Dear all , Please forgive me if my question is too trivial. I know a process z[t] is said to be a Weiner Process if dz[t] ~ N(0, dt). But if dz[t] follows some other distribution like, T-distribution. Skew-Normal distribution etc, then how this process should be called? Thanks and regards,

Not all distributions can be turned into a continuous-time process.If you can do it, you end up with what's generally called a Levy process -- as a term in a SDE, you could write dL(t).The distributions have to be `infinitely divisible'. It can be done for the Student-t distribution; here's a paper that refs a proof of that:http://www.cls.dk/caf/hubalek.pdfHere's a paper of mine on going from (exponential) Levy processes to option values:http://www.optioncity.net/pubs/ExpLevy.pdfThere's also a forum FAQ on Levy processes (search will turn it up).p.s. To finally answer your original question, you can call it a Student-t process. regards,

QuoteOriginally posted by: AlanNot all distributions can be turned into a continuous-time process.If you can do it, you end up with what's generally called a Levy process -- as a term in a SDE, you could write dL(t).The distributions have to be `infinitely divisible'. It can be done for the Student-t distribution; here's a paper that refs a proof of that:http://www.cls.dk/caf/hubalek.pdfHere's a paper of mine on going from (exponential) Levy processes to option values:http://www.optioncity.net/pubs/ExpLevy.pdfThere's also a forum FAQ on Levy processes (search will turn it up).p.s. To finally answer your original question, you can call it a Student-t process. regards,Alan, nicely written paper, but I see the following. "Kou’s derivation stresses theimportance of the memoryless property of the exponential distribution." The exponential distributions are not memory-less, but finite-memory, or have the property that the "infinite past has a negligible effect on the present and thus the future as well"

Thanks. Regarding, memoryless, I won't quibble -- it's Kou's term (and many others), but I have to agree it's an awkward way ofexpressing a Markov-type property.