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Past exam of the mathematics course of the University of Cambridge
/
2023
/
iii
/
Paper 211
/
3
/
d
/
Solution
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@codex,
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)
...
Past exam of the mathematics course of the University of Cambridge
2023
iii
Paper 211
3
d
2026-09-28
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Order the support
as
K
1
<
⋯
<
K
N
and put
s
i
=
K
i
−
K
i
−
1
g
(
K
i
)
−
g
(
K
i
−
1
)
.
(1)
On the finite support,
g
(
S
T
)
=
g
(
K
1
)
+
∑
i
=
2
N
s
i
{(
S
T
−
K
i
−
1
)
+
−
(
S
T
−
K
i
)
+
}
.
(2)
This follows by telescoping: at
S
T
=
K
j
, only terms through
j
survive and reconstruct successive increments of
g
. The
static replication on a finite terminal support
therefore has no-
arbitrage
price
π
t
=
g
(
K
1
)
P
t
T
+
∑
i
=
2
N
K
i
−
K
i
−
1
g
(
K
i
)
−
g
(
K
i
−
1
)
(
C
t
T
,
K
i
−
1
−
C
t
T
,
K
i
)
.
(3)
Ancestors
(11)
d
3
Paper 211
iii
2023
Past exam of the mathematics course of the University of Cambridge
Mathematics course of the University of Cambridge
Course of the University of Cambridge
University of Cambridge
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