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Let $b : [0,T] \times \mathbb{R}^n \rightarrow \mathbb{R}^n, \sigma : [0,T] \times \mathbb{R}^n \rightarrow \mathbb{R}^{n \times m}.$ Consider an SDE $$ dX_t = b(t,X_t)dt + \sigma(t,X_t) dW_t$$ Then a solution exists if the coefficients satisfy integrability and Lipschitz conditions and if $W$ is a $n$-dimensional Brownian motion.

My Question: Do we still have existence and uniqueness of a solution if we replace $W$ by $M = (B, \dots, B)$ with $B$ a Brownian motion.

(In this case $M$ is not a Brownian motion, because the components are not independent.)

White
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    Every SDE run by $M$ can be translated into a SDE run by $W$, changing the matrix function $\sigma$, hence yes, we have all the results you cite. – Did Sep 19 '18 at 10:00

1 Answers1

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As explained by Did the answer is as follows. Define a new drift coefficient $ \gamma$ such that for all $ i \in \{ 1, \dots n \} $ and for all $ t \in [0,T] , x \in \mathbb{R}^n $ $$ \gamma^{i1}(t,x) = \sum_{j=1}^m \sigma^{i,j}(t,x) . $$ And, for all $ i \in \{ 1, \dots n \}, j \in \{2, \dots , m \} $ and $ t \in [0,T] , x \in \mathbb{R}^n $ $$ \gamma^{ij}(t,x) = 0. $$ Let, $W^1 = B$ and choose independent Brownian motions $W_2, \dots, W_m,$ which are also independent of $W^1.$ Let $W = (W_1, \dots, W_n).$ Then, the SDE above is the same as $$ d X_t = b(t,X_t) dt + \gamma(t,X_t) dW_t. $$

There is still one question. For a Brownian motion $W^1$ can we always find other independent Brownian motion $W^2, \dots, W^m$ which are independent of $W^1$ ?

White
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