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Math Help - Conditional probability of joint geometric random variables

  1. #1
    Senior Member I-Think's Avatar
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    Conditional probability of joint geometric random variables

    Suppose that X and Y are independent random variables with the same parameter p.
    What is the value of
    P[X=i|X+Y=n]

    Want to make sure my solution is correct
    P[X=i|X+Y=n]=\frac{P[X=i]P[Y=n-i]}{P[X+Y=n]}
    =\frac{(1-p)^{i-1}p(1-p)^{n-i-1}p}{(1-p)^{n-1}p}
    =p(1-p)

    Correct?
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  2. #2
    MHF Contributor harish21's Avatar
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    Re: Conditional probability of joint geometric random variables

    Quote Originally Posted by I-Think View Post
    Suppose that X and Y are independent random variables with the same parameter p.
    What is the value of
    P[X=i|X+Y=n]

    Want to make sure my solution is correct
    P[X=i|X+Y=n]=\frac{P[X=i]P[Y=n-i]}{P[X+Y=n]}
    =\frac{(1-p)^{i-1}p(1-p)^{n-i-1}p}{(1-p)^{n-1}p}
    =p(1-p)

    Correct?

    careful with your denominator...

    the sum of geometric rvs (for n>=2) is given by:

    P(X+Y=n)=\sum_{x=1}^{n-1} P(X=x, Y=n-x)=\sum_{x=1}^{n-1} P(X=x)\cdot P(Y=n-x)

     \;=\; \sum_{x=1}^{n-1}p(1-p)^{x-1}p(1-p)^{n-x-1}\;=\;....... ,which is your denominator
    Last edited by harish21; December 5th 2011 at 07:22 PM.
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  3. #3
    Senior Member I-Think's Avatar
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    Re: Conditional probability of joint geometric random variables

    So the denominator becomes
    \sum_{x=1}^{n-1} p(1-p)^{x-1} p ({1-p})^{n-x-1}= \sum_{x=1}^{n-1} p^2({1-p})^{n}
    =p^2(1-p)^n

    So the answer is 1?
    Last edited by I-Think; December 5th 2011 at 07:57 PM. Reason: Didn't wrap in TEX tags
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  4. #4
    MHF Contributor harish21's Avatar
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    Re: Conditional probability of joint geometric random variables

    Quote Originally Posted by I-Think View Post
    So the denominator becomes
    \sum_{x=1}^{n-1} p(1-p)^{x-1} p {1-p}^{n-x-1}= \sum_{x=1}^{n-1} p^2{1-p}^{n}
    =p^2(1-p)^n

    So the answer is 1?
    No. try the calculations again.
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  5. #5
    Senior Member I-Think's Avatar
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    Re: Conditional probability of joint geometric random variables

    Round 2
    Denominator = p^2(n-1)(1-p)^{n-2}

    And the final answer is \frac{1}{n-1}?
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  6. #6
    MHF Contributor harish21's Avatar
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    Thumbs up Re: Conditional probability of joint geometric random variables

    Quote Originally Posted by I-Think View Post
    Round 2
    Denominator = p^2(n-1)(1-p)^{n-2}

    And the final answer is \frac{1}{n-1}?
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