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Statistic: Pattern with context dilemma

  1. Oct 23, 2012 #1
    1. The problem statement, all variables and given/known data
    An infinite-length binary text create from these patterns, with probability:
    • "0": 0.8
    • "11": 0.1
    • "100": 0.05
    • "101": 0.05
    ( For example: 0111001001010 ("0" "11" "100" "100" "101" "0") is a valid excerpt from the text)
    I temporary call a,b,c,d for P("0"), P("11"), P("100"), P("101")
    Problem: create a probability table for length-3 patterns (that is, find P(x): x="000","001","010",...,"111", x is excerpted at any point of the text)

    2. Relevant equations
    3. The attempt at a solution
    Some of the patterns can be determined easily: 000, 011, 100, 101, 110 with

    P(000)=a3=0.512
    P(011)=P(110)=a*b=0.08

    However, there are no evidence about 001,010,111. So I think of context, e.g., 001 is create with 2 "0" and anything start with 1 ("11","110","101") or 2 first zero from "100" and final 1 from anything start with 1, hence:

    P(001)=a2*(b+c+d) + c*(b+c+d)=0.138

    With this logic, we have:

    P(010)=a*d+c*d+d*(b+c+d)=0.0525
    P(111)=b*(b+c+d)+(b+d)*b=0.035

    However, that logic must be applied for 000,011,100,101,110 as well:

    P(000)=a3+c*a=0.552
    P(011)=a*b+c*b+d*(b+c+d)=0.095
    P(100)=c+(b+d)*a2=0.146
    P(101)=d+(b+d)*a*(b+c+d)=0.11
    P(110)=b*a=0.08

    And after all, the sum of (1≤x≤8),P(x)=1.2085
    I know that there is something wrong with taking the context to account, but I don't really know any other ways than considering the context.
     
    Last edited: Oct 23, 2012
  2. jcsd
  3. Oct 23, 2012 #2

    haruspex

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    For any bit in the sequence, there are 9 possibilities:
    - the '0' in a '0' pattern
    - one of the two bits in 11 pattern
    - one the 3 in a 100 pattern
    - one the 3 in a 101 pattern
    Can you calculate the relative frequencies of these 9 cases?
    Having done that, suppose you pick a bit as the first of a triplet. For each of the 9 possibilities, can you calculate the probabilities that the next two bits are 00, 01, 10, 11?
     
  4. Oct 24, 2012 #3
    Oh, now I understand my mistake: I don't take relative frequency in account. I calculate again and now the sum is 1. Thank you very much.
     
  5. Oct 24, 2012 #4
    This isn't right unless I have misunderstood.

    000 could be generated by ("0" "0" "0") or ("100" "0")
    011 could be generated by ("0" "11") or ("100" "11") or ("101" "100") or ("101" "11") or ....
     
  6. Oct 24, 2012 #5
    Yes, this was my mistake before, but I have calculated again with relative frequency by haruspex suggested and now the problem has been solve. The new result is 0.536, 0.134, 0.095, 0.085, 89/1500, 41/1500, 7/150, 1/60 of 000, 001, 010, 011, 100, 101, 110, 111 respectively
     
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