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Topic: Lazy hunter catch the rabbit. (Read 13976 times) 

rmsgrey
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Re: Lazy hunter catch the rabbit.
« Reply #25 on: Mar 26^{th}, 2012, 5:47am » 
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on Mar 25^{th}, 2012, 12:22pm, c wrote:(and no, i can't easily swallow your "E(2) should be equal to 2 * E(1)") 
 In order to get two more heads than tails for the first time in a sequence of coin tosses, you have to, at some point have one more heads than tails for the first time. If you break the sequence into two at that point, you have a first "half" which runs until you have one more heads than tails for the first time, and a second "half" which runs until you have one more heads than tails in the second "half" for the first time. The expected time for the whole thing is E(2), and the expected time for each half is E(1) (since the two "halves" are independent) so we get: E(2) = E(1) + E(1) as required.


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c
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Re: Lazy hunter catch the rabbit.
« Reply #26 on: Mar 26^{th}, 2012, 5:59am » 
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That's the same storytelling towr did. But can you prove it? (sorry to dismiss your argument like this, but if i could accept yours, i'd've already accepted his...)


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rmsgrey
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Re: Lazy hunter catch the rabbit.
« Reply #27 on: Mar 26^{th}, 2012, 6:11am » 
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on Mar 26^{th}, 2012, 5:59am, c wrote:That's the same storytelling towr did. But can you prove it? (sorry to dismiss your argument like this, but if i could accept yours, i'd've already accepted his...) 
 Which part do you have a problem with? That the sequences that terminate when HT=2 for the first time are exactly those that are made up of two independent sequences, each of which terminated when HT=1 for the first time? That the second HT=1 sequence is independent of the first? That E(k) is being defined as the expected length of a sequence which terminates when HT=k for the first time? That the expected length of the sequence formed by concatenating two independent sequences is the sum of their individual expected lengths? That the above collectively implies that E(2)=2E(1)?


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c
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Re: Lazy hunter catch the rabbit.
« Reply #28 on: Mar 26^{th}, 2012, 6:13am » 
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"That the expected length of the sequence formed by concatenating two independent sequences is the sum of their individual expected lengths". Thank you for considering my objection.


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rmsgrey
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Re: Lazy hunter catch the rabbit.
« Reply #29 on: Mar 26^{th}, 2012, 7:02am » 
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on Mar 26^{th}, 2012, 6:13am, c wrote:"That the expected length of the sequence formed by concatenating two independent sequences is the sum of their individual expected lengths". Thank you for considering my objection. 
 Actually, the independence isn't necessary here... Letting I and J be the events X=x_{i} and Y=y_{j} respectively: By definition: E(X) = SUM_{i}(x_{i}P(I)) that is, the sum of (the individual outcomes times their probability of occuring). E(X+Y) = SUM_{i,j}((x_{i}+y_{j})P(I&J)) = SUM_{i,j}(x_{i}P(I&J) + y_{j}P(I&J)) = SUM_{i,j}(x_{i}P(I&J)) + SUM_{i,j}(y_{j}P(I&J)) SUM_{i,j}(x_{i}P(I&J)) = SUM_{i}(SUM_{j}(x_{i}P(I)P(JI))) For any given i, since x_{i} and P(I) are independent of j: SUM_{j}(x_{i}P(I)P(JI)) = x_{i}P(I)*SUM_{j}(P(JI)) since, for any given i: SUM_{j}(P(JI)) = 1 we have: SUM_{i,j}(x_{i}P(I&J)) = SUM_{i}(x_{i}P(I)) = E(X) and similarly (provided the sums are asolutely convergent): SUM_{i,j}(y_{j}P(I&J)) = E(Y) so: E(X+Y) = E(X) + E(Y) QED


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