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Let $X_t=\sigma \int_0^t e^{-a(t-s)} dW_s$, where $\sigma , a $ are constants. How can I find the expected value of the product of $X_t, X_s$ For t>s, $\mathbb{E}[X_t, X_s]$, and $\mathbb{E}[X_t, W_s]$ where $W_s$ is brownian.

I am not sure how i can modify ito's isometry so that i can find a simple solution. Thanks

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    Hint: first compute $E(X_t^2)$. Then, for $s>t$, write $X_s$ as $X_t+(X_s-X_t)$.2011-11-08
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    39 minutes. $ $2011-11-08
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    See http://en.wikipedia.org/wiki/Ornstein%E2%80%93Uhlenbeck_process#Formulae_for_moments2011-11-08

1 Answers 1

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Using polarization, Ito's isometry implies $ \mathbb{E}\left( \left( \int_0^t f(s) \mathrm{d} W_s \right) \left( \int_0^t g(s) \mathrm{d} W_s \right) \right) = \int_0^t f(s) g(s) \mathrm{d} s$.

Thus, for $s

Using $W_s = \int_0^s 1 \,\, \mathrm{d} W_\tau$, we similarly get $$ \mathbb{E} \left( X_t W_s\right) = \sigma \int_0^s \mathrm{e}^{-a(t-\tau)}\mathrm{d} \tau = \sigma \mathrm{e}^{-a t} \cdot \frac{ \left(\mathrm{e}^{a s}-1\right)}{a} $$

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    Thank you very much, I got the same answer2011-11-08