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Math Help - Probability for a countable sample space

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    Probability for a countable sample space

    Cosider sample space, S, defined as

    S = {1,2,3,4,5,6,7,8,9,................}

    Experiment is to pick a number randomly from this set (with equal probability).

    Question: How do you define probability measure for this?

    Confusion I have -

    Consider sets A1 = {1}, A2 = {2}, A3 = {3}, A4 = {4}, ..., An = {n},.... so on

    Now let P(An) = x

    P(S) = P(A1)+P(A2)+P(A3)+....+P(An)+..... (under the axiom of countable additivity)

    LHS is 1 but RHS = 0 or infinite, so where is that I'm going wrong?
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  2. #2
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    Quote Originally Posted by aman_cc View Post
    Cosider sample space, S, defined as
    S = {1,2,3,4,5,6,7,8,9,................}
    Experiment is to pick a number randomly from this set (with equal probability).
    Question: How do you define probability measure for this?
    Each event has the probability say p>0.
    It well known that the sum of countable infinite positive constants is inifinte.
    So?
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  3. #3
    MHF Contributor undefined's Avatar
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    I think the setup to the problem is flawed; there is no uniform probability distribution function for this set. I'm not entirely sure though. Here is a reference.
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  4. #4
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    My guess:

    You require:
    \sum_x^\infty {P(x)}  = 1

    P(x) = P(y) ~~~~  \forall x,y >0

    It seems clear (as you have reasoned) that there is no solution. Can you define an appropriate limit and find that? (You will get that in the limit, P(.) -> 0)
    Last edited by SpringFan25; June 18th 2010 at 01:33 PM.
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    @Plato - I'm sorry not able to take your argument fwd.

    @undefined - Thanks for your reference. Yes even I feel it is flawed - but where exactly?
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    MHF Contributor undefined's Avatar
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    Quote Originally Posted by aman_cc View Post
    @undefined - Thanks for your reference. Yes even I feel it is flawed - but where exactly?
    The flaw in my opinion is in assuming that it is possible to "randomly" choose an element of an infinitely countable set such that any element is equally likely to be chosen as any other. The arguments presented show that this assumption is false. Maybe in terms of measure theory there is a better way to say this; I have not studied measure theory.

    I also found this thread on another forum; post #16, which seems to be by the same HallsofIvy who posts on this forum, might be helpful to you.
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  7. #7
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    Thanks. But haven't been able to make much sense. So am I correct in saying - you cannot define a probabiity measure for the set S, which conforms to our intution of 'equally likely'?
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  8. #8
    MHF Contributor matheagle's Avatar
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    the probabilities must go to zero, since the sum is finite

    \sum_{k=1}^{\infty}p_k<\infty
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