singularly continuous

Singularly Continuous
A.3 Random Variable with Neither Density nor Probability Mass Function
Yiling Lai
2008/8/29
1

•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.
2

Constructing the Cantor Set
0
1
3

0 1
0 3
1
3
2
1
3

0 1
0 3
1
3
2
1
C1
3

0 1
0 3
1
3
2
1
C1
3
1
9
1 0 1 3
2
9
2
9
8
9
7
3

0 1
0 3
1
3
2
1
C1
3
1
9
1 0 1 3
2
9
2
9
8
9
7
C2
3

0 1
0 3
1
3
2
1
C1
3
1
9
1 0 1 3
2
9
2
9
8
9
7
C2
C3
3

0 1
0 3
1
3
2
1
C1
3
1
9
1 0 1 3
2
9
2
9
8
9
7
C2
C3
Continue this process…
3

0 1
0 3
1
3
2
1
C1
3
1
9
1 0 1 3
2
9
2
9
8
9
7
C2
C3
Continue this process…



k 1 k C C
3

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.
4

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
0, 1 C


 
 



 



 



 1 ,
9
8
9
7
,
3
2
3
1
,
9
2
9
1
0, 2 C
5

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
0, 1 C


 
 



 



 



 1 ,
9
8
9
7
,
3
2
3
1
,
9
2
9
1
0, 2 C
5
0
3
2
) ( ) ( lim lim  



   
 
k
k
k
k
C C

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
n
n
n Y
Y
6

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
2
3
3
2
1 2
1
    


Y Y Y Y
Y
n
n
n
0 3
1
3
2 1
7

• Case 2.1: Y1=0 and Y2=0 with probability = ¼
• …….
...
3
2
3
2
3
2
3
2
3
3
2
1 2
1
    


Y Y Y Y
Y
n
n
n
0 3
1
9
1
9
2
4
1
p 
4
1
p 
8

• …….
• 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
2
3
3
2
1 2
1
    


Y Y Y Y
Y
n
n
n



n 1 n C C
0 3
1
9
1
9
2
4
1
p 
4
1
p 
8




1 3
2
n
n
n Y
Y

Does Y have a density?
•If Y had a density f…
9

•The complement of the Cantor set:
f=0
9

f=0
•The Cantor set C:
C has zero Lebesgue measure, i.e. P(C)=0
9

f=0
•So f is almost everywhere zero and
9

f=0
•So f is almost everywhere zero and
•The function f would not integrate to one, as is required of a density.
9

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.
10

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.
11

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.
11
Y cannot have a probability mass function.

p=0
Cumulative Distribution Function
• The cumulative distribution function
satisfies
12
p=1/2 p=1/2
0 3
1
3
2 1
0 3
1
9
1
9
2
4
1
p 
4
1
p 

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.
13

A Singularly Continuous Function
14

•A singularly continuous function is a continuous function that has a zero derivative almost everywhere.
•An example is the Cantor function.
16

singularly continuous

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