Intuitively it seems that an infinitely divisible distribution should be continuous but I don't see how it comes into play in the definition of a Levy process or in the Levy-Khinthcine formula?I am particularly concerned with continuity at the origin? Or whether the density function should even be differentiable?Thanks in advance

Last edited by AdareIre on June 7th, 2012, 10:00 pm, edited 1 time in total.

Poisson distribution is infinitely divisible and is a discrete distribution (Poisson process is a Levy process), there are several others (negative binomial for example) so your intuition is off the mark.

Sorry for ambiguity in my previous post.I understand that discrete distributions can be infinitely divisible.Consider a strictly positive density function f(x) defined for all x\in(-\infty,\infty). Can f be infinitely divisible if it has a discontinuity at a certain point ? e.g. say a discontinuity at x = 0 such thatMust f also be differentiable for x\in(-\infty,\infty) to be infinitely divisible?

Last edited by AdareIre on June 7th, 2012, 10:00 pm, edited 1 time in total.

The density of the marginal law of Levy process does not necessarily exist.I don't see anything intuitive about a density being continuous. K. Sato's book discusses distribution properties of the marginal laws of Levy processes. Look at that.

Thanks eh for your comments,My intuition was coming from the perspective that if an inf div density f exists which has a discontinuity at some point. Then there exist a density function f_n such that an n-fold convolution of f_n with itself gives f. I find it hard to see how an n-fold convolution of any function could produce a function with a discontinuity. It seems to me that the resultant density would be smooth.Analogously, suppose the random variable X has a density f and the random variable X_n has a density function f_n as described above. Again it seems strange that the sum of n X_n random variables would have a density function which is not smooth or continuous at least.Unfortunately there is not a copy of Sato in the Library....

QuoteOriginally posted by: AdareIreThanks eh for your comments,My intuition was coming from the perspective that if an inf div density f exists which has a discontinuity at some point. Then there exist a density function f_n such that an n-fold convolution of f_n with itself gives f. I find it hard to see how an n-fold convolution of any function could produce a function with a discontinuity. It seems to me that the resultant density would be smooth.Suppose the support of f is a lattice (say +/- n, n an integer). Then, so will be the convolutions ...

I have worked through some examples with a discontinuity at zero and indeed it does seem to be okay!Suppose f is an infinitely divisible 2-EPT function with no pointmass at zero but is discontinuous at zero. Let f_n be the density function corresponding to the n^th root of the characteristic function of f. As f is discontinuous at zero f_n will have a pointmass at zero.I have no proof of this but it seems to work and make intuitive sense!! Thanks

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