singularly continuous
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Singularly Continuous
A.3 Random Variable with Neither Density nor Probability Mass Function
Yiling Lai
2008/8/29
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Singularly Continuous
• A singularly continuous function is a continuous function that has a zero derivative almost everywhere.
• The distributions which are continuous but with no density function are said to be singular.
• An example is the Cantor function.
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Constructing the Cantor Set 0 1
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Constructing the Cantor Set 0 1
0 3
1
3
2
1
3
Constructing the Cantor Set 0 1
0 3
1
3
2
1 C1
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Constructing the Cantor Set 0 1
0 3
1
3
2
1 C1
3
1
9
10 1
3
2
9
2
9
8
9
7
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Constructing the Cantor Set 0 1
0 3
1
3
2
1 C1
3
1
9
10 1
3
2
9
2
9
8
9
7
C2
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Constructing the Cantor Set 0 1
0 3
1
3
2
1 C1
3
1
9
10 1
3
2
9
2
9
8
9
7
C2
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Constructing the Cantor Set 0 1
0 3
1
3
2
1 C1
3
1
9
10 1
3
2
9
2
9
8
9
7
C2
C3
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Constructing the Cantor Set 0 1
0 3
1
3
2
1 C1
3
1
9
10 1
3
2
9
2
9
8
9
7
C2
C3
Continue this process…
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Constructing the Cantor Set 0 1
0 3
1
3
2
1 C1
3
1
9
10 1
3
2
9
2
9
8
9
7
C2
C3
Continue this process…
1k kCC3
The length of the Cantor Set
• The first set removed is 1/3.
• The second set removed is 2/9.
• The third set removed is 4*1/27=4/27.
• The kth set removed is
• The total length removed is
• So the Cantor set, the set of points not removed, has zero “length”.
• Lebesgue measure of the Cantor set is zero.
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The Probability of Cantor Set • C1 is which has two pieces, each with
probability 1/3, and the whole set C1 has probability 2/3.
• C2 is which has four pieces, each with probability 1/9, and the whole set C1 has probability 4/9.
• At stage k, we have a set Ck that has 2k pieces, each with probability 1/3k, and the whole set Ck has probability (2/3)k.
1,
3
2
3
1,01C
1,
9
8
9
7,
3
2
3
1,
9
2
9
1,02C
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The Probability of Cantor Set • C1 is which has two pieces, each with
probability 1/3, and the whole set C1 has probability 2/3.
• C2 is which has four pieces, each with probability 1/9, and the whole set C1 has probability 4/9.
• At stage k, we have a set Ck that has 2k pieces, each with probability 1/3k, and the whole set Ck has probability (2/3)k.
1,
3
2
3
1,01C
1,
9
8
9
7,
3
2
3
1,
9
2
9
1,02C
5
03
2)()( limlim
k
kk
k
CC
Random Variable
• Fro n=1,2,…, we define
• The Probability for head on each toss is p=1/2.
• Define the random variable
1 3
2
nn
nYY
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p=1/2 p=1/2
• Case 1.1: Y1=0 with probability = 1/2
– Minimum: Y2=Y3=….=0 Y=0
– Maximum: Y2=Y3=….=1 Y=1/3
• Case 1.2: Y1=1 with probability = 1/2
– Minimum: Y2=Y3=….=0 Y=2/3
– Maximum: Y2=Y3=….=1 Y=1
...3
2
3
2
3
2
3
23
3
2
21
1
YYYYY
nn
n
0 3
1
3
21
7
• Case 2.1: Y1=0 and Y2=0 with probability = ¼
• Case 2.2: Y1=0 and Y2=1 with probability = ¼
• …….
...3
2
3
2
3
2
3
23
3
2
21
1
YYYYY
nn
n
0 3
1
9
1
9
2
4
1p
4
1p
8
• Case 2.1: Y1=0 and Y2=0 with probability = ¼
• Case 2.2: Y1=0 and Y2=1 with probability = ¼
• …….
...3
2
3
2
3
2
3
23
3
2
21
1
YYYYY
nn
n
0 3
1
9
1
9
2
4
1p
4
1p
8
• Case 2.1: Y1=0 and Y2=0 with probability = ¼
• Case 2.2: Y1=0 and Y2=1 with probability = ¼
• …….
...3
2
3
2
3
2
3
23
3
2
21
1
YYYYY
nn
n
0 3
1
9
1
9
2
4
1p
4
1p
8
• Case 2.1: Y1=0 and Y2=0 with probability = ¼
• Case 2.2: Y1=0 and Y2=1 with probability = ¼
• …….
• When we consider the first n tosses we see that the random variable Y takes values in the set Cn.
can only take values in the Cantor set .
...3
2
3
2
3
2
3
23
3
2
21
1
YYYYY
nn
n
1n nCC
0 3
1
9
1
9
2
4
1p
4
1p
8
1 3
2
nn
nYY
Does Y have a density?
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Does Y have a density?
• If Y had a density f…
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Does Y have a density?
• If Y had a density f…
• The complement of the Cantor set:
f=0
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Does Y have a density?
• If Y had a density f…
• The complement of the Cantor set:
f=0
• The Cantor set C:
C has zero Lebesgue measure, i.e. P(C)=0
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Does Y have a density?
• If Y had a density f…
• The complement of the Cantor set:
f=0
• The Cantor set C:
C has zero Lebesgue measure, i.e. P(C)=0
• So f is almost everywhere zero and
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Does Y have a density?
• If Y had a density f…
• The complement of the Cantor set:
f=0
• The Cantor set C:
C has zero Lebesgue measure, i.e. P(C)=0
• So f is almost everywhere zero and
• The function f would not integrate to one, as is required of a density.
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Does Y have a probability mass function?
• If Y had a probability mass function…
For some number we would have
• If x is not of the form
– x has a unique base-three expansion
• If x is of the form
– x has two base-three expansions.
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Does Y have a probability mass function? (cont.)
• In either case, there are at most two choices of for which
• In other words, the set has either one or two elements.
• The probability of a set with one element is zero and the probability of a set with two elements is 0+0=0.
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Does Y have a probability mass function? (cont.)
• In either case, there are at most two choices of for which
• In other words, the set has either one or two elements.
• The probability of a set with one element is zero and the probability of a set with two elements is 0+0=0.
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Does Y have a probability mass function? (cont.)
• In either case, there are at most two choices of for which
• In other words, the set has either one or two elements.
• The probability of a set with one element is zero and the probability of a set with two elements is 0+0=0.
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Y cannot have a probability mass function.
p=0
Cumulative Distribution Function
• The cumulative distribution function
satisfies
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p=1/2 p=1/2
0 3
1
3
21
0 3
1
9
1
9
2
4
1p
4
1p
Cumulative Distribution Function
• for every x, F is continuous.
•
• A non-constant continuous function whose derivative is almost everywhere zero is said to be singularly continuous.
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A Singularly Continuous Function
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A Singularly Continuous Function
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A Singularly Continuous Function
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A Singularly Continuous Function
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Singularly Continuous
• A singularly continuous function is a continuous function that has a zero derivative almost everywhere.
• An example is the Cantor function.
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