Stochastic Processes

1. The definition of jump operation on page 4. is correct only when the function is right or left regular. If the function is just regular then of course the function can have in fact three jumps at any point. Two one sided jumps and the jump defined on page 4. Or one can even say that in this case one cannot define the concept jump and one should talk about discontinuity. The simplest way to fix the problem is that one can think about a jump at a point as a function of time and not just a number. The size of a jump is the modulus of discontinuity at the point of discontinuity. Of course if a function is one at a certain point and otherwise it is zero then it has a jump of size one. And the jump is the function itself. The problem with the definition appears in Chapter 2, where the point is that as in Example 2.6 the general regular case is irrelevant, as it always gives the same result as the left-regular case, which is the same as the classical case. So if the integrand is not left-regular we can still define the integral, but we will get a useless concept. The important concept is not the jump itself, but the size of the jump, that is the size of the discontinuity caused by the jump. The main issue is that because of the regularity of the trajectories the number of “big jumps” or “big discontinuities” is finite.

2. In Definition 1.12 3. On page 13. Example 1.22 the proof is given for the almost surely infinite case and not for the almost surely zero case.

4. In the proof of Example 1.18 the last element of the partition is 5. In the proof of Proposition 1.20 one should obviously assume that 6. On page 19, in Definition 1.33 I defined the stopped process with The usual definition in the literature is There is a very simple reason for it: In most of the cases Using this stopping time we want to have the estimation which is not true if the stopping time is zero for some outcomes. (That is why one needs the second definition.) I tried to avoid this problem. That is why in the definition of the local martingales and in the definition of the locally bounded processes I subtracted the value of the process at time zero before localization. I also tried to use the above estimation only for processes which are zero at t=0.

7. On page 19. in the proof of point 6. the process is not left-continuous but just left-regular.

8. On page 33, in the proof of Corollary 1.52, in line 11 “so the last inequality holds”.

9. On page 34 Corollary 1.54: 10. On page 35, in Corollary 1.59 one should emphasize that in the case b is infinity it is assumed, by definition, that the process is defined at b. The domain of definition should contain the right endpoint, even if it is infinity.

11. On page 42, in the proof of Lemma 1.70 in line 4 one should write and not 12. On page 47, Lemma 1.77. One should remark that the set of outcomes for which the limits are finite are also of probability one. To prove this one can use Doob’s inequality to show that almost all trajectories, restricted to the rational numbers, are bounded on bounded intervals.

13. On page 51: in the second line intersection is repeated twice.

14. On page 71, three lines after (1.51) if t>n then 15. On page 75, in the proof of Proposition 1.111 by the zero one law the set of outcomes where the trajectories are bounded is either zero or one. If almost all trajectories are bounded then one can study the process Y(t)=X(t)+t which has unbounded trajectories almost surely. Obviously Y has bounded moments on finite intervals if and only if X has. The estimation on the top of page 76 comes from the fact that on the set one can estimate the integral by 16. On page 96, in the last line of Example 1.137 there is an X instead of M.

ch1.txt · Utolsó módosítás: 2016/09/20 13:04 szerkesztette: medvegyev        