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A simple proof I'm having issues with

  1. Jul 2, 2011 #1
    Hi all,

    The question is: Given that X follows a geometric distribution, prove that

    P(X>j+k | X>k) = P(X>j).

    So this is all I have:

    P(X>k+j) = (1-p)^k+j, (a proof covered in the text)
    = (1-p)^k • (1-p)^j,
    No, what I want to say is,

    since P(X>k) = (1-p)^k = 1,
    then P(X>k+j) = 1 • (1-p)^j,
    = P(x>j).
    But I don't know if I'm allowed to separate the expression (1-p)^k • (1-p)^j into separate probabilities... Thanks in advance!!!
    Last edited: Jul 2, 2011
  2. jcsd
  3. Jul 2, 2011 #2

    Stephen Tashi

    User Avatar
    Science Advisor

    That isn't true.

    The definition of conditional probability tells you that

    [tex] P( X > j + k | X > k) = \frac{ P( (X > j + k) \cap (X > k) )} {P(X > k)} [/tex]

    Set theory tells you that [itex] (X > j + k) \cap (X > k) = (X > j + k) [/itex]
  4. Jul 2, 2011 #3
    Okay, thanks a ton. I can take it from there!
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