I am trying to figure out how a particular SDE can be integrated. The SDE is the normal mean-reverting model:

$dX_t = \theta(\mu - X_t)dt + \sigma dW_t$ (1)

Where $W_t - N(0,t)$. So far, I have found the solution of:

$X_t = X_0e^{-\theta t} + \mu(1-e^{-\theta t}) + e^{-\theta t} \int^t_0 \sigma e^{\theta s} dW_s$ (2)

This is standard and easy enough to find. However, I am looking to decompose this solution into the form.

$X_T = X_0 + (\mu - X_0)(1 - e^{-\theta t})+\sigma \sqrt{\frac{1-e^{-2\theta t}}{2 \theta}}W_t$ (3)

I have found that adding to the right-hand side of (2), $X_0 - X_0$, is a nice trick to get the two left terms on the right-hand side of (3). I am also confident that one needs to find the variance and mean of the integral term in (2) to get the solution for (3). However, I am not sure how to go about doing this.

If anyone can help, I'd be very appreciative. Thanks.


There is a closed solution formula for SDEs of the form $dX_t = (a_t+c_tX_t)dt+(b_t+d_tX_t)dW_t, \, X_0 = \bar{X}$. The solution is $X_t = \mu_t H_t$ with $\mu_t = \bar{X} + \int_0^t \frac{1}{H_s} (a_s-b_sd_s)ds + \int_0^t \frac{1}{H_s} b_s dW_s$

$H_t = \exp\left(\int_0^t (c_s -\frac{1}{2} d_s^2)ds + \int_0^t d_s dW_s \right)$.

You can obtain this solution by first solving the corresponding homogeneous equation and then applying Ito's formula in form of a product rule. It's an analogue to the method of variation of the parameters known from ode theory.

Edit: Sorry I got your question wrong, this step you've already done. To obtain the desired form of the solution you have to calculate $\int_0^t e^{\theta t}dW_s$ and for this you can use the folowing, well known variance result on stochastic integrals: $\mathbb{E}\left[ \left( \int_a^b f(s,\cdot) dW_s \right)^2 \right] = \int_a^b \mathbb{E}[(f(s,\cdot)^2]ds$ which follows from the Ito - isometry for $f\in L^2_{\omega}([a,b])$. By that you get $(\int_0^t e^{\theta s} dW_s)^2 = \int_0^t e^{2\theta s}$, hence $\int_0^t e^{\theta s} dW_s = \sqrt{\frac{1}{2\theta}(e^{2\theta t}-1)}$ and by that you're done.

| cite | improve this answer | |
  • $\begingroup$ Thanks, I appreciate the input. But I'm having a hard time seeing how this helps me get from (2) to the end result (3). Maybe I'm missing something. $\endgroup$ – user5619709 Mar 29 '16 at 23:58
  • $\begingroup$ sorry, I edited my post. I hope this helps you more. $\endgroup$ – CandyOwl Mar 30 '16 at 8:14
  • $\begingroup$ Yes. That is correct. I completely forgot about Ito-isometry and that was the dilemma that I was having. That solved it for me. Thanks! And no worries about getting the question wrong, that was good to know as well. $\endgroup$ – user5619709 Mar 30 '16 at 13:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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