I am confused...So, we can construct a Levy process by specifying an infinitely divisible r.v. as the density of the increments at time scale 1 i.e. L_1.So, I can construct Gamma process as the one having the exponential distribution for L_1.How do I show that the resulting process consists only of jumps i.e. a pure-jump process? In general, having the density of L_1 how one shows that the resulting process is pure-jump?

cont tankov:In infinite activity models one does not need to introduce a Brownian component since the dynamics of jumps is already rich enough to generate nontrivial small time behavior...i can't give you a math explaination (and i'm also interested to it) but i think that levy measure, that is implied, contains all information about the problemby levy measure you know that 'distribution of jumps sizes' does not exist: jumps arrive infinitely often.is it right?bye

Marina, >think that levy measure, that is implied, contains all information about the problemyes, this makes some sense. >by levy measure you know that 'distribution of jumps sizes' does not exist: jumps arrive infinitely often.this statement I didn't quite understand Actually, I think that for my question, one just need to see what charactersitic function looks like. Then by Levy-Khinchine theorem we can see whether Gaussian coefficient is equal to zero, then the component in the Levy-Ito decomposition corresponding to the Brownian motion is equal to zero ---> it's a pure jump process.bye

hi, you are right.i have thougth: levy measure in a jump diffusion process is:a * f(x)where a is >= 0 and f(x) is the probability density function of jumps (size).so if you know f(x) (it is necessary to know the form of characteristic function anyway) you can say, seeing the levy measure, if the process is:continuous( levy measure =0), jump-diffusion (levy measure has form a*f(x)) or a pure jump process (otherwise).but i'm not sure it is correct, what do you think?Anyway to answer to your first post question you shoold know f(x) and if i'm not right also the form of characteristic function.bye

>i have thougth: levy measure in a jump diffusion process is:a * f(x)where a is >= 0 and f(x) is the probability density function of jumps (size).>this is true if you are talking about finite activity (finite number of jumps on every compact interval) Levy process e.g. compound Poisson process. Gamma process has infinite activity. Yes you can write its Levy measure as v(dx)= e^(-x)/x dx , but e^(-x)/x is no a density...Gamma process has infinite activity. >so if you know f(x) (it is necessary to know the form of characteristic function anyway) you can say, seeing the levy measure, if the process is:continuous( levy measure =0), jump-diffusion (levy measure has form a*f(x)) or a pure jump process (otherwise).>yes, you can do this. It's known that there are three popular methods to construct a Levy process:C1) Specify a Levy triplet (b, c, v) = (drift, Gaussian coeff., Levy Measure);C2) Specify an infinitely divisible r.v. as the density of L_1;C3) Time-changing Brownian motion with an independent increasing Levy process T (subordinator).My original question was: Let L_t be a Levy process. Suppose all we know is the density function of L_1 (so we specify Levy process using method C2). How do we find out whether our process is a pure jump process? (see http://math.nyu.edu/faculty/varadhan/sp ... emset2.pdf )For example, in case of Gamma process, we know the density of L_1. Then we can calculate the characteristic function of L_t, and then we would need to find a way to write it in the Levy-Khinchine form. Then by looking at it (using arguments you suggeste) we can figure out what the Levy measure looks like and hence the jump structure of our process.

All good points from everybody. I would only add that, mechanically,if you were given a characteristic exponent function phi(z) and weren't sure ifit was pure jump, you could take z -> infinity. If you find phi(z) grows like z^2, there isa diffusion component -- if it grows at a lesser rate, it's pure jump (with drift).In practice, this is often clear from inspection, but taking the limit would yielda rigorous yes/no test if there was some doubt regards,

cool! this makes sense.