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Math Help - Geometric Random Variable.

  1. #1
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    Geometric Random Variable.

    Hey all i was stuck on this geometric random variable question.



    Part (i)

    (a) pmf =  p(k) = (1-p)^{k-1} \cdot p
    (b) is a simple proof.


    Part (ii)

    (a)  p(X \ge 4) = (1 -\dfrac{1}{3})^{4-1} = (\dfrac{2}{3})^3 = \dfrac {8}{27}

    (b) & (c) i dont have an idea of how to attempt, any help is greatly appricated.

    Thanks guys
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  2. #2
    MHF Contributor matheagle's Avatar
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    2b or not 2b, is that the question, just write out a few cases...

    P(X=2)=P(RR)=(1/3)^2

    P(X=3)=P(BRR)+P(RBR)=2(1/3)^2(2/3)

    P(X=4)=P(BBRR)+P(BRBR)+P(RBBR)=3(1/3)^2(2/3)^2

    It looks like P(X=k)=(k-1)(1/3)^2(2/3)^{k-2}, k=2,3,4....
    Last edited by matheagle; May 22nd 2010 at 08:31 PM.
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  3. #3
    MHF Contributor matheagle's Avatar
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    This is a valid distribution...

     p^2\sum_{k=2}^{\infty}(k-1)(1-p)^{k-2}

     =-p^2{d\over dp}\left(\sum_{k=2}^{\infty}(1-p)^{k-1}\right)

    =-p^2{d\over dp}\left(\sum_{j=0}^{\infty}(1-p)^{j+1}\right)

    =-p^2{d\over dp}\left({1-p\over 1-(1-p))}\right)

    =-p^2{d\over dp}\left(p^{-1}-1\right)

    =-p^2\left(-p^{-2}\right)

    =1
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  4. #4
    MHF Contributor matheagle's Avatar
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    I was waiting for someone to do 2c...
    This is a finite rv, with W=1,2,...,11.

    P(W=1)=P(R)= {5\over 15}={1\over 3}

    P(W=2)=P(BR)= \left({10\over 15}\right)\left({5\over 14}\right)

    P(W=3)=P(BBR)= \left({10\over 15}\right) \left({9\over 14}\right)\left({5\over 13}\right)
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