Now, I would like to provide some examples where a really huge formula can be simplified. Namely in all these examples, we can omit this twirl with indicator and this simplification makes our ranges of formula much more pleasant. The first example which I would like to discuss now is the case when X_t is a process of bounded variation. This means the following, so if we take an interval A, B and divide it into n sub-intervals and consider as the sum X_tk minus X_tk minus one, absolute value of these differences. And then we tend the maximum difference between T_i and T_i minus one to zero. Then, this limit should exist. Okay. You know that for instance, Brownian motion is not a process of bounded variation. It turns out that a Levy process X_t is of bounded variation if and only if the parameter sigma, the second parameter in the Levy triplet, is equal to zero. And as for the Levy measure, the integral of x nu dx, for all x smaller than the one infinite. In this case, it turns out that the integral over R, iux multiplied by this indicator, nu dx, is this. We can simply divide this integral into the difference between two integrals. First one with this term, a second one with this term. If you look attentively in the second term, you'd get that it looks very similar to the first term in this representation. But it equates also iu multiplied by a constant, and if we can divide this integral in the two sub-integrals, we can just combine the first term with the second integral. Okay. This term vanishes because sigma is equal to zero and what we finally have is the following formula: so, a characteristic function of any Levy process with finite variation is equal to exponent, yeah, I should write t multiplied by iu mu-tilde plus integral exponent in the power iux minus one nu dx. This term shall be here, one more bracket. And here, as this mu-tilde is equal to mu minus integral over all x whose absolute values smaller than one, x nu dx. You see that from this condition and immediately follows that mu-tilde is finite. Okay, this is exactly the simplification of the Levy-Khintchine formula for the case when a Levy process has bounded variation. Okay, let me write some more examples. Second example, is the case when X_t is a compound Poisson process. It turns out that a Levy process is a compound Poisson process if and only if sigma is equal to zero and the integral over R, nu dx. This is very often notated as nu of R, this object is finite. Definitely, if these assumptions are fulfilled, then these two assumptions are also fulfilled. In fact, integral x nu dx over all x whose absolute value smaller than one is less or equals in nu of the set absolute value of x smaller than one. If you know that nu of R is finite, then nu of the interval minus 1, 1 is also finite, and therefore, both of these assumptions are fulfilled and since this formula was constructed only via the assumptions that these two kinds of conditions are full hold. Therefore, this formula also holds for the compound Poisson process. Okay. Our third example is the case when X_t is a subordinator. The subordinator is a Levy process which is non-negative or in other words, it's non-decreasing. That is X_t is larger or equal to zero almost surely, or it's exactly the same as to say that X_t is larger or equal X_s almost surely for any t larger than s. Why are these two statements are the same? This is just a corollary from the fact that any Levy process has stationary increments. That is, X_t minus X_s has the same distribution as X_t minus s. Therefore, if the process is non-negative, this means as the right-hand side of this equality is non-negative, then X_t minus X_s is larger or equal to zero. Otherwise, if we know that X_t is larger or equal as an X_s for any t larger than s, and the left-hand side is larger or equal than zero. And from this equality, it immediately follows that X_t minus s is also larger or equal than zero, while as in other words, X is in a negative process. Well, and in terms of Levy triplet, X_t is a subordinator if and only if the following conditions hold: Sigma shall be equal to zero. Levy measure of the negative half line shall be also equal to zero. And Levy measure of the positive half line should possess the following property: integral over x nu dx from zero to one should be finite. If these three conditions hold, then of course the process is of bounded variation. So, it's a very easy exercise. In this case, also, this formula, a simplification of the Levy-Khintchine theorem definitely holds. Okay. Now, we'll have some examples of Levy processes for which the Levy-Khintchine formula can be simplified.
