LINKÖPING UNIVERSITY
Department of Mathematics
Mathematical Statistics
John Karlsson
TAMS29
Stochastic Processes with
Applications in Finance
1. Measure theory: Cantor function, product measure, counting
measure
1.1a
We study the construction of the Cantor function. In the first step we make the
function constant 1/2 on the middle third i.e. on [1/3, 2/3]. In the second step we
make the function constant on the middle thirds of the intervals that are left, i.e.
constant 1/4 on [1/9, 2/9] and constant 3/4 on [7/9, 8/9]. It follows that in the first
step the function is constant on an interval of length 1/3 and in the second step it
is constant on intervals of length 2 · 1/9 + 1 · 1/3. Following the same reasoning we
see that in step n the function is constant on intervals of length
n−1
2
n−1
1
n
n−1 k
1
1
1 X 2
1
n−2
0
+2
·
+ ... + 2 ·
= ·
.
·
3
3
3
3
3
k=0
Since
n−1 k
1 X 2
·
→ 1,
3
3
n → ∞,
k=0
we get that the Lebesgue measure of union of intervals of constancy is 1.
b
In order to show that the Cantor function f is not absolutely continuous, we need
to show that there exists some ε > 0 so that for any δ > 0 we can find a finite
collection π of disjoint intervals π := {(a1 , b1 ), . . . , (an , bn )} such that
|π| :=
n
X
(bi − ai ) < δ,
i=1
and
V (π) :=
n
X
|f (bi ) − f (ai )| > ε,
i=1
Let πn , n ≥ 0 denote a set of intervals. More precisely let
1
2
π0 := {(0, 1)}, π1 :=
0,
,
,1
,
3
3
1
2 3
6 7
8
π2 :=
0,
,
,
,
,
,
,1
,
9
9 9
9 9
9
1
2 3
6 7
8 9
,
,
,
,
π3 :=
0,
,
,
,
27
27 27
27 27
27 27
18 19
20 21
24 25
26
,
,
,
,
,
,
,1 , ,
27 27
27 27
27 27
27
1/4
...
As above |πn | denotes the total length of the intervals in πn i.e. we have
|π0 | = 1 − 0 = 1,
1
2
2
|π1 | =
−0 + 1−
= ,
3
3
3
1
3 2
7 6
8
4
|π2 | =
−0 +
−
+
−
+ 1−
= ,
9
9 9
9 9
9
9
and so on. We see that
n
2
.
3
We note that the intervals missing in each πn are the intervals where Cantor function
is constant. Therefore it follows that the variation V (πn ) of the Cantor function
over each πn is the same i.e.
V (π0 ) = |f (1) − f (0)| = 1 − 0 = 1,
2 1
1 1
(2)
=
− 0 + 1 − = 1,
V (π1 ) = f
− f (0) + f (1) − f
3
3 2
2
|πn | =
(1)
V (π2 ) = . . . = 1.
The claim now follows for e.g. ε = 0.9 since no matter what δ > 0 we choose, by
(1) and (2) we can find a πn such that |πn | < δ and V (πn ) > ε. This holds since
V (πn ) = 1 > 0.9 = ε.
1.2a
We note that for continuously differentiable f and g we have
Z
Z
f dg ≡ f (x)g 0 (x) dx.
Using this and integration by parts we get
Z b
Z b
Z b
Z
f dg +
g df =
f (x)g 0 (x) dx +
a
a
=
a
b
[f (x)g(x)]a
Z
−
b
g(x)f 0 (x) dx
a
b
f 0 (x)g(x) dx +
a
Z
b
g(x)f 0 (x) dx
a
b
= [f (x)g(x)]a = f (b)g(b) − f (a)g(a).
b
Quite complicated. See Hewitt, Edwin (1960), ”Integration by Parts for Stieltjes
Integrals”.
2/4
c
Assume that the discontinuities are at the points c1 , . . . , cn . Let t0 , . . . , tn partition
the interval (a, b) such that a = t0 < c1 < t1 < c2 < . . . < cn < tn = b. We have
Z b
n Z ti
X
... =
... =
Using result from (b) =
a
ti−1
i=1
=
n
X
i=1
f (ti )g(ti ) − f (ti−1 )g(ti−1 ) = f (tn )g(tn ) − f (t0 )g(t0 )
{z
}
|
Telescoping
= f (b)g(b) − f (a)g(a).
1.3a
Given two measure spaces (E1 , F1 , µ1 ) and (E2 , F2 , µ2 ) there is a measure space
(E, F, µ) such that
(i) E = E1 × E2 ,
(ii) µ(A1 × A2 ) = µ1 (A) · µ2 (A2 ) for any A1 ∈ F1 and A2 ∈ F2 .
We call this the product space with the measure µ = µ1 × µ2 . I.e. the elements of
the product space are pairs (x1 , x2 ) where x1 ∈ E1 and x2 ∈ E2 .
b
Since µi is absolutely continuous with respect to νi we have that there exists fi
such that
Z
Z
f2 dν2 , A1 ∈ F1 , A2 ∈ F2 .
f1 dν1 , µ2 (A2 ) =
(3)
µ1 (A1 ) =
A2
A1
By the definition of the product measure we have
Z
Z
(3)
Def
(µ1 × µ2 )(A1 × A2 ) = µ1 (A1 ) · µ2 (A2 ) =
f1 dν1 ·
f2 dν2
A1
A2
Z
Z Z
Z Z
f1 f2 d(ν1 × ν2 ).
=
f2 dν2 f1 dν1 =
f1 f2 dν2 dν1 =
A1
A2
A1
A2
A1 ×A2
It follows
d(µ1 × µ2 )
dµ1 dµ2
= f1 · f2 =
·
.
d(ν1 × ν2 )
dν1 dν2
3/4
1.4
We have E1 × E2 = N2 i.e the pairs (i, j) where i, j ∈ N. The counting measure
is the measure counting how many points are in any given set so the integral with
respect to counting measure is just a sum. We may think of N2 as the 2-dimensional
grid of points where the function values of f are given by
E2
−→
1 0 0
−1 1 0
0 −1 1
0 0 −1
..
.
E1 ↓
0 ...
0
0
1
..
.
Since |f (m, n)| takes values only 0 or 1 we have
Z
(4)
|f | d(µ1 × µ2 ) = Number of non-zero elements
= ∞.
E1 ×E2
The integral
Z
Z
(5)
f (m, n) dµ2 dµ1
E1
E2
corresponds to integrating over E2 first i.e. over each row first and we get
Z Z
f (m, n) dµ2 dµ1 = 1 + 0 + 0 + . . . = 1,
E1
E2
since each row except the first one sums to 0. The integral
Z Z
f (m, n) dµ1 dµ2
(6)
E2
E1
corresponds to integrating over E1 first i.e. over each column first and we get
Z Z
f (m, n) dµ1 dµ2 = 0 + 0 + 0 + . . . = 0,
E2
E1
since each column sums to 0.
NOTE: The integrals does not satisfy the conditions for the Fubini theorem and
we see that it is possible for the integrals (4), (5), and (6) to take different values.
4/4