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MATH 587
Assignment 3 Solutions
November 18, 2004
(1) (a) Let
µ
(
A
)=
µ
(
B
)=0
,

λ
1

(
A
c
)=

λ
2

(
B
c
) = 0. Then
µ
(
A
∪
B
)=0and

λ
1

(
C
)=

λ
2

(
C
)=0
,
and so
λ
1
(
C
)=
λ
2
(
C
) = 0 for every
C
⊂
(
A
∪
B
)
c
.Thu
s
λ
1
+
λ
2
(
C
) = 0 for every
C
⊂
(
A
∪
B
)
c
,
so

λ
1
+
λ
2

((
A
∪
B
)
c
)=0.
(b) Let
µ
(
A
)=0
. I
f
λ
+
1
(
A
)
>
0, then
λ
(
1
B
)
>
0 for some
B
⊂
A
, contradicting
µ
(
B
)=0
. So
λ
+
1
(
A
) = 0 and similarly
λ
−
1
(
A
) = 0, so

λ
1

(
A
)=0.
(c) Let
µ
(
A
)=0
,

λ
2
(
A
c
) = 0. By (b),

λ
1

(
A
)=0.
(d) By (c),
λ
1
⊥
λ
1
. Hence for some
A
∈F
,

λ
1

(
A
)=

λ
1

(
A
c
)=0.
(2) We have
Z
∞
0
px
p
−
1
P
{
X>x
}
dx
=
Z
∞
0
px
p
−
1
Z
I
{
X>x
}
dP dx
=
ZZ
∞
0
px
p
−
1
I
{
X>x
}
dxdP
=
ZZ
X
0
px
p
−
1
dxdP
=
Z
X
p
dP.
(3)
Z
∞
−∞
F
(
x
+
a
)
−
F
(
x
)
dx
=
Z
∞
−∞
Z
I
{
x
≤
X
≤
x
+
a
}
dP dx
=
ZZ
∞
−∞
I
{
x
≤
X
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This note was uploaded on 01/25/2011 for the course STAT 235a at Stanford.
 '07
 RomanVershynin
 Probability

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