Solution to selected problems.

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