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Math Help - a problem related borel cantelli's lemma

  1. #1
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    a problem related borel cantelli's lemma

    A coin is tossed indefinitely. Assume that the tosses are independent and let p be the probability of heads in each toss. For each k = 0, 1, . . . let A_k be the event that k consecutive heads appear (in the 2^k tosses) between the 2^k th toss(included) and 2^k+1 st toss(not included).

    a)Show that if p>= 1/2 , then infinitely many of the events A_k occur with probability 1.

    b)Show that if p < 1/2 , then finitely many of the events A_k occur with probability 1.

    Hint : Use Borel Cantelliís lemmas.
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  2. #2
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    Quote Originally Posted by bogazichili View Post
    A coin is tossed indefinitely. Assume that the tosses are independent and let p be the probability of heads in each toss. For each k = 0, 1, . . . let A_k be the event that k consecutive heads appear (in the 2^k tosses) between the 2^k th toss(included) and 2^k+1 st toss(not included).

    a)Show that if p>= 1/2 , then infinitely many of the events A_k occur with probability 1.

    b)Show that if p < 1/2 , then finitely many of the events A_k occur with probability 1.

    Hint : Use Borel Cantelli’s lemmas.
    Is it clear for you why the question boils down to : "Show that \sum_k P(A_k)<\infty if, and only if p<1/2." ? If not, you should read Borel Cantelli's lemma again.

    The problem is that there is no really simple explicit formula for P(A_k), so you should find bounds instead.

    For instance, A_k is the union of 2^k-k+1 events (''the k coins after i are heads'', for i=2^k,\ldots, 2^{k+1}-k+1). Using P(\bigcup B_i)\leq\sum P(B_i), you get P(A_n)\leq \cdots, and this proves one part of the problem.

    For the other part, here's a hint: use disjoint groups of coins...
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