# Mixing the Ornstein-Uhlenbeck Process and Geometric Brownian Motion

The Ornstein-Uhlenbeck process with mean reversion level 0 is defined as follows:

$$dX_t=a X_t dt + \sigma_1 d W_{1t}. \tag{1}$$

Geometric Brownian motion is defined as follows:

$$dX_t= a X_t dt + \sigma_2 X_t d W_{2t}. \tag{2}$$

Hence, the two processes differ only in the second term, which I call noise term.

Is the SDE that contains both noise terms:

$$dX_t=a X_t dt + \sigma_1 d W_{1t} + \sigma_2 X_t d W_{2t}, \tag{3}$$ where $W_{1t}$ and $W_{2t}$ are independent Wiener processes,

valid? If yes, what is the solution?

The only approach I am familiar with to solve SDEs is to use the formula in "Introduction to stochastic integration" by Kuo, Hui-Hsiung on page 233. While I was successful in re-deriving the solutions of both the Ornstein-Uhlenbeck process and Geometric Brownian motion, I was not able to fit the process of interest (the mixture) into the formula.

• Are $W_{1t}$ and $W_{2t}$ independent standard Brownian motions? – Nawaf Bou-Rabee Dec 19 '16 at 17:32
• Yes that is the case – Julian Karls Dec 20 '16 at 13:38
• Maybe this should be specified in the question? – Nawaf Bou-Rabee Dec 20 '16 at 14:16
• You are right. Sorry for this. I modified the question – Julian Karls Dec 21 '16 at 11:48
• Note that the validity of the SDE (meaning it is well-posed) follows from the fact it is linear. Also note that the main difference between the answer given below and the answer I gave to your previous question with $W_{1t}=W_{2t}$ is in the covariation between $X_t$ and $\Phi_t^{-1}$. – Nawaf Bou-Rabee Dec 23 '16 at 11:54

This is a linear SDE, whose explicit solution is straightforward to obtain. Exactly analogous to linear ODEs, one first finds a fundamental solution, i.e., a solution of the homogeneous version of (3) with the initial condition $X_0=1$, $$\Phi_t = \exp \left( (a - \frac{1}{2} \sigma_2^2) t + \sigma_2 W_{2t} \right)$$ which is a geometric Brownian motion. Given an initial condition $x_0$, then the solution to (3) can be written as: $$X_t = \Phi_t \left( x_0 + \sigma_1 \int_0^t \Phi_s^{-1} dW_{1s} \right) \tag{\star}$$ To prove this, just apply integration-by-parts for Itô processes $$d(A_t B_t) = A_t d B_t + A_t d B_t + [ A, B ]_t$$ with $A_t=X_t$ and $B_t = \Phi_t^{-1}$ to obtain: \begin{align*} & d ( X_t \Phi_t^{-1} ) \\ & = X_t \Phi_t^{-1} \left( (-a+ \sigma_2^2) dt - \sigma_2 d W_{2 t} \right) + \Phi_t^{-1} \left( \vphantom{\frac{1}{2}} a X_t dt + \sigma_1 d W_{1t} + \sigma_2 X_t dW_{2 t} \right) - X_t \Phi_t^{-1} \sigma_2^2 dt \\ &= \sigma_1 \Phi_t^{-1} dW_{1t} \end{align*} and then integrate what remains to get ($\star$). Here we used the fact that the covariation of the two Itô processes $X_t$ and $\Phi_t^{-1}$ is $[X, \Phi^{-1}]_t =-\sigma_2^2 \int_0^t X_s \Phi_s^{-1} ds$. This follows from the fact that $W_{1t}$ and $W_{2t}$ are independent standard Brownian motions.
Note that when $\sigma_1=0$, one recovers a geometric Brownian motion, and when $\sigma_2=0$, one obtains an Ornstein-Uhlenbeck process.
• @michael Why does it look positive? The OU case ($\sigma_2=0$) already shows that the solution is not necessarily positive. Is the notation unclear? – Nawaf Bou-Rabee Dec 21 '16 at 14:12