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Math Help - Probability (Geometric Distribution)?

  1. #1
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    Probability (Geometric Distribution)?

    Can someone help me with this question...im so lost...thanks in advance...

    If X has the geometric distribution with parameter p, show that
    P(X > m + n | X > m) = P(X > n).
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  2. #2
    Moo
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    Hello,

    Well, note that P(X>k)=\sum_{j=k+1}^\infty (1-p)p^j

    And then just write that P(X>m+n|X>m)=\frac{P(X>m+n,X>m)}{P(X>m)}=\frac{P(X  >m+n)}{P(X>m)}

    because if X>m+n happens, X>m happens too.
    Last edited by Moo; October 7th 2009 at 11:39 PM. Reason: thanks
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  3. #3
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by Moo View Post
    Hello,

    Well, note that P(X>k)=\sum_{j=k+1}^\infty (1-p)p^j

    And then just write that P(X>m+n|X>m)=\frac{P(X>m+n,X>m)}{P(X>m)}=\frac{P(X  >m+n,X>m)}{P(X>m)}

    because if X>m+n happens, X>m happens too.

    I believe you meant to drop that last inequality

    P(X>m+n|X>m)=\frac{P(X>m+n,X>m)}{P(X>m)}=\frac{P(X  >m+n)}{P(X>m)}
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  4. #4
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    Quote Originally Posted by rangergiya View Post
    Can someone help me with this question...im so lost...thanks in advance...

    If X has the geometric distribution with parameter p, show that
    P(X > m + n | X > m) = P(X > n).


    try the following link for some explanation memoryless property of geometric distribution

    mean and variance of the geometric distribution
    Last edited by qpmathelp; October 8th 2009 at 08:42 AM.
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