] \ = 0.$$
I am wondering whether I have used the techniques soundly.
stochastic-processes stochastic-calculus stochastic-integrals stochastic-analysis
stochastic-processes stochastic-calculus stochastic-integrals stochastic-analysis
asked Nov 20 at 1:26
user2348674
1558
asked Nov 20 at 1:26
user2348674
1558
asked Nov 20 at 1:26
user2348674
1558
asked Nov 20 at 1:26
user2348674
1558
asked Nov 20 at 1:26
user2348674
1558
1558
$mathbb{E}(dW_s)$ doesn't make any sense...
– saz
Nov 20 at 7:21
@saz $mathbb{E}[dW_s]$ is expectation of the infinitesimal increment in the Wiener process which is equal to zero. What would be wrong with that?
– user2348674
Nov 21 at 0:20
If it equals zero, then you are already done after the first "=". To make your argumentation sound, you will need to prove rigorously the first "=".
– saz
Nov 21 at 6:54
@saz can't it be done via just writing the integral as an infinite sum of product of increment and the cubed brownian motion? These turn out to be independent, so the expectation can be split
– Makina
Nov 24 at 4:01
1
Well, yes, I didn't say that it is impossible. There are also general statements which say that $M_t := int_0^t f(s) , dW_s$ is a martingale under certain assumptions on $f$, this implies in particular $mathbb{E}(M_t)=mathbb{E}(M_0)=0$.The best approach really depends on your background.
– saz
Nov 24 at 7:03
|
show 1 more comment
$mathbb{E}(dW_s)$ doesn't make any sense...
– saz
Nov 20 at 7:21
@saz $mathbb{E}[dW_s]$ is expectation of the infinitesimal increment in the Wiener process which is equal to zero. What would be wrong with that?
– user2348674
Nov 21 at 0:20
If it equals zero, then you are already done after the first "=". To make your argumentation sound, you will need to prove rigorously the first "=".
– saz
Nov 21 at 6:54
@saz can't it be done via just writing the integral as an infinite sum of product of increment and the cubed brownian motion? These turn out to be independent, so the expectation can be split
– Makina
Nov 24 at 4:01
1
Well, yes, I didn't say that it is impossible. There are also general statements which say that $M_t := int_0^t f(s) , dW_s$ is a martingale under certain assumptions on $f$, this implies in particular $mathbb{E}(M_t)=mathbb{E}(M_0)=0$.The best approach really depends on your background.
– saz
Nov 24 at 7:03
$mathbb{E}(dW_s)$ doesn't make any sense...
– saz
Nov 20 at 7:21
$mathbb{E}(dW_s)$ doesn't make any sense...
– saz
Nov 20 at 7:21
@saz $mathbb{E}[dW_s]$ is expectation of the infinitesimal increment in the Wiener process which is equal to zero. What would be wrong with that?
– user2348674
Nov 21 at 0:20
@saz $mathbb{E}[dW_s]$ is expectation of the infinitesimal increment in the Wiener process which is equal to zero. What would be wrong with that?
– user2348674
Nov 21 at 0:20
If it equals zero, then you are already done after the first "=". To make your argumentation sound, you will need to prove rigorously the first "=".
– saz
Nov 21 at 6:54
If it equals zero, then you are already done after the first "=". To make your argumentation sound, you will need to prove rigorously the first "=".
– saz
Nov 21 at 6:54
@saz can't it be done via just writing the integral as an infinite sum of product of increment and the cubed brownian motion? These turn out to be independent, so the expectation can be split
– Makina
Nov 24 at 4:01
@saz can't it be done via just writing the integral as an infinite sum of product of increment and the cubed brownian motion? These turn out to be independent, so the expectation can be split
– Makina
Nov 24 at 4:01
1
1
Well, yes, I didn't say that it is impossible. There are also general statements which say that $M_t := int_0^t f(s) , dW_s$ is a martingale under certain assumptions on $f$, this implies in particular $mathbb{E}(M_t)=mathbb{E}(M_0)=0$.The best approach really depends on your background.
– saz
Nov 24 at 7:03
Well, yes, I didn't say that it is impossible. There are also general statements which say that $M_t := int_0^t f(s) , dW_s$ is a martingale under certain assumptions on $f$, this implies in particular $mathbb{E}(M_t)=mathbb{E}(M_0)=0$.The best approach really depends on your background.
– saz
Nov 24 at 7:03
|
show 1 more comment
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$mathbb{E}(dW_s)$ doesn't make any sense...
– saz
Nov 20 at 7:21
@saz $mathbb{E}[dW_s]$ is expectation of the infinitesimal increment in the Wiener process which is equal to zero. What would be wrong with that?
– user2348674
Nov 21 at 0:20
If it equals zero, then you are already done after the first "=". To make your argumentation sound, you will need to prove rigorously the first "=".
– saz
Nov 21 at 6:54
@saz can't it be done via just writing the integral as an infinite sum of product of increment and the cubed brownian motion? These turn out to be independent, so the expectation can be split
– Makina
Nov 24 at 4:01
1
Well, yes, I didn't say that it is impossible. There are also general statements which say that $M_t := int_0^t f(s) , dW_s$ is a martingale under certain assumptions on $f$, this implies in particular $mathbb{E}(M_t)=mathbb{E}(M_0)=0$.The best approach really depends on your background.
– saz
Nov 24 at 7:03