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seen Jun 16 '11 at 2:58

Apr
9
awarded  Enlightened
Apr
9
awarded  Nice Answer
Mar
19
comment Joint moments of Brownian motion
By the discussion before, we can suppose that $t_1<t_2<...<t_n.$ put $\xi_1=W_{t_1}, \xi_2=W_{t_2}-W_{t_1},...,\xi_n=W_{t_n}-W_{t_{n-1}}$, which are independent Gaussian variables. Then, we have $$E[W_{t_1}W_{t_2}...W_{t_n}]=E[\xi_1(\xi_1+\xi_2)...(\xi_1+\xi_2+...+\xi_n)],$$ Which could be done. For the case three, it could be written as $$\xi_1(\xi_1+\xi_2)(\xi_1+\xi_2+\xi_3)=\sum_{i=1}^{3!}{\xi_1\xi_{[\frac{i}{3}]}‌​}\xi_{i\%3}, \xi_0=\xi_3$$. I try to treat it as a growth of tree for each term like binary code in computer science ; but I don't get the explicit formula.
Mar
19
comment How to calculate $E[(\int_0^t{W_sds})^n], n \geq 2$
@Didier: Thank you. Where could I find the theorem?
Mar
19
comment How to calculate $E[(\int_0^t{W_sds})^n], n \geq 2$
@TheBridge: I am also curious about the distribution of $\int_0^t{e^{W_s}ds}$.
Mar
19
comment How to calculate $E[(\int_0^t{W_sds})^n], n \geq 2$
By the discussion before, we can suppose that $t_1<t_2<...<t_n.$ put $\xi_1=W_{t_1}, \xi_2=W_{t_2}-W_{t_1},...,\xi_n=W_{t_n}-W_{t_{n-1}}$, which are independent Gaussian variables. Then, we have $$E[W_{t_1}W_{t_2}...W_{t_n}]=E[\xi_1(\xi_1+\xi_2)...(\xi_1+\xi_2+...+\xi_n)],$$ Which could be done. But I don't know the explicit formula.
Mar
18
comment How to calculate $E[(\int_0^t{W_sds})^n], n \geq 2$
Yes. You are right. It can be done; but it is not obvious for the general formula. I will try again for the general case.
Mar
18
comment How to calculate $E[(\int_0^t{W_sds})^n], n \geq 2$
Let say for triple case, $$1=\sum_{\sigma(t1,t2,t3)}I_{t_1\leq t_2\leq t_3},$$ where $\sigma(t1,t2,t3)$ means all the permutation of $t_1,t_2,t_3$. Hence, we could get the integral by symmetry. Am I right?
Mar
18
revised How to calculate $E[(\int_0^t{W_sds})^n], n \geq 2$
Update the integrand
Mar
18
awarded  Commentator
Mar
18
comment How to calculate $E[(\int_0^t{W_sds})^n], n \geq 2$
Thanks. Your approach is much easier and easy to extend to high dimension.
Mar
18
comment How to calculate $E[(\int_0^t{W_sds})^n], n \geq 2$
@Didier Piau: Thanks. I should study limit theory later.^_^
Mar
18
comment How to calculate $E[(\int_0^t{W_sds})^n], n \geq 2$
@Didier Piau: $X_t$ is the limit of a serial of Gaussian random variables. We need to show that the limit is Gaussian.
Mar
18
asked How to calculate $E[(\int_0^t{W_sds})^n], n \geq 2$
Mar
14
awarded  Teacher
Mar
13
comment Is this a martingale?
@George: Thanks. With your comments, it is easy to prove $Z_t$ is square-integrable martingale over interval $(0,\frac{1}{2}).$ I will figure out the remaining parts.
Mar
13
comment Is this a martingale?
@Byron: Thanks. Now I get it.
Mar
13
comment Is this a martingale?
@Byron: I think $Z_t$ does not go to $0$ as $t\rightarrow +\infty$, if $Z_t$ is a martingale, since $E[Z_t]=E[Z_0]=1.$ Right?
Mar
13
awarded  Editor
Mar
13
revised Is this a martingale?
edited body