Solution to selected problems.

Chapter 3. Semimartingales and Decomposable Processes
1. Let {T i } n i=1 be a set of predictable stopping time. For each i, T i has an announcing sequence
{T i,j } ∞ j=1 . Let S k := max i T i,k and R k := min i T i,k . Then S k , R k are stopping time. {S k } and {R k }
make announcing sequences of maximum and minimum of {T i } i .
2. Let T n = S + (1 − 1/n) for each n. Then T n is a stopping time and T n ↑ T as n ↑ ∞. Therefore
T is a predictable stopping time.
3. Let S, T be two predictable stopping time. Then S ∧ T , S ∨ T are predictable stopping time
as shown in exercise 1. In addition, Λ = {S ∨ T = S ∧ T }. Therefore without loss of generality,
we can assume that S ≤ T . Let {T n } be an announcing sequence of T . Let R n = T n + n1 {Tn ≥S}.
Then R n is a stopping time since {R n ≤ t} = {T n ≤ t} ∩ ({t − T n ≤ n} ∪ {T n < S}) ∈ F t . R n is
increasing and lim R n = T Λ = S Λ .
4. For each X ∈ L, define a new process X n by X n = ∑ k∈N X k/2 n1 [k/2 n ,(k+1)/2 n ). Since each
summand is an optional set (See exercise 6) X n is an optional process. As a mapping on the
product space, X is a pints limit of X n . Therefore X is optional. Then by the definition P ⊂ O.
5. Suffice to show that all càdlàg processes are progressively measurable. (Then we can apply
monotone class argument.) For a càdlàg process X on [0, t], define a new process X n by putting
Xu n = X k/2 n for u ∈ [(k − 1)t/2 n , kt/2 n ), k = {1, . . . , 2 n }. Then on Ω × [0, t],
[ k − 1
{X n ∈ B} = ∪ k∈N+
[{ω : X k/2 n(ω) ∈ B} ×
2 n , k )]
2 n ∈ F t ⊗ B([0, t]) (50)
since X is adapted and X n is progressively measurable process. Since X is càdlàg , {X n } converges
pints to X, which therefore is also F t ⊗ B([0, t]) measurable.
6. (1) (S, T ]: Since 1 (S,T ] ∈ L, (S.T ] = {(s, ω) : 1 (S,T ] = 1} ∈ P by definition.
(2) [S, T ): Since 1 [S,T ) ∈ D, [S.T ) = {(s, ω) : 1 [S,T ) = 1} ∈ O by definition.
(3) (S, T ): Since [S, T ] = ∪ n [S.T + 1/n), [S, T ] is an optional set. Then (S, T ) = [0, T ) ∩ [0, S] c
and (S, T ) is optional as well.
(4) [S, T ) when S, T is predictable : Let {S n } and {T n } be announcing sequences of S and T .
Then [S, T ) = ∩ m ∪ n (S m , T n ]. Since (S m , T n ] is predictable by (1) for all m, n, [S, T ) is predictable.
7. Pick a set A ∈ F Sn . Then A = A ∩ {S n < T } ∈ F T − by theorem 7 and the definition of
{S n } n . (Note: The proof of theorem 7 does not require theorem 6). Since this is true for all n,
∨ n F Sn ⊂ F T − . To show the converse let Π = {B ∩ {t < T } : B ∈ F t }. Then Π is closed with
respect to finite intersection. B ∩ {t < T } = ∪ n (B ∩ {t < S n }). Since (B ∩ {t < S n }) ∩ {S n ≤ t} =
∅ ∈ F t ,B ∩ {t < S n } ∈ F Sn . Therefore B ∩ {t < T } ∈ ∨ n F Sn and Π ⊂ ∨ n F Sn . Then by Dynkin’s
theorem, F T − ⊂ ∨ n F Sn .
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