# Simulation of interacting Ornstein-Uhlenbeck processes

I would like to simulate the following system of interacting OU processes on $[0,T]$:

$$dX_t^1=(X_t^2-X_t^1)\,dt+\sigma_1 \,dW_t^1,\quad X_0^1=x_1$$

$$dX_t^2=(X_t^1-X_t^2)\,dt+\sigma_2 \,dW_t^2,\quad X_0^2=x_2$$

where $W^1$ and $W^2$ are two independent Wiener processes. I am aware of the fact there is a closed-form formula for the classical one-dimensional OU process, but in the case of two interacting processes as above I don't believe this is possible. As a result, my first idea was to naively apply the Euler-Maruyama scheme, as follows:

$$X_{t_{k+1}}^1=X_{t_k}^1+(X_{t_k}^2-X_{t_k}^1)h+\sigma_1(W_{t_{k+1}}^1-W_{t_k}^1)$$

$$X_{t_{k+1}}^2=X_{t_k}^2+(X_{t_k}^1-X_{t_k}^2)h+\sigma_2(W_{t_{k+1}}^2-W_{t_k}^2)$$

with $X_{t_0}^1=x_1$, $X_{t_0}^2=x_2$, $t_k=\frac{kT}{N}$, N being the number of subdivisions, and $h=\frac{T}{N}$.

However, I am not sure whether the Euler-Maruyama scheme can be applied to this interacting case, and whether or not the scheme would effectively converge in this specific context. Any ideas or references to literature would be greatly appreciated, thanks.

You can simply try the transformation $$A_t = \frac{X_t^1 + X_t^2}{\sqrt{2}}, \quad B_t = \frac{X_t^1 - X_t^2}{\sqrt{2}}$$ The equations will get much simpler.

• ok thanks, but how does this help me simulate the system? do you mean there is a closed-form expression which can be obtained? – user223935 Jun 30 '15 at 3:29
• Of course there is a close-form expression. After the transformation the equations are decoupled into two simple one-dimensional OU processes. – corindo Jun 30 '15 at 3:31
• I have applied the transformation as in my answer above, but I don't see how one can obtain a system of decoupled processes? – user223935 Jun 30 '15 at 18:42

The transformation you suggested gives:

$dA_t=\frac{1}{\sqrt{2}}(\sigma_1 dW_t^1+\sigma_2 dW_t^2)$

$dB_t=\frac{1}{\sqrt{2}}(2X_t^2-2X_t^1+\sigma_1 dW_t^1-\sigma_2 dW_t^2)$

How is this supposed to help me?

• You can notice that the $2X_t^2 - 2X_t^1$ at the right hand side of your second equation is equal to $-2\sqrt{2} B_t$. – corindo Jul 1 '15 at 1:05
• ok I see now thanks, do you think it is possible to find a similar transformation in the 3-dimensional (or even N-dimensional) case described in my answer below? – user223935 Jul 1 '15 at 2:22
• That would not be possible in general. In this case it works because the corresponding matrix is diagonalizable. For the non diagonalizable case you have to solve it like linear equations. And I have no idea whether the Euler scheme converge or not. – corindo Jul 1 '15 at 2:56