MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.


share|cite|improve this question
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. – Ilya Jan 27 '12 at 12:22
@Ilya: Thank you. Indeed, it was difficult to identify the very nature of this process. – Jon Jan 27 '12 at 13:25
@ 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. – The Bridge Jun 13 '12 at 15:55
@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. – Jon Jun 13 '12 at 16:37
@TheBridge is right, naturally: sign(X) is not sign(W). – Did Jul 18 '12 at 8:48

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.