5
$\begingroup$

I have the following SDE: $$dY_{t}=A\left(\frac{W_{t}^{1}}{\sqrt{t}},\frac{Y_{t}}{\sqrt{t}}\right)dW_{t}^{1}+B\left(\frac{W_{t}^{1}}{\sqrt{t}},\frac{Y_{t}}{\sqrt{t}}\right)dW_{t}^{2}$$

where $W_{t}^{1}$ and $W_{t}^{1}$ are two independent brownian motions and $$A(x,y)=a\frac{\Phi(y)\Phi(-y)e^{0.5(y^2-x^2)}+\Phi(x)\Phi(-x)e^{0.5(x^2-y^2)}}{1+a(1-2\Phi(x))(1-2\Phi(y))}, \ \ \ \ \ \ \ \ B=\sqrt{1-A^2}$$

It seems unlikely, but I was wondering if there may be an explicit solution to this stochastic differential equation in terms of $W_{t}^{1}$ and $W_{t}^{2}$? Perhaps it can be reduced to a linear SDE with a suitable function?

Any help would be greatly appreciated.

$\endgroup$
1
$\begingroup$

When $W_1$ and $W_2$ are independent, $Y$ is equal in law to a Brownian Motion:

  1. It is obviously a local martingale under its natural filtration, and
  2. From Ito isometry, $$\mathbf{E}Y_t^2 = \mathbf{E} \int_0^t A(...)^2 + B(...)^2 ds = t.$$ Then, from Levy's theorem (http://almostsure.wordpress.com/2010/04/13/levys-characterization-of-brownian-motion/), these two conditions imply that $Y$ is a Brownian Motion.
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.