Statistics 851 (Fall 2013)
September 11, 2013
Prof. Michael Kozdron
Lecture #4: There is no uniform probability on
([0
,
1]
,
2
[0
,
1]
)
Our goal for today is to prove the first of the claims made last lecture, namely that there
does not exist a uniform probability on the sample space [0
,
1] with the
σ
algebra 2
[0
,
1]
.
Suppose that
P
is our candidate for the uniform probability on ([0
,
1]
,
2
[0
,
1]
). Motivated by
our experience with elementary probability, it is desirable for such a uniform probability to
satisfy
P
{
[
a, b
]
}
=
b
−
a
for any interval [
a, b
]
⊆
[0
,
1]. In other words, the probability of
any interval is just its length. In fact, if 0
≤
a < b
≤
1, then the uniform probability should
satisfy
P
{
[
a, b
]
}
=
P
{
(
a, b
)
}
=
P
{
[
a, b
)
}
=
P
{
(
a, b
]
}
=
b
−
a.
In particular,
P
{
a
}
= 0
for every 0
≤
a
≤
1
.
Furthermore, the uniform probability should also satisfy countable additivity since this is
one of the axioms for probability. That is, if 0
≤
a
1
< b
1
<
· · ·
< a
n
< b
n
<
· · ·
≤
1, then
the uniform probability should also satisfy
P
∞
i
=1
[
a
i
, b
i
]
=
∞
i
=1
P
{
[
a
i
, b
i
]
}
=
∞
i
=1
(
b
i
−
a
i
)
.
For instance, the probability that the outcome is in the interval [0
,
1
/
4] is 1
/
4, the probability
the outcome is in the interval [1
/
3
,
1
/
2] is 1/6, and the probability that the outcome is in
either the interval [0
,
1
/
4] or [1
/
3
,
1
/
2] should be 1
/
4 + 1
/
6 = 5
/
12. That is,
P
{
[0
,
1
/
4]
∪
[1
/
3
,
1
/
2]
}
=
P
{
[0
,
1
/
4]
}
+
P
{
[1
/
3
,
1
/
2]
}
=
1
4
+
1
6
=
5
12
.
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 Fall '14
 Statistics, Probability, Equivalence relation, equivalence class, Congruence relation, uniform probability, Prof. Michael Kozdron

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