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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$\begingroup$ 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). $\endgroup$– horchlerJul 16, 2013 at 18:29
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$\begingroup$ @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 $Y_t = ln(X_t)$ $\endgroup$– MayouAug 21, 2013 at 16:23
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2 Answers
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Another Solution
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 $$X_t=U_tV_t=e^{-(\theta+\frac{1}{2}\sigma^2)t+\sigma W_t}\left(X(0)+\mu\theta\int_{0}^{t}e^{(\theta+\frac{1}{2}\sigma^2)s-\sigma W_s}\right)$$ $$X_t=e^{-(\theta+\frac{1}{2}\sigma^2)t+\sigma W_t}+\mu\theta\int_{0}^{t}e^{-(\theta+\frac{1}{2}\sigma^2)(t-s)+\sigma (W_t-W_s)}ds$$
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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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$\begingroup$ Fine solution +1 $\endgroup$– user16651Jun 12, 2016 at 22:27