Solution to selected problems.

Solution to selected problems.
Chapter 1. Preliminaries
1. ∀A ∈ F S , ∀t ≥ 0, A ∩ {T ≤ t} = (A ∩ {S ≤ t}) ∩ {T ≤ t}, since {T ≤ t} ⊂ {S ≤ t}. Since
A ∩ {S ≤ t} ∈ F t and {T ≤ t} ∈ F t , A ∩ {T ≤ t} ∈ F t . Thus F S ⊂ F T .
2. Let Ω = N and F = P(N) be the power set of the natural numbers. Let F n = σ({2}, {3}, . . . , {n+
1}), ∀n. Then (F n ) n≥1 is a filtration. Let S = 3 · 1 3 and T = 4. Then S ≤ T and
{
{3} if n = 3
{ω : S(ω) = n} =
∅ otherwise
{
Ω if n = 4
{ω : T (ω) = n} =
∅ otherwise
Hence {S = n} ∈ F n , {T = n} ∈ F n , ∀n and S, T are both stopping time. However {ω : T − S =
1} = {ω : 1 {3} (ω) = 1} = {3} /∈ F 1 . Therefore T − S is not a stopping time.
3. Observe that {T n ≤ t} ∈ F t and {T n < t} ∈ F t for all n ∈ N , t ∈ R + , since T n is stopping
time and we assume usual hypothesis. Then
(1) sup n T n is a stopping time since ∀t ≥ 0, {sup n T n ≤ t} = ∩ n {T n ≤ t} ∈ F t .
(2) inf n T n is a stopping time since {inf n T n < t} = ∪{T n < t} ∈ F t
(3) lim sup n→∞ is a stopping time since lim sup n→∞ = inf m sup n≥m T n (and (1), (2).)
(4) lim inf n→∞ is a stopping time since lim inf n→∞ = sup m inf n≥m T n (and (1), (2).).
4. T is clearly a stopping time by exercise 3, since T = lim inf n T n . F T ⊂ F Tn , ∀n since T ≤ T n ,
and F T ⊂ ∩ n F Tn by exercise 1. Pick a set A ∈ ∩ n F Tn .∀n ≥ 1, A ∈ F Tn and A ∩ {T n ≤ t} ∈ F t .
Therefore A ∩ {T ≤ t} = A ∩ (∩ n {T n ≤ t}) = ∩ n (A ∩ {T n ≤ t}) ∈ F t . This implies ∩ n F Tn ⊂ F T
and completes the proof.
5. (a) By completeness of L p space, X ∈ L P . By Jensen’s inequality, E|M t | p = E|E(X|F t )| p ≤
E [E(|X| p |F t )] = E|X| p < ∞ for p > 1.
(b) By (a), M t ∈ L p ⊂ L 1 . For t ≥ s ≥ 0, E(M t |F s ) = E(E(X|F t )|F s ) = E(X|F s ) = M s a.s. {M t }
is a martingale. Next, we show that {M t } is continuous. By Jensen’s inequality, for p > 1,
E|M n t − M t | p = E|E(M n ∞ − X|F t )| p ≤ E|M n ∞ − X| p , ∀t ≥ 0. (1)
It follows that sup t E|Mt n − M t | p ≤ E|M∞ n − X| p → 0 as n → ∞. Fix arbitrary ε > 0. By
Chebychev’s and Doob’s inequality,
(
)
P sup |Mt n − M t | > ε ≤ 1 ( ) p p
t
ε p E(sup |Mt n − M t | p sup
) ≤
t E|Mt n − M t | p
t
p − 1
ε p → 0. (2)
Therefore M n converges to M uniformly in probability. There exists a subsequence {n k } such that
M nk converges uniformly to M with probability 1. Then M is continuous since for almost all ω, it
is a limit of uniformly convergent continuous paths.
1

6. Let p(n) denote a probability mass function of Poisson distribution with parameter λt. Assume
λt is integer as given.
∑λt
E|N t − λt| =E(N t − λt) + 2E(N t − λt) − = 2E(N t − λt) − = 2 (λt − n)p(n)
=2e −λt λt
∑
(λt − n) (λt)n
n!
n=0
=2e −λt (λt) λt
(λt − 1)!
( ∑λt
= 2λte −λt (λt) n
−
n!
n=0
n=0
λt−1
∑
n=0
)
(λt) n
n!
(3)
7. Since N has stationary increments, for t ≥ s ≥ 0,
E(N t − N s ) 2 = EN 2 t−s = V ar(N t−s ) + (EN t−s ) 2 = λ(t − s)[1 + λ(t − s)]. (4)
As t ↓ s (or s ↑ t), E(N t − N s ) 2 → 0. N is continuous in L 2 and therefore in probability.
8. Suppose τ α is a stopping time. A 3-dimensional Brownian motion is a strong Markov process
we know that P (τ α < ∞) = 1. Let’s define W t := B τα +t −B τα . W is also a 3-dimensional Brownian
motion and its argumented filtration Ft W is independent of F τα+ = F τα . Observe ‖W 0 ‖ = ‖B τα ‖ =
α. Let S = inf t {t > 0 : ‖W t ‖ ≤ α}. Then S is a Ft W stopping time and {S ≤ s} ∈ Fs W . So {S ≤ s}
has to be independent of any sets in F τα . However {τ α ≤ t} ∩ {S ≤ s} = ∅, which implies that
{τ α ≤ t} and {S ≤ s} are dependent. Sine {τ α ≤ t} ∈ F τα , this contradicts the fact that F τα and
Ft
W are independent. Hence τ α is not a stopping time.
9. (a) Since L 2 -space is complete, it suffices to show that S n t = ∑ n
i=1 M i t is a Cauchy sequence
w.r.t. n. For m ≥ n, by independence of {M i },
E(S n t − S m t ) 2 = E
( m
∑
i=n+1
M i t
) 2
=
m∑
i=n+1
E ( Mt
i ) 2
∑
m 1 = t
i 2 . (5)
i=n+1
, Since ∑ ∞
i=1 1/i2 < ∞, as n, m → ∞, E(S n t − S m t ) 2 → ∞. {S n t } n is Cauchy and its limit M t is
well defined for all t ≥ 0.
(b)First, recall Kolmogorov’s convergence criterion: Suppose {X i } i≥1 is a sequence of independent
random variables. If ∑ i V ar(X i) < ∞, then ∑ i (X i − EX i ) converges a.s.
For all i and t, △M i t = (1/i)△N i t and △M i t > 0. By Fubini’s theorem and monotone convergence
theorem,
∑
△M s = ∑
s≤t s≤t
∞∑
△Ms i =
i=1
∞∑ ∑
△Ms i =
i=1 s≤t
∞∑ ∑
i=1 s≤t
△N i s
i
=
∞∑
i=1
N i t
i . (6)
Let X i = (Nt i −t)/i. Then {X i } i is a sequence of independent random variables such that EX i = 0,
V ar(X i ) = 1/i 2 and hence ∑ ∞
∑ i=1 V ar(X i) < ∞. Therefore Kolmogorov’s criterion implies that
∞
i=1 X i converges almost surely. On the other hand, ∑ ∞
i=1 t/i = ∞. Therefore, ∑ s≤t △M s =
∑ ∞
i=1 N t i /i = ∞.
2

10. (a) Let N t = ∑ i 1 i (N i t − t) and L t = ∑ i 1 i (Li t − t). As we show in exercise 9(a), N, M are
well defined in L 2 sense. Then by linearity of L 2 space M is also well defined in L 2 sense since
M t = ∑ i
1 [
(N
i
i t − t) − (L i t − t) ] = ∑ i
1
i (N i t − t) − ∑ i
1
i (Li t − t) = N t − L t . (7)
Both terms in right size are martingales change only by jumps as shown in exercise 9(b). Hence
M t is a martingale which changes only by jumps.
(b) First show that given two independent Poisson processes N and L, ∑ s>0 △N s△L s = 0 a.s.,
i.e. N and L almost surely don’t jump simultaneously. Let {T n } n≥1 be a sequence of jump times
of a process L. Then ∑ s>0 △N s△L s = ∑ n △N T n
. We want to show that ∑ n △N T n
= 0 a.s. Since
△N Tn ≥ 0, it is enough to show that E△N Tn = 0 for ∀n ≥ 1.
Fix n ≥ 1 and let µ Tn be a induced probability measure on R + of T n . By conditioning,
E(△N Tn ) = E [E (△N Tn |T n )] =
∫ ∞
0
E (△N Tn |T n = t) µ Tn (dt) =
∫ ∞
0
E (△N t ) µ Tn (dt), (8)
where last equality is by independence of N and T n . It follows that E△N Tn = E△N t . Since
△N t ∈ L 1 and P (△N t = 0) = 1 by problem 25, E△N t = 0, hence E△N Tn = 0.
Next we show that the previous claim holds even when there are countably many Poisson processes.
assume that there exist countably many independent Poisson processes {N i } i≥1 . Let A ⊂ Ω be a
set on which more than two processes of {N i } i≥1 jump simultaneously. Let Ω ij denotes a set on
which N i and N j don’t jump simultaneously. Then P (Ω ij ) = 1 for i ≠ j by previous assertion.
Since A ⊂ ∪ i>j Ω c ij , P (A) ≤ ∑ i>j P (Ωc ij ) = 0. Therefore jumps don’t happen simultaneously
almost surely.
Going back to the main proof, by (a) and the fact that N and L don’t jump simultaneously, ∀t > 0,
∑
|△M s | = ∑ |△N s | + ∑ |△L s | = ∞ a.s. (9)
s≤t
s≤t
s≤t
11. Continuity: We use notations adopted in Example 2 in section 4 (P33). Assume E|U 1 | < ∞.
By independence of U i , elementary inequality, Markov inequality, and the property of Poisson
process, we observe
lim P (|Z t − Z s | > ɛ) = lim
s→t
≤ lim
s→t
∑
k
P (
≤ lim
s→t
E|U 1 |
ɛ
k∑
i=1
s→t
∑
k
P (|Z t − Z s | > ɛ |N t − N s = k)P (N t − N s = k)
∑
|U i | > ɛ)P (N t − N s = k) ≤ lim
[kP (|U 1 | > ɛ ]
s→t k ) P (N t − N s = k)
∑
k 2 E|U 1 |
P (N t − N s = k) = lim {λ(t − s)} = 0
s→t ɛ
k
k
(10)
3

Independent Increment: Let F be a distribution function of U. By using independence of
{U k } k and strong Markov property of N, for arbitrary t, s : t ≥ s ≥ 0,
)
E
(e iu(Z t−Z s )+ivZ s
(
=E E
(
=E
=E
= E
(e iu P N t
k=Ns+1 U k+iv P )
Ns
k=1 U k
(e iu P N t
k=Ns+1 U k+iv P Ns
k=1 U k
|F s
))
e iv P Ns
k=1 U k
E
(
e iv P Ns
k=1 U k
)
E
(e iu P N t
)) (
k=Ns+1 U k
|F s = E
(e iu P N t
)
k=Ns+1 U k
= E ( e ivZs) E
This shows that Z has independent increments.
e iv P Ns
k=1 U k
E
(
e iu(Zt−Zs)) .
(e iu P N t
))
k=Ns+1 U k
(11)
Stationary increment: Since {U k } k are i.i.d and independent of N t ,
E ( e iuZ ) t
=E ( E ( [ (∫ ) ]
e iuZ t
)) N(t)
|N t = E e iux F (dx) = ∑ (λt) n e −λt
n!
n≥0
{ (∫
= exp −tλ
)}
[1 − e iux ]F (dx) .
(∫
) n
e iux F (dx)
(12)
By (11) and (12), set v = u,
(
E e iu(Zt−Zs)) =E ( e iuZt) /E ( e iuZs) { (∫
)}
= exp −(t − s)λ [1 − e iux ]F (dx)
=E
(e ) (13)
iu(Z t−s)
Hence Z has stationary increments.
12. By exercise 12, a compound Poisson process is a Lévy process and has independent stationary
increments.
(
∞∑ n∑
)
E|Z t − λtEU 1 | ≤E(E(|Z t ||N t )) + λtE|U 1 | ≤ E |U i ||N t = n P (N t = n) + λtE|U 1 |
n=1
(14)
i=1
=E|U 1 |EN t + λtE|U 1 | = 2λtE|U 1 | < ∞,
For t ≥ s, E(Z t |F s ) = E(Z t − Z s + Z s |F s ) = Z s + E(Z t−s ). Since
(
∞∑ n∑
)
EZ t = E(E(Z t |N t )) = E U i |N t = n P (N t = n) = λtEU 1 (15)
n=1 i=1
E(Z t − EU 1 λt|F s ) = Z s − EU 1 λs a.s. and {Z t − EU 1 λt} t≥0 is a martingale.
13. By Lévy decomposition theorem and hypothesis, Z t can be decomposed as
∫
∫
Z t = xN t (·, dx) + t(α − xν(dx)) = Z t ′ + βt (16)
R
|x|≥1
4

where Z t ′ = ∫ R xN t(·, dx), β = α− ∫ |x|≥1 xν(dx). By theorem 43, E(eiuZ′ t ) =
∫R (1−eiux )ν(dx). Z t ′ is
a compound Poisson process (See problem 11). Arrival rate (intensity) is λ since E( ∫ R N t(·, dx)) =
t ∫ R ν(dx) = λt. Since Z t is a martingale, EZ t ′ = −βt. Since EZ t ′ = E( ∫ R xN t(·, dx)) = t ∫ R xν(dx),
β = − ∫ R xν(dx). It follows that Z t = Z t ′ − λt ∫ R x ν λ
(dx) is a compensated compound Poisson
process. Then problem 12 shows that EU 1 = ∫ R x ν λ
(dx). It follows that the distribution of jumps
µ = (1/λ)ν.
14. Suppose EN t < ∞. At first we show that Z t ∈ L 1 .
(
Nt
)
)
∑
∞∑
E|Z t | ≤ E |U i | = E |U i | P (N t = n) =
Then
≤ sup
i
i=1
E|U i |
∞∑
n=0
E(Z t |F s ) = E(Z s +
i=1
n=0
( n∑
i=1
nP (N t = n) = EN t sup E|U i | < ∞.
i
∞∑
n=0 i=1
∞∑
∞∑
U i 1 {s

16. Let F t be natural filtration of B t satisfying usual hypothesis. By stationary increments
property of Brownian motion and symmetry,
W t = B 1−t − B 1
d = B1 − B 1−t
d = B1−(1−t) = B t (21)
This shows W t is Gaussian. W t has stationary increments because W t − W s = B 1−s − B 1−t
d = Bt−s .
Let G t be a natural filtration of W t . Then G t = σ(−(B 1 − B 1−s ) : 0 ≤ s ≤ t). By independent
increments property of B t , G t is independent of F 1−t . For s > t, W s −W t = −(B 1−t −B 1−s ) ∈ F 1−t
and hence independent of G t .
17. a) Fix ε > 0 and ω such that X·(ω) has a sample path which is right continuous, with left
limits. Suppose there exists infinitely many jumps larger than ε at time {s n } n≥1 ∈ [0, t]. (If there
are uncountably many such jumps, we can arbitrarily choose countably many of them.) Since [0, t]
is compact, there exists a subsequence {s nk } k≥1 converging to a cluster point s ∗ ∈ [0, 1]. Clearly
we can take further subsequence converging to s ∗ ∈ [0, 1] monotonically either from above or from
below. To simplify notations, Suppose ∃{s n } n≥1 ↑ s ∗ . ( The other case {s n } n≥1 ↓ s ∗ is similar.)
By left continuity of X t , there exists δ > 0 such that s ∈ (s ∗ − δ, s ∗ ) implies |X s − X s ∗ −| < ε/3 and
|X s− − X s ∗ −| < ε/3. However for s n ∈ (s ∗ − δ, s ∗ ),
|X sn − X s ∗ −| = |X sn − − X s ∗ − + △X sn | ≥ |△X sn | − |X sn − − X s ∗ −| > 2ε
3
This is a contradiction and the claim is shown.
b) By a), for each n there is a finitely many jumps of size larger than 1/n. But J = {s ∈ [0, t] :
|△X s | > 0} = ∪ ∞ n=1 {s ∈ [0, t] : |△X s| > 1/n}. We see that cardinality of J is countable.
18. By corollary to theorem 36 and theorem 37, we can immediately see that J ε and Z − J ε are
Lévy processes. By Lévy -Khintchine formula (theorem 43), we can see that ψ J εψ Z−J ε = ψ Z . Thus
J ε and Z − J ε are independent. ( For an alternative rigorous solution without Lévy -Khintchine
formula, see a proof of theorem 36 .)
19. Let T n = inf{t > 0 : |X t | > n}. Then T n is a stopping time. Let S n = T n 1 {X0 ≤n}. We have
{S n ≤ t} = {T n ≤ t, X 0 ≤ n} ∪ {X 0 > n} = {T n ≤ t} ∪ {X 0 > n} ∈ F t and S n is a stopping time.
Since X is continuous, S n → ∞ and X S n
1 {Sn >0} ≤ n, ∀n ≥ 1. Therefore X is locally bounded.
20. Let H be an arbitrary unbounded random variable. (e.g. Normal) and let T be a positive
random variable independent of H such that P (T ≥ t) > 0, ∀t > 0 and P (T < ∞) = 1. (e.g.
Exponential). Define a process Z t = H1 {T ≥t} . Z t is a càdlàg process and adapted to its natural
filtration with Z 0 = 0. Suppose there exists a sequence of stopping times T n ↑ ∞ such that Z Tn is
bounded by some K n ∈ R. Observe that
{
H T n ≥ T > t
Z Tn
t =
0 otherwise
Since Z T n
is bounded by K n , P (Z T n
> K n ) = P (H > K n )P (T n ≥ T > t) = 0. It follows that
P (T n ≥ T > t) = 0, ∀n and hence P (T n ≤ T ) = 1. Moreover P (∩ n {T n ≤ T }) = P (T = ∞) = 1.
This is a contradiction.
(22)
(23)
6

21. a) let a = (1 − t) −1 ( ∫ 1
t Y (s)ds) and Let M t = Y (ω)1 (0,t) (ω) + a1 [t,1) (ω). For arbitrary
B ∈ B([0, 1]), {ω : M t ∈ B} = ((0, t) ∩ {Y ∈ B}) ∪ ({a ∈ B} ∩ (t, 1)). (0, t) ∩ {Y ∈ B} ⊂ (0, t) and
hence in F t . {a ∈ B} ∩ (t, 1) is either (t, 1) or ∅ depending on B and in either case in F t . Therefore
M t is adapted.
Pick A ∈ F t . Suppose A ⊂ (0, t). Then clearly E(M t : A) = E(Y : A). Suppose A ⊃ (t, 1). Then
( ∫ 1 1
)
E(M t : A) = E(Y : A ∩ (0, t)) + E
Y (s)ds : (t, 1)
1 − t t
(24)
=E(Y : A ∩ (0, t)) +
∫ 1
t
Y (s)ds = E(Y : A ∩ (0, t)) + E(Y : (t, 1)) = E(Y : A).
Therefore for ∀A ∈ F t , E(M t : A) = E(Y : A) and hence M t = E(Y |F t ) a.s.
b) By simple calculation EY 2 = 1/(1 − 2α) < ∞ and Y ∈ L 2 ⊂ L 1 . It is also straightforward to
compute M t = (1 − t) −1 ∫ 1
t Y (s)ds = (1 − α)−1 Y (t) for ω ∈ (t, 1).
c) From b), M t (ω) = Y (ω)1 (0,t) (ω) + 1/(1 − α) −1 Y (t)1 (t,1) (ω). Fix ω ∈ (0, 1). Since 0 < α < 1/2,
Y/(1 − α) > Y and Y is a increasing on (0, 1). Therefore,
(
) ( )
Y (t)
sup M t = sup ∨ sup Y (ω) = Y (ω)
Y (ω)
∨ Y (ω) = (25)
0 ε} ⊂ (0, ε] for all ε > 0. Then T (ω) ≤ ε on (ε, 1) for all ε > 0 and it
follows that T ≡ 0. This is contradiction. Therefore there exists ε 0 such that {T > ε 0 } ⊃ (ε 0 , 1).
Fix ε ∈ (0, ε 0 ). Then {T > ε} ⊃ {T > ε 0 } ⊃ (ε 0 , 1). On the other hand, since T is a stopping time,
{T > ε} ⊂ (0, ε] or {T > ε} ⊃ (ε, 1). Combining these observations, we know that {T > ε} ⊃ (ε, 1)
for all ε ∈ (0, ε 0 ). ∀ε ∈ (0, ε 0 ), there exists δ > 0 such that ε − δ > 0 and {T > ε − δ} ⊃ (ε − δ, 1).
Especially T (ε) > ε − δ. Taking a limit of δ ↓ 0, we observe T (ε) ≥ ε on (0, ε 0 ).
c) Using the notation in b), T (ω) ≥ ω on (0, ε 0 ) and hence M T (ω) = ω −1/2 on (0, ε 0 ). Therefore,
EMT 2 ≥ E(1/ω : (0, ε 0)) = ∫ ε 0
0 ω−1 dω = ∞. Since this is true for all stopping times not identically
equal to 0, M cannot be locally a L 2 martingale.
d) By a), |M t (ω)| ≤ ω −1/2 ∨ 2 and M has bounded path for each ω. If M T 1 {T >0} were a bounded
random variable, then M T would be bounded as well since M 0 = 2. However, from c) M T /∈ L 2
unless T ≡ 0 a.s. and hence M T 1 {T >0} is unbounded.
23. Let M be a positive local martingale and {T n } be its fundamental sequence. Then for t ≥ s ≥
0, E(Mt
Tn 1 {Tn >0}|F s ) = M T n
s 1 {Tn >0}. By applying Fatou’s lemma, E(M t |F s ) ≤ M s a.s. Therefore
7

a positive local martingale is a supermartingale. By Doob’s supermartingale convergence theorem,
positive supermartingale converges almost surely to X ∞ ∈ L 1 and closable. Then by Doob’s
optional stopping theorem E(M T |F S ) ≤ M S a.s. for all stopping times S ≤ T < ∞. If equality
holds in the last inequality for all S ≤ T , clearly M is a martingale since deterministic times
0 ≤ s ≤ t are also stopping time. Therefore any positive honest local martingale makes a desired
example. For a concrete example, see example at the beginning of section 5, chapter 1. (p. 37)
24. a) To simplify a notation, set F 1 = σ(Z T − , T ) ∨ N . At first, we show G T − ⊃ F 1 . Since
N ⊂ G 0 , N ⊂ G T − . Clearly T ∈ G T − since {T ≤ t} = (Ω ∩ {t < T }) c ∈ G T − .
∀t > 0, {Zt T − ∈ B} = ({Z t ∈ B} ∩ {t < T }) ∪ ({Z T − ∈ B} ∩ {t ≥ T } ∩ {T < ∞}).
{Z t ∈ B} ∩ {t < T } ∈ G T − by definition of G T − . Define a mapping f : {T < ∞} → {T < ∞} × R +
by f(ω) = (ω, T (ω)). Define a σ-algebra P 1 consists of subsets of Ω × R + that makes all left
continuous processes with right limits (càglàd processes) measurable. Especially {Z t− } t≥0 is P-
measurable i.e, {(ω, t) : Z t (ω) ∈ B} ∈ P. Without proof, we cite a fact about P. Namely,
f −1 (P) = G T − ∩ {T < ∞}. Since Z T (ω)− (ω) = Z − ◦ f(ω), {Z T − ∈ B} ∩ {T < ∞} = f −1 ({(ω, t) :
Z t (ω) ∈ B}) ∩ {T < ∞} ∈ G T − ∩ {T < ∞}. Note that {T < ∞} = ∩ n {n < T } ∈ G T − . Then
Z T − ∈ G T − and G T − ⊃ F 1 .
Next We show G T − ⊂ F 1 . G 0 ⊂ F 1 , since Z T −
0 = Z 0 . Fix t > 0. We want to show that for all
A ∈ G t , A ∩ {t < T } ∈ F 1 . Let Λ = {A : A ∩ {t < T } ∈ F 1 }. Let
Π = {∩ n i=1{Z si ≤ x i } : n ∈ N + , 0 ≤ s i ≤ t, x i ∈ R} ∪ N (27)
Then Π is a π-system and σ(Π) = G t . Observe N ⊂ F 1 and
(∩ n i=1{Z si ≤ x i }) ∩ {t < T } = (∩ n i=1{Z T −
s i
≤ x i }) ∩ {t < T } ∈ F 1 , (28)
Π ⊂ Λ. By Dynkin’s theorem (π − λ theorem), G t = σ(Π) ⊂ Λ hence the claim is shown.
b) To simplify the notation, let H σ(T, Z T ) ∧ N . Then clearly H ⊂ G T since N ⊂ G T , T ∈ G T ,
and Zt
T ∈ G T for all t. It suffices to show that G T ⊂ H. Let
{
}
L = A ∈ G ∞ : E[1 A |G T ] = E[1 A |H]
(29)
Observe that L is a λ-system (Dynkin’s system) and contains all the null set since so does G ∞ . Let
⎧
⎫
⎨ n⋂
⎬
C = {Z tj ∈ B j } : n ∈ N, t j ∈ [0, ∞), B j ∈ B(R)
⎩
⎭ . (30)
j=1
Then C is a π-system such that σ(C) ∨ N = G ∞ . Therefore by Dynkin’s theorem, provided that
C ⊂ L, σ(C) ⊂ L and thus G ⊂ L. For arbitrary A ∈ G T ⊂ G ∞ , 1 A = E[1 A |G T ] = E[1 A |H] ∈ H and
hence A ∈ H.
It remains to show that C ⊂ L. Fix n ∈ N. Since H ⊂ G T , it suffices to show that
[
∏ n ]
∣
E
∣G T ∈ H. (31)
j=1
1 {Ztj ∈B j }
1 P is called a predictable σ-algebra. Its definition and properties will be discussed in chapter 3.
8

For this, let t 0 = 0 and t n+1 = ∞ and write
[
∏ n
E
j=1
1 {Ztj ∈B j }
] n+1
∣ ∑
∣G T =
∏
1 {T ∈[tk−1 ,t k )}
k=1
j 0, ∀t > 0, {|△Z t | > ε} = ∪ n ∩ n≥m {|Z t − Z T −1/n | > ε}.
Therefore,
P (|△Z t | > ε) ≤ lim inf
n→∞ P (|Z t − Z T −1/n | > ε) = 0 (33)
Since this is true for all ε > 0, P (|△Z t | > 0) = 0, ∀t.
26. To apply results of exercise 24 and 25, we first show following almost trivial lemma.
Lemma Let T be a stopping time and t ∈ R + . If T ≡ t, then G T = G t and G T − = G t− .
Proof. G T − = {A ∩ {T > s} : A ∈ G s } = {A ∩ {t > s} : A ∈ G s } = {A : A ∈ G s , s < t} = G t− . Fix
A ∈ G t . A ∩ {t ≤ s} = A ∈ G t ⊂ G s (t ≤ s), or ∅ ∈ G s (t > s) and hence G t ⊂ G T . Fix A ∈ G T ,
A ∩ {t ≤ s} ∈ G s , ∀s > 0. Especially, A ∩ {t ≤ t} = A ∈ G t and G T ⊂ G t . Therefore G T = G t .
Fix t > 0. By exercise 25, Z t = Z t− a.s.. By exercise 24 (d), G t = G t− since t is a bounded stopping
time.
27. Let A ∈ F t . Then A ∩ {t < S} = (A ∩ {t < S}) ∩ {t < T } ∈ F T − since A ∩ {t < S} ∈ F t .
Then F S− ⊂ F T − .
Since T n ≤ T , F Tn− ⊂ F T − for all n as shown above. Therefore, ∨ n F Tn− ⊂ F T − . Let A ∈ F t .
A ∩ {t < T } = ∪ n (A ∩ {t < T n }) ∈ ∨ n F Tn − and F T − ⊂ ∨ n F Tn −. Therefore, F T − = ∨ n F Tn −.
28. Observe that the equation in theorem 38 depends only on the existence of a sequence of simple
functions approximation f1 Λ ≥ 0 and a convergence of both sides in E{ ∑ j a jN Λ j
j
} = t ∑ j a jν(Λ j ).
For this, f1 Λ ∈ L 1 is enough. (Note that we need f1 Λ ∈ L 2 to show the second equation.)
9

29. Let M t be a Lévy process and local martingale. M t has a representation of the form
∫
∫
M t = B t + x [N t (·, dx) − tν(dx)] + αt + xN t (·, dx) (34)
{|x|≤1}
{|x|>1}
First two terms are martingales. Therefore WLOG, we can assume that
∫
M t = αt + xN t (·, dx) (35)
{|x|>1}
M t has only finitely many jumps on each interval [0, t] by exercise 17 (a). Let {T n } n≥1 be a sequence
of jump times of M t . Then P (T n < ∞) = 1, ∀n and T n ↑ ∞. We can express M t by a sum of
compound Poisson process and a drift term (See Example in P.33):
M t =
∞∑
U i 1 {t≥Ti } − αt. (36)
i=1
Since M t is local martingale by hypothesis, M is a martingale if and only if M t ∈ L 1 and E[M t ] = 0,
∀t. There exists a fundamental sequence {S n } such that M S n
is a martingale, ∀n. WLOG, we
can assume that M S n
is uniformly integrable. If U 1 ∈ L 1 , M t ∈ L 1 for every α and M t becomes
martingale with α ∗ = EN t EU 1 /t. Furthermore, for any other α, M t can’t be local martingale since
for any stopping time T , Mt
T = L T t + (α ∗ − α)(t ∧ T ) where L t is a martingale given by α = α ∗ and
Mt T can’t be a martingale if α ≠ α ∗ . It follows that if M t is only a local martingale, U 1 /∈ L 1 .
By exercise 24,
since M T 1−
t
Therefore,
F T1 − = σ(M T 1− , T 1 ) ∨ N = σ(T 1 ) ∨ N (37)
= M t 1 {t 0 : Z t ≥ z}. Then T z is a stopping time and P (T z < ∞) = 1. Let’s define a
coordinate map ω(t) = Z t (ω). Let R = inf{s < t : Z s ≥ z}. We let
{
{
1 s ≤ t , ω(t − s) < z − y
Y s (ω) =
, Y ′ 1 s ≤ t , ω(t − s) > z + y
s(ω) =
(40)
0 otherwise
0 otherwise
10

So that
Y R ◦ θ R (ω) =
{
1 R ≤ t , Z t < z − y
0 otherwise
, Y ′ R ◦ θ R (ω) =
{
1 R ≤ t , Z t > z + y
0 otherwise
(41)
Strong Markov property implies that on {R < ∞} = {T z ≤ t} = {S t ≥ z},
E 0 (Y R ◦ θ R |F R ) = E ZR Y R , E 0 (Y ′ R ◦ θ R |F R ) = E ZR Y ′ R (42)
Since Z R ≥ z and Z is symmetric,
E a Y s = E a 1 {Zt−s z+y} = E a Y ′
s, ∀a ≥ z, s < t. (43)
By taking expectation,
P 0 (T z ≤ t, Z t < z − y) = E(E 0 (Y R ◦ θ R |F R ) : R < ∞) = E(E ZR Y R : R < ∞)
≤E(E ZR Y ′ R : R < ∞) = E(E 0 (Y ′ R ◦ θ R |F R ) : R < ∞) = P 0 (T z ≤ t, Z t > z + y)
=P (Z t > z + y)
31. We define T z , R, ω(t) as in exercise 30. We let
{
1 s ≤ t , ω(t − s) ≥ z
Y s (ω) =
0 otherwise
(44)
(45)
Since Z R ≥ z and Z is symmetric,
E a Y s = E a 1 {Zt−s ≥z} ≥ 1 2
∀a ≥ z, s < t. (46)
By the same reasoning as in exercise 30, taking expectation yields
P 0 (Z t ≥ z) = P 0 (T z ≤ t, Z t ≥ z) = E(E 0 (Y R ◦ θ R |F R ) : R < ∞) = E(E ZR Y R : R < ∞)
≥E( 1 2 : R < ∞) = 1 2 P (R < ∞) = 1 2 P (S t ≥ z)
(47)
11

Chapter 2. Semimartingales and Stochastic Integrals
1. Let x 0 ∈ R be a discontinuous point of f. Wlog, we can assume that f is a right continuous
function with left limit and △f(x 0 ) = d > 0. Since inf t {B t = x 0 } < ∞ and due to Strong Markov
property of B t , we can assume x 0 = 0. Almost every Brownian path does not have point of decrease
(or increase) and it is a continuous process. So B· visit x 0 = 0 and changes its sign infinitely many
times on [0, ɛ] for any ɛ > 0. Therefore, Y t = f(B t ) has infinitely many jumps on any compact
interval almost surely. Therefore,
∑
(△Y s ) 2 = ∞ a.s.
s≤t
2. By dominated convergence theorem,
lim E Q[|X n − X| ∧ 1] = lim E P
n→∞ n→∞
[
|X n − X| ∧ 1 · dQ
dP
]
= 0
|X n − X| ∧ 1 → 0 in L 1 (Q) implies that X n → X in Q-probability.
3. B t is a continuous local martingale by construction and by independence between X and Y ,
[B, B] t = α 2 [X, X] t + (1 − α 2 )[Y, Y ] t = t
So by Lévy theorem, B is a standard 1-dimensional Brownian motion.
[X, B] t = αt,
[Y, B] t = √ 1 − α 2 t
4. Assume that f(0) = 0. Let M t = B f(t) and G t = F f(t) where B· is a one-dimensional standard
Brownian motion and F t is a corresponding filtration. Then
E[M t |G s ] = E [ B f(t) |F f(s)
]
= Bf(s) = M s
Therefore M t is a martingale w.r.t. G t . Since f is continuous and B t is a continuous process, M t is
clearly continuous process. Finally,
[M, M] t = [B, B] f(t) = f(t)
If f(0) > 0, then we only need to add a constant process A t to B f(t) such that 2A t M 0 +A 2 t = −B 2 f(0)
for each ω to get a desired result.
5. Since B is a continuous process with △B 0 = 0, M is also continuous. M is local martingale
since B is a locally square integrable local martingale.
[M, M] t =
∫ t
0
H 2 s ds =
∫ t
0
1ds = t
So by Lévy ’s characterization theorem, M is also a Brownian motion.
12

6. Pick arbitrary t 0 > 1. Let Xt n = 1 (t0 − 1 ,∞)(t) for n ≥ 1, X t = 1 [t0 ,∞), Y t = 1 [t0 ,∞). Then
n
X n , X, Y are finite variation processes and Semimartingales. lim n Xt
n = X t almost surely. But
lim
n
[X n , Y ] t0 = 0 ≠ 1 = [X, Y ] t0
7. Observe that
[X n , Z] = [H n , Y · Z] = H n · [Y, Z], [X, Z] = [H, Y · Z] = H · [Y, Z]
and [Y, Z] is a semimartingale. Then by the continuity of stochastic integral, H n → H in ucp
implies, H n · [Y, Z] → H · [Y, Z] and hence [X n , Z] → [XZ] in ucp.
8. By applying Ito’s formula,
[f n (X) − f(X), Y ] t =[f n (X 0 ) − f(X 0 ), Y ] t +
+ 1 2
∫ t
0
∫ t
(f” n (X s ) − f” n (X s ))d[[X, X], Y ] s
=(f n (X 0 ) − f(X 0 ))Y 0 +
0
∫ t
0
(f ′ n(X s ) − f ′ (X s ))d[X, Y ] s
(f ′ n(X s ) − f ′ (X s ))d[X, Y ] s
Note that [[X, X], Y ] ≡ 0 since X and Y are continuous semimartingales and especially [X, X] is a
finite variation process. As n → ∞, (f n (X 0 )−f(X 0 ))Y 0 → 0 for arbitrary ω ∈ Ω. Since [X, Y ] has a
path of finite variation on compacts, we can treat f n (X) · [X, Y ], f(X) · [X, Y ] as Lebesgue-Stieltjes
integral computed path by path. So fix ω ∈ Ω. Then as n → ∞,
sup |f n (B s ) − f(B s )| = sup |f n (x) − f(x)| → 0
0≤s≤t
inf s≤t B s≤x≤sup s≤t B s
since f n ′ → f uniformly on compacts. Therefore
∫ t
∣ ∣∣∣ ∣ (f n(X ′ s ) − f ′ (X s ))d[X, Y ] s ≤ sup |f n (B s ) − f(B s )|
0≤s≤t
0
∫ t
0
d|[X, Y ]| s −→ 0
9. Let σ n be a sequence of random partition tending to identity. By theorem 22,
∑ (
) (
)
[M, A] = M 0 A 0 + lim M T i+1 n − M
Ti
n A T i+1 n − A
Ti
n
n→∞
i
∣
≤ 0 + lim
sup
n→∞ i
∣
∣M T i+1 n − M
Ti
n
∣ ∑ i
∣
∣A T n i+1 − A
T n i
∣ = 0
since M is a continuous process and ∑ ∣
i ∣A T i+1 n − A
Ti
n ∣ < ∞ by hypothesis. Similarly
∑ (
) (
)
[A, A] = M0 2 + lim A T i+1 n − A
Ti
n A T i+1 n − A
Ti
n
n→∞
Therefore,
≤ 0 + lim
sup
n→∞ i
i
∣
∣A T i+1 n − A
Ti
n
∣ ∑ i
[X, X] = [M, M] + 2[M, A] + [A, A] = [M, M]
∣
∣A T n i+1 − A
T n i
∣ = 0
13

10. X 2 is P -semimartingale and hence Q-semimartingale by Theorem 2. By Corollary of theorem
15, (X − · X) Q is indistinguishable form (X − · X) P . Then by definition of quadric variation,
[X, X] P = X 2 − (X − · X) P = X 2 − (X − · X) Q = [X, X] Q
up to evanescent set.
11. a) Λ = [−2, 1] is a closed set and B has a continuous path, by Theorem 4 in chapter 1,
T (ω) = inf t : B t /∈ Λ is a stopping time.
b) M t is a uniformly integrable martingale since M t = E[B T |F t ]. Clearly M is continuous.
Clearly N is a continuous martingale as well. By Theorem 23, [M, M] t = [B, B] T t = t ∧ T and
[N, N] t = [−B, −B] T t = t ∧ T . Thus [M, M] = [N, N]. However P (M t > 1) = 0 ≠ P (N t > 1). M
and N does not have the same law.
12. Fix t ∈ R + and ω ∈ Ω. WLOG we can assume that X(ω) has a càdlàg path and ∑ s≤t |△X s(ω)| <
∞. Then on [0, t], continuous part of X is bounded by continuity and jump part of X is bounded
by hypothesis. So {X s } s≤t is bounded. Let K ⊂ R be K = [inf s≤t X s (ω), sup s≤t X s (ω)]. Then f,
f ′ , f” is bounded on K. Since ∑ s≤t {f(X s) − f(X s− ) − f ′ (X s− )△X s } is an absolute convergent
series, (see proof of Ito’s formula (Theorem 32), it suffices to show that ∑ s≤t {f ′ (X s− )△X s } < ∞.
By hypothesis,
∑ ∣
s≤t
∣f ′ (X s− )△X s
∣ ∣ ≤ sup
x∈K
|f ′ (x)| ∑ s≤t
|△X s | < ∞
Since this is true for all t ∈ R + and almost all ω ∈ Ω, the claim is shown.
13. By definition, stochastic integral is a continuous linear mapping J X : S ucp → D ucp . (Section
4). By continuity, H n → H under ucp topology implies H n · X → H · X under ucp topology.
14. Let  = 1 + A to simplify notations. Fix ω ∈ Ω and let Â−1 · X = Z. Then Z ∞ (ω) < ∞ by
hypothesis and  · Z = X. Applying integration by parts to X and then device both sides by Â
yields
X t
 = Z t − 1 Â
∫ t
0
Z s dÂs
Since Z ∞ < ∞ exists, for any ɛ > 0, there exists τ such that |Z t − Z ∞ | < ε for all t ≥ τ. Then for
t > τ,
∫
1 t
Z s dÂs = 1 ∫ τ
Z s dÂs + 1 ∫ t
 t −
(Z s − Z ∞ )dÂs + Z Âτ
∞
 t  t  t
0
0
τ
Let’s evaluate right hand side. As t → ∞, the first term goes to 0 while the last term converges
to Z ∞ by hypothesis. By construction of τ, the second term is smaller than ε(A t − A τ )/A t , which
converges to ε. Since this is true for all ε > 0,
1
lim
t→∞ Â
∫ t
0
Z s dÂs = Z ∞
Thus lim t→∞ X t /Ât = 0.
14
 t

15. If M 0 = 0 then by Theorem 42, there exists a Brownian motion B such that M t = B [M,M]t .
By the law of iterated logarithm,
lim
t→∞
M t B [M,M]t B τ
= lim = lim
[M, M] t t→∞ [M, M] t τ→∞ τ = 0
If M 0 ≠ 0, Let X t = M t − M 0 . Then [X, X] t = [M, M] t − M 2 0 . So [X, X] t → ∞ and
M t
[M, M] t
= X t + M 0
[X, X] t + M 2 0
→ 0
17. Since integral is taken over [t − 1/n, t] and Y is adapted, Xt n is also an adapted process. For
t > s ≥ 1/n such that |t − s| ≤ 1/n,
∫ t ∫ t−1/n
|Xt n − Xs n | = n
Y τ dτ − Y τ dτ
≤ 2n(t − s)
∣ s
s−1/n ∣ sup |Y τ |
s− 1 n ≤τ≤t
Since X is constant on [0, t] and [1, ∞), let π be a partition of [t, 1] and M = sup s−
1
n ≤τ≤t |Y τ | < ∞.
Then for each n, total variation of X n is finite since
sup |Xt n i+1
− Xt n i
| ≤ 2nM ∑
π
π
(t i+1 − t i ) = 2nM(1 − t)
Therefore X n is a finite variation process and in particular a semimartingale. By the fundamental
theorem of calculus and continuity of Y , Y s = lim n→∞ n ∫ s
s−1/n Y sdu = lim n→∞ Xt n for each t > 0.
At t = 0, lim n X n = 0 = Y 0 . So X n → Y for all t ≥ 0. However Y need not be a semimartingale
since Y t = (B t ) 1/3 where B t is a standard Brownian motion satisfies all requirements but not a
semimartingale.
18. B t has a continuous path almost surely. Fix ω such that B t (ω) is continuous. Then lim n→∞ Xt n (ω) =
B t (ω) by Exercise 17. By Theorem 37, the solution of dZt
n = Zs−dX n s n , Z 0 = 1 is
= exp
(X t n − 1 )
2 [Xn , X n ] t = exp(Xt n ),
Z n t
since X n is a finite variation process. On the other hand, Z t = exp(B t − t/2) and
lim
n→∞ Zn t
= exp(B t ) ≠ Z t
19. A n is a clearly càdlàg and adapted. By periodicity and symmetry of triangular function,
∫ π
2
0
|dA n s | = 1 n
∫ π
2
Since total variation is finite, A n t
lim
sup
n→∞
0≤t≤ π 2
0
|d sin(ns)| = 1 n · n ∫ π
2n
0
d(sin(ns)) =
∫ π
2
0
d(sin(s)) = 1
is a semimartingale. On the other hand,
|A n t | = lim
n→∞
1
n = 0 15

20. Applying Ito’s formula to u ∈ C 2 (R 3 − {0}) and B t ∈ R 3 \{0}, ∀t, a.s.
u(B t ) = u(x) +
∫ t
0
∇u(B s ) · dB s + 1 2
∫ t
0
△u(B s )ds = 1
‖x‖ + 3∑
i=1
∫ t
0
∂ i u(B s )dB i s
and M t = u(B t ) is a local martingale. This solves (a). Fix 1 ≤ α ≤ 3. Observe that E(u(B 0 ) α ) <
∞. Let p, q be a positive number such that 1/p + 1/q = 1. Then
∫
E x (u(B t ) α 1 1
‖x−y‖2
) =
R 3 ‖y‖ α e− 2t dy
(2πt) 3/2
∫
∫
1 1
‖x−y‖2
1 1
‖x−y‖2
=
{B(0;1)∩B c (x;δ)} ‖y‖ α e− 2t dy +
(2πt) 3/2
R 3 \{B(0;1)∩B c (x;δ)} ‖y‖ α e− 2t dy
(2πt) 3/2
∫
≤
sup
y∈{B(0;1)∩B c (x;δ)}
+
1
(2πt) 3 2
( ∫
R 3 \{B(0;1)∩B c (x;δ)}
e − ‖x−y‖2
2t ·
B(0;1)
1
‖y‖ α dy
1
‖y‖ αp dy ) 1/p ( ∫
R 3 \{B(0;1)∩B c (x;δ)}
(
1
‖x−y‖2
e− 2t
(2πt) 3/2
) q
dy) 1/q
Pick p > 3/α > 1. Then the first term goes to 0 as t → ∞. In particular it is finite for all t ≥ 0.
The first factor in the second term is finite while the second factor goes to 0 as t → ∞ since
∫
1 1
(2πt) 3/2 (2πt)
3(q−1)/2
e−
‖x−y‖2
2t/q
dy =
∫
1
(2πt) 3(q−1)/2
1 1
‖x−y‖2
q 3/2 e− 2t/q
dy =
(2πt/q) 3/2
1 1
(2πt) 3(q−1)/2 q 3/2
and the second factor is finite for all t ≥ 0. (b) is shown with α = 1. (c) is shown with α = 2. It
also shows that
lim
t→∞ Ex (u((B t ) 2 ) = 0
which in turn shows (d).
21. Applying integration by parts to (A ∞ − A·)(C ∞ − C . ),
(A ∞ − A·)(C ∞ − C . ) = A ∞ C ∞ −
∫ ·
Since A, C are processes with finite variation,
[A, C] t = ∑
△A s △C s
∫ t
0
(A ∞ − A s− )dC s =
Substituting these equations,
0

22. Claim that for all integer k ≥ 1 and prove by induction.
(A ∞ − A s ) k = k!
∫ ∞
s
dA s1
∫ ∞
s 1
dA s2 . . .
∫ ∞
s p−1
dA sp (48)
For k = 1, equation (48) clearly holds. Assume that it holds for k = n. Then
(k + 1)!
∫ ∞
s
dA s1
∫ ∞
s 1
dA s2 . . .
∫ ∞
s p−1
dA sk+1 = (k + 1)
∫ ∞
s
(A ∞ − A s1 )dA s1 = (A ∞ − A s ) k+1 , (49)
since A is non-decreasing continuous process, (A ∞ − A) · A is indistinguishable from Lebesgue-
Stieltjes integral calculated on path by path. Thus equation (48) holds for n = k + 1 and hence for
all integer n ≥ 1. Then setting s = 0 yields a desired result since A 0 = 0.
24. a)
∫ t
0
∫
1
t
1 − s dβ s =
=
=
0
∫ t
0
∫ t
0
∫
1
t
1 − s dB s −
0
∫
1
t
1 − s dB s − B 1
∫
1
t
1 − s dB s − B 1
1 B 1 − B s
1 − s 1 − s ds
0
0
∫
1
t
(1 − s) 2 ds +
= B [ (
t 1
1
1 − t − 1 − s
]t
, B − B 1
1 − t − 1 1
= B t − B 1
1 − t
+ B 1
Arranging terms and we have desired result.
0
B s
0 (1 − s) 2 ds
∫
1
t
( ) 1
(1 − s) 2 ds + B s d
0 1 − s
)
b) Using integration by arts, since [1 − s, ∫ s
0 (1 − u)−1 dβ u ] t = 0,
∫ t
∫
1
t
∫
X t = (1 − t)
0 1 − s dβ 1
t ∫ s
s = (1 − s)
0 1 − s dβ 1
s +
0 0 1 − u dβ u(−1)ds
∫ t
(
=β t + − X )
s
ds,
1 − s
as required. The last equality is by result of (a).
27. By Ito’s formula and given SDE,
d(e αt X t ) = αe αt X t dt + e αt dX t = αe αt X t dt + e αt (−αX t dt + σdB t ) = σe αt dX t
Then integrating both sides and arranging terms yields
∫ t
)
X t = e
(X −αt 0 + σ e αs dB s
0
17

B
28. By the law of iterated logarithm, lim sup t t→∞ t
= 0 a.s. In particular, for almost all ω there
exists t 0 (ω) such that t > t 0 (ω) implies B t (ω)/t < 1/2 − ɛ for any ɛ ∈ (0, 1/2). Then
{ (
lim E(B Bt
t) = lim exp t
t→∞ t→∞ t − 1 )}
≤ lim e −ɛt = 0, a.s.
2 t→∞
29. E(X) −1 = E(−X + [X, X]) by Corollary of Theorem 38. This implies that E(X) −1 is the
solution to a stochastic differential equation,
E(X) −1
t = 1 +
∫ t
0
E(X) −1
s− d(−X s + [X, X] s ),
which is the desired result. Note that continuity assumed in the corollary is not necessary if we
assume △X s ≠ −1 instead so that E(X) −1 is well defined.
30. a) By Ito’s formula and continuity of M, M t = 1 + ∫ t
0 M sdB s . B t is a locally square
integrable local martingale and M ∈ L. So by Theorem 20, M t is also a locally square integrable
local martingale.
b)
[M, M] t =
and for all t ≥ 0,
∫ t
0
E([M, M] t ) =
M 2 s ds =
∫ t
0
∫ t
0
e 2B s−s ds
E[e 2B s
]e −s ds =
∫ t
0
e s ds < ∞
Then by Corollary 3 of Theorem 27, M is a martingale.
c) Ee Bt is calculated above using density function. Alternatively, using the result of b),
Ee Bt = E(M t e t 2 ) = e
t
2 EM0 = e t 2
31. Pick A ∈ F such that P (A) = 0 and fix t ≥ 0. Then A ∩ {R t ≤ s} ∈ F s since F s contains all
P -null sets. Then A ∈ G t = F Rt . If t n ↓ t, then by right continuity of R, R tn ↓ R t . Then
G t = F Rt = ∩ n F Rt n
= ∩ nG tn = ∩ s≥t G s
by the right continuity of {F t } t and Exercise 4, chapter 1. Thus {G t } t satisfies the usual hypothesis.
32. If M has a càdlàg path and R t is right continuous, ¯Mt has a càdlàg path. ¯Mt = M Rt ∈ F Rt =
G t . So ¯M t is adapted to {G t }. For all 0 ≤ s ≤ t, R s ≤ R t < ∞. Since M is uniformly integrable
martingale, by optional sampling theorem,
¯M s = M Rs = E(M Rt |F Rs ) = E( ¯M t |G s ),
a.s.
So ¯M is G-martingale.
18

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 .
19

8. Let S, T be stopping times such that S ≤ T . Then F Sn ⊂ F T . and ∨ n F Sn ⊂ F T . By the
same argument as in exercise 7, F T − ⊂ ∨ n F Sn . Since F T − = F T by hypothesis, we have desired
result. (Note: {S n } is not necessarily an announcing sequence since S n = T is possible. Therefore
∨F Sn ≠ F T − in general.)
9. Let X be a Lévy process, G be its natural filtration and T be stopping time. Then by
exercises 24(c) in chapter 1, G T = σ(G T − , X T ). Since X jumps only at totally inaccessible time
(a consequence of theorem 4), X T = X T − for all predictable stopping time T . Therefore if T
is a predictable time, G T = σ(G T − , X T ) = σ(G T − , X T − ) = G T − since X T − ∈ G T − . Therefore a
completed natural filtration of a Lévy process is quasi left continuous.
10. As given in hint, [M, A] is a local martingale. Let {T n } be its localizing sequence, that is
[M, A] T n
· is a martingale for all n. Then E([X, X] T n
t ) = E([M, M] T n
t )+E([A, A] T n
t ). Since quadratic
variation is non-decreasing non-negative process, first letting n → ∞ and then letting t → ∞, we
obtain desired result with monotone convergence theorem.
11. By the Kunita-Watanabe inequality for square bracket processes, ([X + Y, X + Y ]) 1/2 ≤
([X, X]) 1/2 + ([Y, Y ]) 1/2 , It follows that [X + Y, X + Y ] ≤ 2 ([X, X] + [Y, Y ]). This implies that
[X + Y, X + Y ] is locally integrable and the sharp bracket process 〈X + Y, X + Y 〉 exists. Since
sharp bracket process is a predictable projection (compensator) of square bracket process, we obtain
polarization identity of sharp bracket process from one of squared bracket process. Namely,
〈X, Y 〉 = 1 (〈X + Y.X + Y 〉 − 〈X, X〉 − 〈Y, Y 〉) (51)
2
Then the rest of the proof is exactly the same as the one of theorem 25, chapter 2, except that we
replace square bracket processes by sharp bracket processes.
12. Since a continuous finite process with finite variation always has a integrable variation, without
loss of generality we can assume that the value of A changes only by jumps. Thus A can be
represented as A t = ∑ s≤t △A s. Assume that C· = ∫ ·
0 |dA s| is predictable. Then T n = inf{t > 0 :
C t ≥ n} is a predictable time since it is a debut of right closed predictable set. Let {T n,m } m be an
announcing sequence of T n for each n and define S n by S n = sup 1≤k≤n T k,n . Then S n is a sequence
of stopping time increasing to ∞, S n < T n and hence C Sn ≤ n. Thus EC Sn < n and C is locally
integrable. To prove that C is predictable, we introduce two standard results.
lemma Suppose that A is the union of graphs of a sequence of predictable times. Then there
exists a sequence of predictable times {T n } such that A ⊂ ∪ n [T n ] and [T n ] ∩ [T m ] = ∅ for n ≠ m.
Let {S n } n be a sequence of predictable stopping times such that A ⊂ ∪ n [S n ]. Put T 1 = S 1 and for
n ≥ 2, B n = ∩ n−1
k=1 [S k ≠ S n ], T n = (S n ) Bn . Then B n ∈ F Sn−, T n is predictable, [T n ] ∩ [T m ] = ∅
when n ≠ m, and A = ∪ n≥1 [T n ]. (Note: By the definition of the graph, [T n ] ∩ [T m ] = ∅ even if
P (T n = T m = ∞) > 0 as long as T n and T m are disjoint on Ω × R + )
20

lemma Let X t be a càdlàg adapted process and predictable. Then there exists a sequence of
strictly positive predictable times {T n } such that [△X ≠ 0] ⊂ ∪ n [T n ].
Proof. Let S 1/k
1/k
n+1 = inf{t : t > Tn (ω), |X 1/k S
− X t | > 1/k or |X 1/k
n
S
− X t− | > 1/k}. Then since
n
X is predictable, we can show by induction that {S n } n≥1 is predictable stopping time. In addition
[△X ≠ 0] ⊂ ∪ n,k≥1 [Sn
1/k ]. Then by previous lemma, there exists a sequence of predictable stopping
times {T n } such that ∪ n,k≥1 [Sn 1/k ] ⊂ ∪ n [T n ]. Then [△X ≠ 0] ⊂ ∪ n [T n ]
Proof of the main claim: Combining two lemmas, we see that {△A ≠ 0} is the union
of a sequence of disjoint graphs of predictable stopping times. Since A t = ∑ s≤t △A s is absolute
convergent for each ω, it is invariant with respect to the change of the order of summation. Therefore
A t = ∑ n △A S n
1 {Sn ≤t} where S n is a predictable time. △A Sn 1 {Sn ≤t} is a predictable process since
S n is predictable and A Sn ∈ F Sn−. This clearly implies that |△A Sn |1 {Sn≤t} is predictable. Then
C is predictable as well.
Note:
As for the second lemma, following more general claim holds.
lemma Let X t be a càdlàg adapted process. Then X is predictable if and only if X satisfies
the following conditions (1). There exists a sequence of strictly positive predictable times {T n } such
that [△X ≠ 0] ⊂ ∪ n [T n ]. (2). For each predictable time T , X T 1 {T

Poisson process,
⎡
C∑
t−s
E[N t − N s |F s ] = E ⎣
=µ
i=1
U i
⎤
⎦ =
∞∑
kP (C t−s = k) = µλ(t − s)
k=1
⎡
⎤
C ∞∑ ∑ t−s
E ⎣ U i |C t−s = k⎦ P (C t−s = k)
k=1
Therefore N t − µλt is a martingale and the claim is shown.
i=1
(53)
16. By direct computation,
1
1 − e −λh
λ(t) = lim P (t ≤ T < t + h|T ≥ t) = lim
h→0 h h→0 h
A case that T is Weibull is similar and omitted.
= λ. (54)
17. Observe that P (U > s) = P (T > 0, U > s) = exp{−µs}. P (T ≤ t, U > s) = P (U >
s) − P (T > t.U > s). Then
∫ t+h
P (t ≤ T < t + h, t ≤ U) t
e −µt (λ + θt)e −(λ+θt)u du
P (t ≤ T < t + h|T ≥ t, U ≥ t) = =
P (t ≤ T, t ≤ U)
exp{−λt − µt − θt 2 }
and
= −1[exp{−λ(t + h) − θt(t + h)} − exp{−λt − θt2 }]
exp{−λt − θt 2 }
= 1 − exp{−(λ + θt)h}
λ # 1 − exp{−(λ + θt)h}
(t) = lim
= λ + θt (56)
h→0 h
18. There exists a disjoint sequence of predictable times {T n } such that A = {t > 0 :
△〈M, M〉 t ≠ 0} ⊂ ∪ n [T n ]. (See the discussion in the solution of exercise 12 for details.) In
addition, all the jump times of 〈M, M〉· is a predictable time. Let T be a predictable time such
that 〈M, M〉 T ≠ 0. Let N = [M, M] − 〈M, M〉. Then N is a martingale with finite variation
since 〈M, M〉 is a compensator of [M, M]. Since T is predictable, E[N T |F T − ] = N T − . On the
other hand, since {F t } is a quasi-left-continuous filtration, N T − = E[N T |F T − ] = E[N T |F T ] = N T .
This implies that △〈M, M〉 T = △[M, M] T = (△M T ) 2 . Recall that M itself is a martingale. So
M T = E[M T |F T ] = E[M T |F T − ] = M T − and △M T = 0. Therefore △〈M, M〉 T = 0 and 〈M, M〉 is
continuous.
19. By theorem 36, X is a special semimartingale. Then by theorem 34, it has a unique
decomposition X = M + A such that M is local martingale and A is a predictable finite variation
process. Let X = N + C be an arbitrary decomposition of X. Then M − N = C − A. This implies
that A is a compensator of C. It suffices to show that a local martingale with finite variation
is locally integrable. Set Y = M − N and Z = ∫ t
0 |dY s|. Let S n be a fundamental sequence of
Y and set T n = S n ∧ n ∧ inf{t : Z t > n}. Then Y Tn ∈ L 1 (See the proof of theorem 38) and
Z Tn ≤ n + |Y Tn | ∈ L 1 . Thus Y = M − N is has has a locally integrable variation. Then C is a sum
of two process with locally integrable variation and the claim holds.
22
(55)

20. Let {T n } be an increasing sequence of stopping times such that X Tn is a special semimartingale
as shown in the statement. Then by theorem 37, Xt∧T ∗ n
= sup s≤t |X T n
s | is locally integrable.
Namely, there exists an increasing sequence of stopping times {R n } such that (Xt∧T ∗ n
) R n
is integrable.
Let S n = T n ∧ R n . Then S n is an increasing sequence of stopping times such that (Xt ∗ ) Sn
is integrable. Then Xt
∗ is locally integrable and by theorem 37, X is a special semimartingale.
21. Since Q ∼ P, dQ/dP > 0 and Z > 0. Clearly M ∈ L 1 (Q) if and only if MZ ∈ L 1 (P). By
generalized Bayes’ formula.
E Q [M t |F s ] = E P[M t Z t |F s ]
E P [Z t |F s ]
Thus E P [M t Z t |F s ] = M s Z s if and only if E Q [M t |F s ] = M s .
= E P[M t Z t |F s ]
Z s
, t ≥ s (57)
22. By Rao’s theorem. X has a unique decomposition X = M + A where M is (G, P)-local
martingale and A is a predictable process with path of locally integrable variation. E[X ti+1 −
X ti |G ti ] = E[A ti+1 − A ti |G ti ]. So
[ n∑
]
sup E |E[A ti − A ti +1|G ti ]| < ∞ (58)
τ
i=0
Y t = E[X t |F t ] = E[M t |F t ] + E[A t |F t ]. Since E[E[M t |F t ]|F s ] = E[M t |F s ] = E[E[M t |G s ]|F s ]] =
E[M s | fil s ], E[M t |F t ] is a martingale. Therefore
[ n∑
]
n∑
Var τ (Y ) = E |E[E[A ti |F ti ] − E[A ti +1|F ti +1]|F ti ]| = E [|E[A ti − A ti +1|F ti ]|] (59)
i=1
i=1
For arbitrary σ-algebra F, G such that F ⊂ G and X ∈ L 1 ,
E (|E[X|F]|) = E (|E (E[X|G]|F) |) ≤ E (E (|E[X|G]||F)) = E (|E[X|G]|) (60)
Thus for every τ and t i , E [|E[A ti − A ti +1|F ti ]|] ≤ E [|E[A ti − A ti +1|G ti ]|]. Therefore Var(X) < ∞
(w.r.t. {G t })implies Var(Y) < ∞ (w.r.t {F t }) and Y is ({F t }, P)-quasi-martingale.
23. We introduce a following standard result without proof.
Lemma. Let N be a local martingale. If E([N, N] 1/2
∞ ) < ∞ or alternatively, N ∈ H 1 then
N is uniformly integrable martingale.
This is a direct consequence of Fefferman’s inequality. (Theorem 52 in chapter 4. See chapter 4
section 4 for the definition of H 1 and related topics.) We also take a liberty to assume Burkholder-
Davis-Gundy inequality (theorem 48 in chapter 4) in the following discussion.
Once we accept this lemma, it suffices to show that a local martingale [A, M] ∈ H 1 . By Kunita-
Watanabe inequality, [A, M] 1/2
∞ ≤ [A, A] 1/4
∞ [M, M] 1/4
∞ . Then by Hölder inequality,
( )
E [A, M] 1/2
∞
(
≤ E
[A, A] 1/2
∞
) 1 (
2
E
[M, M] 1/2
∞
23
)
. (61)

(
By hypothesis E [A, A] 1/2
∞
) 1
2
< ∞. By BDG inequality and the fact that M is a bounded martingale,
E([M, M] 1/2 ) ≤ c 1 E[M ∗ ∞] < ∞ for some positive constant c 1 . This complete the proof.
24. Since we assume that the usual hypothesis holds throughout this book (see page 3), let
Ft
0 = σ{T ∧ s : s ≤ t} and redefine F t by F t = ∩ ɛ Ft+ɛ 0 ∨ N . Since {T < t} = {T ∧ t < t} ∈ F t , T
is F-stopping time. Let G = {G t } be a smallest filtration such that T is a stopping time, that is a
natural filtration of the process X t = 1 {T ≤t} . Then G ⊂ F.
For the converse, observe that {T ∧ s ∈ B} = ({T ≤ s} ∩ {T ∈ B}) ∪ ({T > s} ∩ {s ∈ B}) ∈ F s
since {s ∈ B} is ∅ or Ω, {T ∈ B} ∈ F T and in particular {T ≤ s}∩{T ∈ B} ∈ F s and {T > s} ∈ F s .
Therefore for all t, T ∧s, (∀s ≤ t) is G t measurable. Hence Ft 0 ⊂ G t . This shows that G ⊂ F. (Note:
we assume that G satisfies usual hypothesis as well. )
25. Recall that the {(ω, t) : △A t (ω) ≠ 0} is a subset of a union of disjoint predictable times and
in particular we can assume that a jump time of predictable process is a predictable time. (See the
discussion in the solution of exercise 12). For any predictable time T such that E[△Z T |F T − ] = 0,
△A T = E[△A T |F T − ] = E[△M T |F T − ] = 0 a.s. (62)
26. Without loss of generality, we can assume that A 0 = 0 since E[△A 0 |F 0− ] = E[△A 0 |F 0 ] =
△A 0 = 0. For any finite valued stopping time S, E[A S ] ≤ E[A ∞ ] since A is an increasing process.
Observe that A ∞ ∈ L 1 because A is a process with integrable variation. Therefore A {A S } S is
uniformly integrable and A is in class (D). Applying Theorem 11 (Doob-Meyer decomposition) to
−A, we see that M = A − Ã is a uniformly integrable martingale. Then
0 = E[△M T |F T − ] = E[△A T |F T − ] − E[△ÃT |F T − ] = 0 − E[△ÃT |F T − ]. a.s. (63)
Therefore à is continuous at time T .
27. Assume A is continuous. Consider an arbitrary increasing sequence of stopping time {T n } ↑ T
where T is a finite stopping time. M is a uniformly integrable martingale by theorem 11 and the
hypothesis that Z is a supermartingale of class D. Then ∞ > EZ T = −EA T and in particular
A T ∈ L 1 . Since A T ≥ A Tn for each n. Therefore by Doob’s optional sampling theorem, Lebesgue’s
dominated convergence theorem and continuity of A yields,
lim
n
E[Z T − Z Tn ] = lim
n
E[M T − N Tn ] − lim
n
E[A T − A Tn ] = −E[lim
n
(A T − A Tn )] = 0. (64)
Therefore Z is regular.
Conversely suppose that Z is regular and assume that A is not continuous at time T . Since
A is predictable, so is A − and △A. In particular, T is a predictable time. Then there exists an
announcing sequence {T n } ↑ T . Since Z is regular,
0 = lim
n
E[Z T − Z Tn ] = lim E[M T − M Tn ] − lim E[A T − A Tn ] = E[△A T ]. (65)
Since A is an increasing process and △A T ≥ 0. Therefore △A T = 0 a.s. This is a contradiction.
Thus A is continuous.
24

31. Let T be an arbitrary F µ stopping time and Λ = {ω : X T (ω) ≠ X T − (ω)}. Then by
Meyer’s theorem, T = T Λ ∧ T Λ c where T Λ is totally inaccessible time and TΛ c is predictable time.
By continuity of X, Λ = ∅ and T Λ = ∞. Therefore T = T Λ c. It follows that all stopping times are
predictable and there is no totally inaccessible stopping time.
32. By exercise 31, the standard Brownian space supports only predictable time since Brownian
motion is clearly a strong Markov Feller process. Since O = σ([S, T [: S, T are stopping time and S ≤
T ) and P = σ([S, T [: S, T are predictable times and S ≤ T ), if all stopping times are predictable
O = P.
35. E(M t ) ≥ 0 and E(M t ) = exp [B τ t − 1/2(t ∧ τ)] ≤ e. So it is a bounded local martingale and
hence martingale. If E(−M) is a uniformly integrable martingale, there exists E(−M ∞ ) such that
E[E(−M ∞ )] = 1. By the law of iterated logarithm, exp(B τ − 1/2τ)1 {τ=∞} = 0 a.s. Then
[ (
E[E(−M ∞ )] = E exp −1 − τ ) ]
1
2 {τ