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Stochastic Calculus for Finance II
Chapter 1
財金所 潘政宏
1.1 Probability Space
The set of all sample points is called sample
space, say Ω .
Def1.1.1:Let F be a collection of subsets of
nonempty set Ω. We say that F is a
-algebra if
(1) the empty set ∈ F.
c
(2) A ∈ F, then A ∈ F.
(3) A1,A2,A3,… ∈ F, then n 1 An ∈ F.
1.1 Probability Space
(Ω, F) is called a measurable space.
Def1.1.2:Given (Ω, F) , a probability measure is a
real-valued function P: F →[0,1] and
(1) ∀A ∈ F, P(A) ∈[0,1].
(2) A1,A2,A3,… is a sequence of disjoint sets
in F, then P A P A
i 1
i
i 1
i
(3) P(Ω)=1
(Ω, F ,P) is called a probability space.
1.2 Random variables
Def1.2.1: Let (Ω, F ,P) be a probability space.
X : Ω →R is a random variable if
-1
B F
X
∀B ∈ B,
(Ω, F)
(R, B)
X
X-1 (B)
∈F
B
∈B
1.6 Change of Measure
Thm1.6.1:
Let (Ω, F, P) be a probability space and let Z be an almost surely
nonnegative random variable with EZ=1. For A ∈ F define
Then is a probability measure. Furthermore, if X is a nonnegative
random variable, then
If Z is almost surely strictly positive, we also have
for every nonnegative random variable Y.
1.6 Change of Measure
d P( )
P( A) d P( )
dP( ) Z ( )dP( )
A
A dP( )
A
EX X ( )d P( ) X ( )
d P( )
dP( ) X () Z ()dP() E[ XZ ]
dP( )
dP( )
1
Y
EY Y ( )dP( ) Y ( )
d P( ) Y ( )
d P( ) E[ ]
Z ( )
Z
d P( )
1.6 Change of Measure
Def1.6.3:
Let Ω be a nonempty set and F a σ-algebra of
subsets of Ω. Two probability measures P and
on (Ω, F) are said to be equivalent if they agree
which sets in F have probability zero.
i.e.
1.6 Change of Measure
Example:
Ω={1,2,3,4}
F= ({1},{2},{3},{4})
1
1
1
P({1}) , P({2}) , P({3}) , P({4}) 0
2
3
6
1
1
1
P({1}) , P({2}) , P({3}) , P({4}) 0
4
2
2
1
1
1
P({1}) , P({2}) , P({3}) 0, P({4})
4
2
2
1.6 Change of Measure
Thm1.6.7 (Radon-Nikodým):
Let P and be equivalent probability measures defined on
(Ω, F). Then there exists an almost surely positive random
variable Z such that EZ=1 and
1.6 Change of Measure
EX:X~N(0,1) with respect to P.
Y=X+θ~N(θ,1) with respect to P.
Find such that Y~N(0,1) with respect to
y
Sol:Let
1
2
.
2
2
dP
Z
dP
e
1
e
2
dy
( y )
2
e
2
y
2
2
dy
then Z>0 and
EZ e
y
2
2
dP e
y
2
2
1
e
2
( y )2
2
dy
1
e
2
y2
2
dy 1