Stat 310A/Math 230A Theory of Probability
Midterm Solutions
Andrea Montanari
November 1, 2010
The midterm was long! This will be taken into account in the grading. We will assign points proportion
ally to the number of questions answered (e.g. Problem 1 counts for 4 questions) and then rescale upwards
the grades distribution.
This is also a good time to discuss any difficulty you encountered with the instructors.
Problem 1
Let
λ
2
be the Lebesgue measure on (
R
2
,
B
R
2
). We know already that it is invariant under translation i.e.
that
λ
2
(
B
+
x
) =
λ
2
(
B
) for any Borel set
B
and
x
∈
R
2
(whereby
B
+
x
=
{
y
∈
R
2
:
y

x
∈
B
}
).
(a)
Show that it is invariant under rotations as well, i.e. that for any
α
∈
[0
,
2
π
], and any Borel set
B
⊆
R
2
,
λ
2
(
R
(
α
)
B
) =
λ
2
(
B
) (whereby
R
(
α
) denotes a rotation by an angle
α
and
R
(
α
)
B
=
{
x
∈
R
2
:
R
(

α
)
x
∈
B
}
).
Solution :
Throughout the solution we will use the fact that
λ
2
=
λ
1
×
λ
1
whence we obtain the action
of
λ
2
on rectangles:
λ
2
(
A
1
×
A
2
) =
λ
1
(
A
1
)
λ
1
(
A
2
).
Also, for
J
1
, J
2
⊆
R
two intervals, let
T
J
1
,J
2
be any
triangle with two sides equal to
J
1
(parallel to the first axis) and
J
2
(equal to the second axis). from the
addittivity of
λ
2
it follows immediately that
λ
2
(
T
J
1
,J
2
) =

J
1
 · 
J
2

/
2.
(We use here the fact that for a
segment
S
=
{
x
0
+
x
1
λ
:
λ
∈
[
a, b
]
}
,
x
0
, x
1
∈
R
2
,
λ
2
(
S
) = 0, which can be proved by covering
S
with
squares.)
α
β
Consider next a rectangle
A
= [0
, a
)
×
[0
, b
), and let
A
0
=
R
(
α
)
A
.
Using again addittivity (see figure
above) it follows that, for
β
=
π/
2

α
:
λ
2
(
A
0
) = (
a
cos
α
+
b
cos
β
)(
a
sin
α
+
b
sin
β
)

a
2
sin
α
cos
α

b
2
sin
β
cos
β
=
ab
(cos
α
sin
β
+ cos
β
sin
α
) =
ab
sin(
α
+
β
) =
ab .
Hence
λ
2
(
A
) =
λ
2
(
R
(
α
)
A
) and by translation invariance this holds for any
A
= [
a
1
, a
2
)
×
[
b
1
, b
2
) (not
necessarily with a corner at the origin).
Since the
π
system
P
=
{
A
= [
a
1
, a
2
)
×
[
b
1
, b
2
) :
a
1
< a
2
, b
1
< b
2
}
generates the Borel
σ
algebra, and
recalling that
λ
2
is
σ
finite, this proves the claim by Caratheodory uniqueness theorem.
(b)
For
s
∈
R
+
, and
B
⊆
R
2
Borel, let
sB
≡ {
x
∈
R
2
:
s

1
x
∈
B
}
. Prove that
λ
2
(
sB
) =
s
2
λ
2
(
B
).
1
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Solution :
The proof is analogous to the previous one.
Let
μ
be the measure defined by
μ
(
B
)
≡
s

2
λ
2
(
sB
). For
A
= [
a
1
, a
2
)
×
[
b
1
, b
2
),
a
1
< a
2
, b
1
< b
2
, we have
sA
= [
sa
1
, sa
2
)
×
[
sb
1
, sb
2
), whence
μ
(
A
) =
1
s
2
λ
2
(
sA
) =
1
s
2
(
sa
2

sa
1
)(
sb
2

sb
1
) = (
a
2

a
1
)(
b
2

b
1
) =
λ
2
(
A
)
.
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 '11
 MONTANARI,A.
 Probability, measure, Lebesgue measure, Lebesgue integration, lim sup xn, lim inf xn

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