# OU process and computing $\int_{t}^{t+\Delta t}e^{\theta s}dWs$.

I am trying to compute a forward equation of an Ornstein-Uhlenbeck process defined in the line below to use in MATLAB to make pretty pictures

$$dX_{t} = \theta(X_{t}-\mu)dt + \sigma dW_{t}$$

by solving the differential equation using variation of parameters on

$$dX_{t+\Delta t} = \theta(X_{t+\Delta t}-\mu) d(t+h) + \sigma dW_{t+h}.$$

After some work, I end up with the following:

$$X_{t+\Delta t} = e^{\theta \Delta t}X_{t} + \mu(1-e^{\theta\Delta t}) + \sigma e^{\theta(t+\Delta t)}\int_{t}^{t+\Delta t}e^{-\theta s}dW_{s}.$$

From what I've seen online, everything seems fine except the last integral is supposed to look something like

$$\sigma e^{\theta(t+\Delta t)}\int_{t}^{t+\Delta t}e^{-\theta s}dW_{s} = \sigma \sqrt{-\frac{1-e^{-2\theta \Delta t}}{2\theta}}Z_{t},\quad Z_{t}\sim N(0,1)$$

but I don't know how to integrate this with respect to $W_{s}$. I do not know a lot about stochastic calculus and this isn't an assignment so I'm really struggling trying to figure out what tool I need to do to finish this calculation.

One observation I made is that the result I am supposed to get looks like the square root of the result of applying the Ito isometry to the $\int \cdot \,dW_{s}$ term, but I don't understand where the $Z_{t}$ would come from or why the Ito isometry is even relevant to my task if I am not trying to compute the expectation or variance.

My other observation is that if this were a non-stochastic integral, at this point I would probably want to approximate the integral with Gaussian Quadrature or something, but I'm not sure if that is the right way to think of it either.

If you know a bit of stochastic calculus theory and gaussian process you might very simply deduce the result you are trying to prove.

First notice that your integrand with respect to the Brownian motion is determistic in such case a stochastic integral is called a Wiener integral and has the very nice property to be a gaussian (centered) process.

This readily means that $Y_t=\int_{t}^{t+\Delta t}e^{-\theta s}dW_{s}$ is a gaussian random variable (with mean equal to 0). So there is only one thing left unknown then, to wit the second moment of this random variable. Now using Itô's isometry you get :

$$E[Y_t^2]=E[\int_{t}^{t+\Delta t}e^{-2.\theta s}ds]=\int_{t}^{t+\Delta t}e^{-2.\theta s}ds=...$$

The very tractable calculus is left to you.

Best regards

• I understand now, thank you – JessicaK Oct 23 '15 at 3:00