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Let $B\left(s\right)$ be a brownian motion and $\sigma\left(s\right)$ be the nondeterministic function. The following equation then holds $$ \mathbb{E}\left(\exp\left(i\int_{0}^{t}\sigma\left(s\right)\, e^{-a\left(t-s\right)}dB\left(s\right)\right)\right)=\mathbb{E}\left(\exp\left(-\frac{1}{2}\int_{0}^{t}\sigma^{2}\left(s\right)\, e^{-2a\left(t-s\right)}ds\right)\right) $$

How can i prove this equality? i'm quite stuck here.


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Have you tried Ito's formula? – Davide Giraudo Aug 1 '12 at 10:09
Cheers. no,i haven't. According to my reference here, he get this equality by introducing a sigma-algebra $\zeta_{t}=\sigma\left\{ \sigma^{2}\left(s\right),\,0\leq s\leq t\right\} \vee\mathcal{F}_{t} $. I dont understand how he prove this, and i try to find another way to prove, like martingale... – alfred chen Aug 1 '12 at 19:29
Formally you can certainly write down a Taylor series and use ito isometry for the even terms and argue why the odd terms vanish – Kai Sikorski Apr 2 '14 at 6:07

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