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The limit of random variable is not defined

  1. Nov 16, 2015 #1
    Let ##X_i## are i.i.d. and take -1 and +1 with probability 1/2 each. How to prove ##\lim_{n\rightarrow\infty}{\sum_{i=1}^{n}{X_i} }##does not exsits (even infinite limit) almost surely.
    My work:
    I use cauchy sequence to prove it does not converge to a real number.
    But I do not how to prove it does not converge to infinity. Can some one give hints?
     
    Last edited: Nov 16, 2015
  2. jcsd
  3. Nov 16, 2015 #2

    andrewkirk

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    Denote ##\sum_{k=1}^j X_k## by ##S_j##

    Then If the sum ##S_j## converges in probability to infinity then
    $$\forall M>0\ \forall \epsilon>0:\ \exists N>0 \textrm{ such that } j\geq N\Rightarrow Pr(S_j>M)>1-\epsilon\ \ \ \ \ (1)$$

    For this not to be the case we negate the previous line to obtain:

    $$\exists M>0\ \exists \epsilon>0:\ \forall N>0: \exists j\geq N\Rightarrow Pr(S_j>M)<1-\epsilon\ \ \ \ \ (2)$$

    That's what has to be proved. One way to do that (not necessarily the quickest) is to use the central limit theorem. Note that ##S_j=2B_j-j## where ##B_j## is binomial with parameters ##j,0.5##, which has mean ##\frac{j}{2}## and variance ##\frac{j}{4}##. The Central Limit Theorem tells us that ##B_j## converges in probability to a random variable with distribution ##N(\frac{j}{2},\frac{j}{4})##, so that ##S_j## converges in probability to a RV ##Z_j## with distn ##N(0,j)##.

    It's easy enough to prove that the sequence of RVs ##(Z_j)_{j=1}^\infty## satisfies (2). You then just need to put that together with the convergence in probability of ##S_j## to ##Z_j## to get the required result.
     
    Last edited: Nov 16, 2015
  4. Nov 16, 2015 #3
    ##\sum_{j=1}^n X_j## by ##S_j##?
     
  5. Nov 16, 2015 #4

    andrewkirk

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    Make that ##S_n##. I'll correct it above.
     
  6. Nov 16, 2015 #5
    What is the relation between almost surely and in probability? How to use here?
     
    Last edited: Nov 16, 2015
  7. Nov 17, 2015 #6

    andrewkirk

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    That's correct. Almost surely is a stronger type of convergence than in probability.
     
  8. Nov 17, 2015 #7
    I have difficulty about combining the final results, is there any inequality involved?
     
    Last edited: Nov 17, 2015
  9. Nov 17, 2015 #8
    I got you about (2) [<1/2]
     
  10. Nov 17, 2015 #9

    andrewkirk

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    Hmm. On reflection, making the connection from the convergence in distribution to a normal, to a conclusion that the set of points ##\omega\in\Omega## for which ##\lim_{j\to\infty}S_j(\omega)=\infty## has zero measure is not as straightforward as I initially thought. I will think more about it when I get a decent slab of free time.
     
  11. Nov 18, 2015 #10

    WWGD

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    Isn't this a random walk based, say at 0 , in the Real line? I don't know how to describe convergence in a random walk.
     
  12. Nov 18, 2015 #11

    andrewkirk

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    Yes. That's how I've been thinking about it.

    The formal statement of the problem, as I understand it, is a request to prove that $$Pr(\lim_{j\to\infty} S_j=\infty)=0$$

    where we interpret ##\lim_{j\to\infty}S_j=\infty## to mean that ##\forall M\in\mathbb{N}\ \exists n\in\mathbb{N}\ ## such that ##j>n\Rightarrow S_j>M##. All the instances of ##S_j## in that statement should really be written ##S_j(\omega)## where ##\omega## is an element of the sample space ##\Omega## of infinite sequences, in order to clarify that the statements are about specific sequences ##(S_j(\omega))_{j\in\mathbb{N}}##rather than about the random variables ##S_j##.

    If we label by ##A## the set of all sequences/walks ##S_j## that have that property, which we call 'diverging to infinity' then we are trying to prove that ##Pr(A)=0##.

    I'm starting to think that my Normal Approximation suggestion won't help though.

    Another possible avenue of attack would be if we could show that ##A##is countable. Then we would know it must have measure zero, since the set ##\Omega## of all sequences is uncountable. We can also think about the sequences as binary expansions of real numbers in [0,1] and I'm wondering if any known results about the measure of subsets of that interval might help,
     
  13. Nov 19, 2015 #12

    Samy_A

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  14. Nov 19, 2015 #13
  15. Nov 19, 2015 #14

    Samy_A

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    Yes, that's what I meant. Could that work, as ##(V=\infty) \cap A =\emptyset## (I think)?
     
  16. Nov 19, 2015 #15
    Define ##B=\{\omega\in \Omega:V(\omega)=\infty\}##? And show ##A\cap B##=0$?
     
  17. Nov 19, 2015 #16

    Samy_A

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    If ##\omega\in A## then ##\forall M\in\mathbb{N}\ \exists n\in\mathbb{N}\ ## such that ##\forall j>n\Rightarrow S_j(\omega)>M##.
    Take ##M>0##. Doesn't that imply that ##\omega## can return to 0 at most ##n## times, so ##\omega \not\in B##?
     
    Last edited: Nov 19, 2015
  18. Nov 19, 2015 #17
    I think You are right!
     
  19. Nov 19, 2015 #18

    WWGD

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    Isn't it also possible to use a recursion in this case , whose solution gives a closed form for the general probability?
     
  20. Nov 19, 2015 #19
    What do you mean by "recursion"?
     
  21. Nov 19, 2015 #20

    WWGD

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