# close form for stochastic integral

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 $$dX_t = \theta(\mu - X_t)\; dt + \sigma X_t \; dW_t$$

• 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

We should look for a solution of the form $$X(t)=U(t)V(t)$$ where $$dU_t=-\theta\,U_tdt+\sigma\,U_t\,dW_t$$ and $$dV_t=\alpha(t)dt+\beta(t)dW_t$$ $$U$$ is a geometric Brownian motion, therefore $$U(t)=U(0)\,e^{-(\theta+\frac{1}{2}\sigma^2)t+\sigma W_t}$$ let $$U(0)=1$$, this yields $$V(0)=X(0)$$. Now we should find $$\alpha(t)$$ and $$\beta(t)$$. $$dX_t=U_tdV_t+V_tdU_t+d[U,V](t)$$ we have $$dX_t=(\alpha (t)U_t-\theta\,X_t+\sigma\beta(t)U_t)dt+(\beta(t)U_t+\sigma\,X_t)dW_t$$ thus $$\beta(t)=0$$ and $$\alpha(t)U_t=\mu\,\theta$$, as a result $$dV_t=\frac{\mu\theta}{U_t}dt$$ in the other words $$V_t=V_0+\mu\theta\int_{0}^{t}\frac{1}{U_s}ds$$ finally $$X_t=U_tV_t=e^{-(\theta+\frac{1}{2}\sigma^2)t+\sigma W_t}\left(X(0)+\mu\theta\int_{0}^{t}e^{(\theta+\frac{1}{2}\sigma^2)s-\sigma W_s}\right)$$ $$X_t=e^{-(\theta+\frac{1}{2}\sigma^2)t+\sigma W_t}+\mu\theta\int_{0}^{t}e^{-(\theta+\frac{1}{2}\sigma^2)(t-s)+\sigma (W_t-W_s)}ds$$