Solution to selected problems.

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