8
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ 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). $\endgroup$
    – horchler
    Jul 16, 2013 at 18:29
  • $\begingroup$ @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 $Y_t = ln(X_t)$ $\endgroup$
    – Mayou
    Aug 21, 2013 at 16:23

2 Answers 2

9
$\begingroup$

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 $$X_t=U_tV_t=e^{-(\theta+\frac{1}{2}\sigma^2)t+\sigma W_t}\left(X(0)+\mu\theta\int_{0}^{t}e^{(\theta+\frac{1}{2}\sigma^2)s-\sigma W_s}\right)$$ $$X_t=e^{-(\theta+\frac{1}{2}\sigma^2)t+\sigma W_t}+\mu\theta\int_{0}^{t}e^{-(\theta+\frac{1}{2}\sigma^2)(t-s)+\sigma (W_t-W_s)}ds$$

$\endgroup$
7
$\begingroup$

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*}

$\endgroup$
1
  • $\begingroup$ Fine solution +1 $\endgroup$
    – user16651
    Jun 12, 2016 at 22:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.