Supplement to " Testing Whether Jumps Have Finite or Infinite Activity

نویسندگان

  • Yacine Aït-Sahalia
  • JEAN JACOD
چکیده

When  ≥ 2 these processes are finite-valued, but not necessarily so when   2. In this case we have to be careful: if  = inf( : () =∞) and 0 = inf( :  0() = ∞) then both processes () and  0() are null at 0 and finite-valued on [0 ) and [0 0) respectively, the first continuous and the second one càdlàg with jumps not bigger than 1, and they are infinite on (∞) and (0∞). However, if 0   ∞ or 0  0 ∞ we may have () or  0()0 finite or infinite, and the left limits satisfy ()− = () and  0()0− ≤  0()0 ≤  0()0− + 1. Note that  0(0) is the number of jumps over [0 ], whereas (0) = R  0 (R) . Lemma 3. For any random time  and any  ≥ 0, the two sets {() =∞} and { 0() =∞} are a.s. equal. Proof. Set  = inf( : () ≥ ) and 0  = inf( :  0() ≥ ), and consider the sets  = {()  ∞} and 0 = { 0()  ∞}. Observe that () ≤  and  ()0 ≤  + 1, which entail  = ∪≥1{ ≤ } and 0 = ∪≥1{ ≤ 0 }. Moreover () is the predictable compensator of  0(), so for any stopping time  we have E(() ) = E( 0() ). We deduce that  0()  ∞ and ()0   ∞ a.s., and the almost sure equality  = 0 is then obvious.

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تاریخ انتشار 2011