Piecewise constant local martingales with bounded numbers of jumps

A piecewise constant local martingale $M$ with boundedly many jumps is a uniformly integrable martingale if and only if $M_\infty^-$ is integrable.


Main theorem
Let (Ω, F , (F t ) t≥0 , P) denote a filtered probability space with t≥0 F t ⊂ F . In Section 2, we shall prove the following theorem.
Theorem 1. Assume for some N ∈ N 0 and some stopping times 0 ≤ ρ 1 ≤ · · · ≤ ρ N we have a local martingale M of the form where J m is F ρm -measurable for each m = 1, · · · , N . If then M is a uniformly integrable martingale.
In (1.2), we could replace the limit inferior by a limit since M only has finitely many jumps and hence converges to a random variable M ∞ . Hence, (1.2) is equivalent to E[M − ∞ ] < ∞.
Corollary 2. Suppose the notation and assumptions of Theorem 1 hold, but with (1.2) replaced by Then M is a martingale.
Proof. Fix a deterministic time T ≥ 0 and consider the local martingale M = M T ; that is, M is the local martingale M stopped at time T . Then M satisfies the conditions of Theorem 1, with J m replaced by J m 1 {ρm≤T } for each m = 1, · · · , N . Hence, M is a uniformly integrable martingale. Since T was chosen arbitrarily the assertion follows. Jacod and Shiryaev (1998) prove the following special case of Theorem 1.
Proposition 3. Fix N ∈ N 0 and assume we have a discrete-time filtration G = (G m ) m=0,1,··· ,N and a G- Note that Proposition 3 follows from Theorem 1. Indeed, define the continuous-time process M and the filtration (F t ) t≥0 by M t = Y [t]∧n and F t = G [t]∧n , respectively, where [t] denotes the largest integer smaller than or equal to t. Then M is a local martingale as in (1.1), with N replaced by N + 1. To see this, set ρ m = m − 1 and J m = Y m−1 − Y m−2 with Y −1 := 0, for each m = 1, · · · , N + 1. Applying Theorem 1 then yields Proposition 3.

Proofs of Theorem 1
In the following, we will provide two proofs of Theorem 1. The first one assumes Proposition 3 is already shown and reduces the more general situation of Theorem 1 to the discrete-time setup of Proposition 3. The second proof does not assume Proposition 3, but instead provides a direct argument based on an induction.
Proof I, relying on Proposition 3. Let us set ρ = 0 and ρ N +1 = ∞ and let (τ n ) n∈N denote a localization sequence of M such that M τn is a uniformly integrable martingale for each n ∈ N. For any stopping time τ we may define a sigma algebra and, furthermore, lim n↑∞ σ n = ∞. We now fix n ∈ N and prove that Y σn is a G-martingale, which then yields that Y is a G-local martingale. To this end, we have, for each m = 0, · · · , N , Hence, Y is indeed a G-local martingale. The assumptions of the theorem yield that E[Y − N ] < ∞; hence Y a G-uniformly integrable martingale by Proposition 3. Now, fix t ≥ 0 and A ∈ F t . Then we get E Proof II, relying on an induction argument. We proceed by induction over N . The case N = 0 is clear. Hence, let us assume the assertion is proven for some N ∈ N 0 and consider the assertion with N replaced by N + 1. Let (τ n ) n∈N denote a corresponding localization sequence such that M τn is a uniformly integrable martingale for each n ∈ N.
Step 1 : In the first step, we want to argue that the nondecreasing sequence ( τ n ) n∈N , given by is also a localization sequence for M . To this end, fix k ∈ N and consider the process

Then we have
Next, we argue that M is also a local martingale, again with localization sequence (τ n ) n∈N . Indeed, for n ∈ N, t, h ≥ 0, and A ∈ F t note that where we used the definition of M , {ρ 1 ≤ τ k ≤ t} ∈ F t , A ∩ {ρ 1 ≤ τ k } ∩ {τ k > t} ∈ F τ k , and the martingale property of M τn . Alternatively, we could have observed that M · = · 0 1 {ρ 1 ≤τ k <s} dM s (using the fact that 1 {ρ 1 ≤τ k } 1 ]]τ k ,∞[[} is bounded and predictable since it is adapted and leftcontinuous). Hence, M is a local martingale of the form satisfying (2.1), and the induction hypothesis yields that M is a uniformly integrable martingale. This again yields that is also a uniformly integrable martingale, proving the claim that ( τ n ) n∈N is a localization sequence for M .
Step 2 : We want to argue that M t ∈ L 1 for each t ∈ [0, ∞]. To this end, fix t ∈ [0, ∞] and note Here, the inequality in (2.2) is an application of Fatou's lemma. The equality in (2.3) relies on the fact that for any uniformly integrable martingale The inequality in (2.4) uses that (M τn ) − is a uniformly integrable submartingale, thanks to Jensen's inequality, for each n ∈ N. The inequality in (2.5) (which is, actually, an equality) uses the fact that M τn ∈ {0, M ∞ }, for each n ∈ N, by construction of the localization sequence ( τ n ) n∈N . Finally, the inequality in (2.6) holds by assumption.
Step 3 : We now argue that M is a uniformly integrable martingale. To this end, fix t ≥ 0 and A ∈ F t . Observe that (2.9) We obtained the equality in (2.7) since τ n = ∞ on the event { τ n ≥ ρ 1 }, and since the first term on the left-hand side is zero by the dominated convergence theorem and the second one thanks to the form of M . In (2.8), we used the martingale property of M τn in the first term and the fact that M τn = 0 on the event { τ n < ∞} in the second term, for each n ∈ N. Finally, we exchanged limit and expectation in (2.9) again by an application of the dominated convergence theorem. This then concludes the proof.
The next example illustrates that the assumptions of Corollary 2 are not sufficient to guarantee that M is a uniformy integrable martingale, even if there is only one jump possible, that is, even if N = 1. The example is adapted from Ruf (2015), where it is used as a counterexample for a different conjecture.
Example 5. Let ρ be an N ∪ {∞}-valued random variable with This then yields that