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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.

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  • $\begingroup$ Are $W_{1t}$ and $W_{2t}$ independent standard Brownian motions? $\endgroup$ – Nawaf Bou-Rabee Dec 19 '16 at 17:32
  • $\begingroup$ Yes that is the case $\endgroup$ – Julian Karls Dec 20 '16 at 13:38
  • $\begingroup$ Maybe this should be specified in the question? $\endgroup$ – Nawaf Bou-Rabee Dec 20 '16 at 14:16
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    $\begingroup$ You are right. Sorry for this. I modified the question $\endgroup$ – Julian Karls Dec 21 '16 at 11:48
  • $\begingroup$ 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 $\Phi_t^{-1}$. $\endgroup$ – Nawaf Bou-Rabee Dec 23 '16 at 11:54
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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.

This solution is adapted from more general results for linear SDEs which may be found in, e.g., Chapter 5 of Lawrence C. Evans' AMS book entitled An Introduction to Stochastic Differential Equations.

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  • $\begingroup$ (*) looks positive but this will not have a positive solution. $\endgroup$ – user83457 Dec 21 '16 at 14:04
  • $\begingroup$ @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? $\endgroup$ – Nawaf Bou-Rabee Dec 21 '16 at 14:12
  • $\begingroup$ I'm sorry, that's multiplication in (*), isn't it $\endgroup$ – user83457 Dec 21 '16 at 14:50
  • $\begingroup$ Yes, it is multiplication. $\endgroup$ – Nawaf Bou-Rabee Dec 21 '16 at 14:50

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