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I am new to stochastic calculus. Can I know how to compute the close-form solution for $$\int_0^t \exp(\alpha s - \sigma W_s) \; ds$$ and $$\int_0^t \exp(\alpha s - \sigma W_s) \; dW_s.$$ I encounter that when trying to solve for the following SDE $$dX_t = \theta(\mu - X_t)\; dt + \sigma X_t \; dW_t$$

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If the SDE is written correctly, that is not an Ornstein-Uhlenbeck process and your integrals don't seem to match it either. An O-U process has additive noise (i.e., diffusion function is not a function of the state variable) while the SDE as written has multiplicative noise. Also, an O-U process definitely does have a known analytical solution (see Doob, Ann. Math. 43, 1942). – horchler Jul 16 '13 at 18:29
    
@n.c. Your comment isn't accurate unfortunately. As "horchler" pointed out, the Ornstein-Uhlenbeck process does NOT have multiplicative noise, unlike the process posted in this question. To appropriately solve this SDE, consider applying Ito's Lemma on $Y_t = ln(X_t)$ – Mayou Aug 21 '13 at 16:23

To solve this equation, let \begin{align*} M_t = e^{(\theta + \frac{1}{2}\sigma^2 ) t - \sigma W_t}. \end{align*} Then \begin{align*} dM_t = M_t\Big[\big(\theta +\sigma^2\big) dt - \sigma dW_t\Big]. \end{align*} Moreover, \begin{align*} d(M_t X_t) &= M_t dX_t + X_t dM_t + d\langle M, X \rangle_t\\ &=\theta\,\mu\, M_t dt. \end{align*} Then, \begin{align*} M_t X_t &= X_0 + \theta\,\mu\,\int_0^t M_s ds. \end{align*} That is, \begin{align*} X_t &= X_0 e^{-(\theta + \frac{1}{2}\sigma^2 ) t + \sigma W_t} + \theta\,\mu\,e^{-(\theta + \frac{1}{2}\sigma^2 ) t + \sigma W_t}\int_0^t e^{(\theta + \frac{1}{2}\sigma^2 ) s - \sigma W_s} ds\\ &=X_0 e^{-(\theta + \frac{1}{2}\sigma^2 ) t + \sigma W_t} + \theta\,\mu\,\int_0^t e^{-(\theta + \frac{1}{2}\sigma^2 ) (t-s) + \sigma(W_t - W_s)} ds. \end{align*}

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