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I am working on the following problem, and struggling with it. Can anyone help?

Let $$H=e^{\int_0^T B_s\,ds}$$ where $T>0$. Show first $E[H^2]<\infty$. Then find an adapted process ($\epsilon_t$) with $$E\left[\int^T_0\epsilon_S^2\,ds\right]<\infty$$ such that $$H(\omega)=H_0+\int^T_0\epsilon_t(\omega) \, dB_t(\omega)~~~~~ \mathbb{P}\text{-a.s.}$$ where $H_0=E[H]$.

I did the first step. $E[H^2]=E\left[e^{2\int^T_0 B_s \, ds}\right]=e^{2\int^T_0E[B_s]\,ds}=e^{\int^T_00\,ds}<\infty$. Not sure whether it is correct or wrong. The teacher said we only need to play with the Ito's formula to get the answer. Can someone help? Thanks!

Stupid
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  • first step not correct, use that $\int^TB_s ds$ is Gaussian. For second, apply Ito to $ Y_s = E(H-H_0 | B_t, t<s)$, the point of which is, that's the martingale you are representing. – mike Apr 12 '13 at 13:23
  • @mike Do you mind to explain more? I do not quite follow you... – Stupid Apr 13 '13 at 05:38
  • I suppose you understand the first part. A simple example for the second is suppose you want to represent $e^{B_1}$. The martingale whose value at 1 is $e^{B_1} = E(e^{B_1} | \mathcal F _s ) = e^{B_s} e^{\frac {(1-s)^2}2}$ Write that as a stochastic integral by applying Ito to $ e^{B_s} e^{\frac {(1-s)^2}2}$ and I think you find out $e^{B_1} = e^{\frac 1 2 } + \int e^{B_s+ (1-s)^2/2} dB_s$ – mike Apr 14 '13 at 15:08
  • Oh, yes, yes, I get it. Thanks a lot!!! @mike – Stupid Apr 15 '13 at 01:22

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