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Math Help - Limit superior proof

  1. #1
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    Limit superior proof

    Hi there,

    I'm working through a basic proof in a probability theory book, and I'm stuck trying to work out whether one of the statements is a typo, or I'm missing something:

    A_k is an arbitrary indexed set

    If
    \omega\in\limsup_{n\rightarrow\infty}=\bigcap_{n=1  }^{\infty}\bigcup_{k=n}^{\infty}A_k
    then since for every n, \omega\in\cup_{k\ge{n}}A_k, for all n there exists some k_n\ge{n} such that \omega\in{A_k}_n and therefore
    \sum_{j=1}^{\infty}1_A_{j}(\omega)\ge\sum_{n}{1_{{  A_k}_n}}(\omega)=\infty

    And at this point I'm confused: it seems to be that by construction, the indicator function in the second sum should always return 1, but the first does not necessarily. Therefore while they are both infinite sums, the first one may be smaller than the second, not the other way around.
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  2. #2
    Super Member girdav's Avatar
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    Re: Limit superior proof

    \omega\in \limsup A_k means that the set \{k\in\Bbb N, \omega\in A_k\} is infinite. For example, if A_{2k}=\{0\} and A_{2k+1}=\{1\} then 0\in\limsup_kA_k. The last two sums are both infinite.
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  3. #3
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    Re: Limit superior proof

    Thanks for you reply.

    So I don't think my question was clear enough. I absolutely understand that both sums are infinite. However I don't agree with the direction of the inequality between them: if I've understood the construction, the indicator of the second sum should always equal 1, whereas the first sum's indicator, whilst having an infinite number of 1's, can also return zeros. Is my understanding correct and therefore, this aspect of the proof wrong?
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  4. #4
    Super Member girdav's Avatar
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    Re: Limit superior proof

    In the sum of the RHS, only the term of index k_n, whereas in the RHS, we sum all the terms. Since they are non-negative, we have this inequality.
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  5. #5
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    Re: Limit superior proof

    I'm still not quite sure I understand.

    I read the sum on the RHS to mean a sum over all n - since we've stated there exists a {k_n}\ge{n} for every n. Thus both sums are over the same domain (i.e. all positive integers).

    Moreover, the actual {k_n} index could very well be the same for multiple terms of the (RHS) sum - whatever it takes to ensure that \omega\in{{A_k}_n} for all n.

    Thus the RHS indicator always returns ones, the LHS may or may not, but since they're over the same domain, the LHS is smaller.

    Sorry if I'm being dumb on this - if I've misunderstood the notation, could you point to the specific thing I've misread?

    Thanks
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  6. #6
    Super Member girdav's Avatar
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    Re: Limit superior proof

    Let I:=\{k, \omega\in A_k\}\subset \Bbb N. Then the inequality is \sum_{k\in \Bbb N}1_{A_k}(\omega)\geq \sum_{k\in I}1_{A_k}(\omega).
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  7. #7
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    Re: Limit superior proof

    Hi, sorry I still understand how your explanation resolves this:

    Quote Originally Posted by girdav View Post
    Let I:=\{k, \omega\in A_k\}\subset \Bbb N. Then the inequality is \sum_{k\in \Bbb N}1_{A_k}(\omega)\geq \sum_{k\in I}1_{A_k}(\omega).
    1. I don't see how \sum_{n}{1_{{A_k}_n}}(\omega) equates to your RHS sum. Doesn't \sum_{n} mean sum over all n?
    2. Even if my sum is the same as your RHS sum, to me your sum very clearly seems to be an equality: both indicators return an exactly equal number of 1's.
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  8. #8
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    Re: Limit superior proof

    Quote Originally Posted by entropyslave View Post
    Moreover, the actual {k_n} index could very well be the same for multiple terms of the (RHS) sum - whatever it takes to ensure that \omega\in{{A_k}_n} for all n.
    I agree with the OP because of this fact. Unlike \sum_{j=1}^{\infty}1_{A_{j}}(\omega), the right-hand side \sum_{n=1}^\infty{1_{  A_{k_n}}}(\omega) may contain 1_{A_n}(\omega) several times for some values of n. For example, if \omega\in A_n iff n\equiv0\pmod{3}, we could take k_n=3\lceil n/3\rceil. Then the right-hand side is 1_{A_3}(\omega)+1_{A_3}(\omega)+ 1_{A_3}(\omega)+1_{A_6}(\omega)+ 1_{A_6}(\omega)+1_{A_6}(\omega)+\dots. Because of this repetition, we can't say that the sequence k_1, k_2, ... is a subsequence of 1, 2, ...

    The comparison would be correct if we defined k_n to be some index such that \omega\in A_{k_n} and k_n>k_1, ..., k_n>k_{n-1}. Then k_1, k_2, ... is a subsequence of 1, 2, ..., so \sum_{j=1}^{\infty}1_{A_{j}}(\omega)\ge\sum_{n=1}^  \infty{1_{A_{k_n}}(\omega).

    Edit: Changed k_n=\lceil n/3\rceil to k_n=3\lceil n/3\rceil.
    Last edited by emakarov; July 30th 2012 at 08:59 AM.
    Thanks from entropyslave
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  9. #9
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    Re: Limit superior proof

    Quote Originally Posted by emakarov View Post
    The comparison would be correct if we defined k_n to be some index such that \omega\in A_{k_n} and k_n>k_1, ..., k_n>k_{n-1}. Then k_1, k_2, ... is a subsequence of 1, 2, ..., so \sum_{j=1}^{\infty}1_{A_{j}}(\omega)\ge\sum_{n=1}^  \infty{1_{A_{k_n}}(\omega).
    Thanks for your message and good example.

    As an aside, could you help me understand why your correction example (above) is not simply always an equality? As far as I see it, you've defined the subsequence of k_n to be exactly the indexes that the indicator in the left sum returns 1, and excluded the indices that the left indicator return zero. Thus, (I think) your two sums are equal.
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  10. #10
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    Re: Limit superior proof

    Quote Originally Posted by entropyslave View Post
    As an aside, could you help me understand why your correction example (above) is not simply always an equality? As far as I see it, you've defined the subsequence of k_n to be exactly the indexes that the indicator in the left sum returns 1, and excluded the indices that the left indicator return zero. Thus, (I think) your two sums are equal.
    In post #8, k_n is defined simply as some index such that \omega\in A_{k_n} and k_n is greater than all previous k_j's. The sequence k_n doesn't have to cover all indices j such that \omega\in A_j. In the situation where \omega\in A_n iff n\equiv0\pmod{3}, we can define k_1 = 30, k_2 = 60, etc.

    By the way, I made a mistake defining k_n=\lceil n/3\rceil; it should be 3\lceil n/3\rceil. Then k_1=k_2=k_3=3; k_4=k_5=k_6=6, etc.
    Thanks from entropyslave
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  11. #11
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    Re: Limit superior proof

    Wonderful, I get it now, thanks. So if you added that k_1 was the lowest index, and each k_n was the next highest with \omega\in{A_k}_n we should have equality.

    And well spotted, I had indeed missed that mistake.
    Last edited by entropyslave; July 30th 2012 at 09:13 AM. Reason: missed something
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