Solving a Mean-Reverting SDE with a Stochastic Integrating Factor

# Computing closed-form solutions for stochastic integrals


# Computing closed-form solutions for stochastic integrals












I am new to stochastic calculus. I would like to compute the closed-form solution for

$$ \int_0^t \exp \left( \alpha s - \sigma W_s \right) \; {\rm d}s \tag{1}$$

$$ \int_0^t \exp \left( \alpha s - \sigma W_s \right) \; {\rm d} W_s \tag{2} $$

which I encountered when trying to solve the following stochastic differential equation (SDE)

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

How to compute these closed-form solutions?

## Answer by user16651 (score 10)

https://quant.stackexchange.com/a/27585

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

## Answer by Gordon (score 8)

https://quant.stackexchange.com/a/25988

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*}