1
$\begingroup$

Given Tanaka sde

$$dX_t=[a{\rm sign}(X_t)+b]dW_t$$

is there associated a diffusion process and so a Kolmogorov (Fokker-Planck) equation? What is this equation?

References answering the question are welcome.

Thanks.

$\endgroup$
11
  • $\begingroup$ Well, it's not quite an SDE in the sense that the right-hand side does not depend on $X$, so you can write the explicit solution $$ X_t = X_0+\int\limits_0^t[a\sign W_s +b ]\mathrm dW_s. $$ On the other hand, the Kolmogorov equation is defined only for Markov processes, I guess. I am not sure that the process $X$ is Markov. $\endgroup$
    – SBF
    Jan 27, 2012 at 12:22
  • $\begingroup$ @Ilya: Thank you. Indeed, it was difficult to identify the very nature of this process. $\endgroup$
    – Jon
    Jan 27, 2012 at 13:25
  • 1
    $\begingroup$ @ Jon : a remark as noted by Ilya, your sde is not Tanaka's sde, the right equation is $dX_t=sign(X_t)dW_t$. Best regards. $\endgroup$
    – The Bridge
    Jun 13, 2012 at 15:55
  • $\begingroup$ @The Bridge: There is such a big difference that I cannot see it and this is not the remark by Ilya at all. As you can note, if I take the two constants to be $a=1$ and $b=0$ you are back to Tanaka sde otherwise it is just the sum of a Tanaka plus a normal Wiener. $\endgroup$
    – Jon
    Jun 13, 2012 at 16:37
  • $\begingroup$ @TheBridge is right, naturally: sign(X) is not sign(W). $\endgroup$
    – Did
    Jul 18, 2012 at 8:48

0

Your Answer

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