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Math Help - Naming the distribution

  1. #1
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    Naming the distribution

    I am given the scenario where a biased coin, with probability p (0,1) of landing on heads is tossed repeatedly. For each n=1,2,..... let Xn be the total number of heads seen after the nth toss.
    What is the distribution of Xn?


    Is Xn a geometric distribution?

    Thanks
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  2. #2
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    Re: Naming the distribution

    Quote Originally Posted by yellowcarrotz View Post
    I am given the scenario where a biased coin, with probability p (0,1) of landing on heads is tossed repeatedly. For each n=1,2,..... let Xn be the total number of heads seen after the nth toss. What is the distribution of Xn?
    I hope that I am reading you correctly.
    \left( {\forall n \geqslant 1} \right)\left[ {\;X_n  = k,\,k = 0,1, \cdots ,n} \right].
    In other words, X_n can have n+1 values.

    The probability distribution is binomal: \mathcal{P}\left( {X_n  = k} \right) = \binom{n}{k}p^k(1-p)^{n-k}

    Did I understand what you are asking?
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  3. #3
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    Re: Naming the distribution

    Quote Originally Posted by Plato View Post
    I hope that I am reading you correctly.
    \left( {\forall n \geqslant 1} \right)\left[ {\;X_n  = k,\,k = 0,1, \cdots ,n} \right].
    In other words, X_n can have n+1 values.

    The probability distribution is binomal: \mathcal{P}\left( {X_n  = k} \right) = \binom{n}{k}p^k(1-p)^{n-k}

    Did I understand what you are asking?
    You've read correctly.

    The geometric distribution will be X's distribution if X is the number of tosses you had to do in ordrer to get your first tail.
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  4. #4
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    Re: Naming the distribution

    Thanks, i understand why it isnt geometric now

    The last part of the question says now suppose for k=1,2,.. ...Ak is the event that the first head is seen on the kth toss. Fix n>=1. Show that for all k (1,2,.....n)

    P(Ak| Xn=1)=1/n

    Would I start of by writing P(Ak| Xn=1) = P(Ak ; Xn=1) / P(Xn=1)
    =P(Ak n Xn=1)/ P(Xn=1)

    to calculate P(Xn=1) would I plug this into the binomial formula above, then I'm not sure what to do abou P(Ak n Xn=1)?

    thanks for your help
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  5. #5
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    Re: Naming the distribution

    that intersection means that you had only one head in n tosses (Xn=1)
    AND that head appeared in the k-th toss (Ak)
    SO looking at the n independent tosses, you
    had tail, tail, tail..., head on k-th, tail,...tail.
    that event has probability q n-1 times and p once.
    The ratio is

    {q^{n-1}p\over {n\choose 1}q^{n-1}p}={1\over n}
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  6. #6
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    Re: Naming the distribution

    thanks very much, ive got it
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