Solution to selected problems.

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