4
$\begingroup$

While studying a portfolio optimization problem, I came across the process $$dX(t) = X(t)\,\Big(\,\big(\mu - \alpha\,1_{\{X(t)\,\geq\,C\}}\big)\,dt + \sigma\,dW(t) \Big)$$ which has a discountinuous drift: once the process crosses a pre-defined and fixed threshold $C>0$, it obtains an additional negativ drift of $-\alpha$.

I would like to understand if the SDE above has a solution and/or if it is possible to obtain the distribution/moments of the process at a given point in time.

Has anybody seen something similar or point me to relevant results in the literature? Any comments, ideas and suggestions are highly appreciated. Thank you!

EDIT 1

We can assume $\mu>0$ and $\alpha>0$.

EDIT 2

I would like to propose a solution approach using Girsanov's theorem: $$ \begin{eqnarray*} dX(t) &=& X(t)\,\Big(\,\big(\mu - \alpha\,1_{\{X(t)\,\geq\,C\}}\big)\,dt + \sigma\,dW(t) \Big) \\ &=& X(t)\,\Big(\,\mu\,dt + \sigma\,\Big(dW(t) - \frac{\alpha}{\sigma}\,1_{\{X(t)\,\geq\,C\}}\,dt\Big) \Big) \\ &=& X(t)\,\Big(\,\mu\,dt + \sigma\,d\tilde{W}(t) \Big) \\ \end{eqnarray*} $$ with Brownian motion $\tilde{W}(t)=W(t)+\int_{0}^{t}\gamma(s)\,ds$, associated measure $$ \frac{d\tilde{\mathbb{P}}}{d\mathbb{P}} \bigg|_{\mathcal{F}_{t}} = e^{-\int_{0}^{t}\gamma(s)\,dW(s) - \frac{1}{2}\int_{0}^{t}\gamma(s)^{2}ds} $$ and $\gamma(t):=-\frac{\alpha}{\sigma}1_{\{X(t)\,\geq\,C\}}$. The SDE under $\tilde{\mathbb{P}}$ has a (strong) solution, which would imply a (strong) solution for the original problem under $\mathbb{P}$.

The pressing question is of course if this change of measure is well-defined. Novikov's condition clearly holds, but I might have missed a measurability/integrability criterion.

$\endgroup$
2
  • $\begingroup$ I guess $\mu>0$, but is $\alpha$ < or >$\mu$ ? $\endgroup$ Jul 7, 2016 at 8:00
  • $\begingroup$ @JeanDuchon Yes, $\mu>0$. $\alpha$ does not stand in a natural realtion to $\mu$, but if either of the cases $\alpha\leq\mu$ or $\alpha\geq\mu$ is easier to solve, we can simply assume this to be true. $\endgroup$
    – Mark
    Jul 12, 2016 at 7:03

1 Answer 1

2
$\begingroup$

You are absolutely on the right track, and in fact, your change of measure is well-defined. In order to fully leverage this change of measure, one must eliminate the time integrals that appear in it. This is nicely done in Brownian Motion and Stochastic Calculus by Ioannis Karatzas and Steven E. Shreve (1991); see in particular page 439 of Chapter 6 entitled Paul Levy's Theory of Brownian Local Time.

Minor point: I think your derivation would be more transparent if you apply Ito's formula to $\log X(t)$ and then use Girsanov.

$\endgroup$
1
  • $\begingroup$ That you for your comments and the useful reference! $\endgroup$
    – Mark
    Aug 24, 2016 at 7:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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