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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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    $\begingroup$ Hint: first compute $E(X_t^2)$. Then, for $s>t$, write $X_s$ as $X_t+(X_s-X_t)$. $\endgroup$ – Did Nov 8 '11 at 5:52
  • $\begingroup$ 39 minutes. $ $ $\endgroup$ – Did Nov 8 '11 at 6:30
  • $\begingroup$ See en.wikipedia.org/wiki/… $\endgroup$ – user940 Nov 8 '11 at 15:51
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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 $s<t$ $$ \begin{eqnarray} \mathbb{E} \left( X_s X_t \right) &=& \mathbb{E} \left( \left( \sigma \int_0^t \mathrm{e}^{-a(s - \tau)} \mathbf{1}\left(\tau \le s\right) \mathrm{d}W_\tau \right) \left( \sigma \int_0^t \mathrm{e}^{-a(t - \tau)} \mathrm{d}W_\tau \right) \right) \\ &=& \sigma^2 \int_0^{\min(s,t)} \mathrm{e}^{-a (s+t-2 \tau)}\mathrm{d} \tau = \sigma^2 e^{-a (s+t)} \cdot \frac{e^{2 a \min (s,t)}-1}{2 a} = \sigma^2 e^{-a t} \cdot \frac{\sinh(a s)}{a} \end{eqnarray} $$

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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  • $\begingroup$ Thank you very much, I got the same answer $\endgroup$ – Tasha Chen Nov 8 '11 at 6:08

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