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Probability - Geometric Random Variable

  1. Aug 12, 2009 #1
    1. The problem statement, all variables and given/known data

    Let X be a random variable with distribution function px(x) defined by:

    px(0) = a and px(x) = Px(-x) = ((1-a)/2)*p*(1-p)^(x-1), x = 1,2...

    where a and p are two constants between 0 and 1, and px(0) is meant to be the probability that X=0

    a) What is the mean of X?
    b) Use the variance of a geometric random variable to compute the variance of X.

    3. The attempt at a solution

    Okay so for a) I just used the geometric random variable formula, but made ((1-a)/2) = b. Since the mean of a geometric random variable with probability of success p is E(X) = 1/p, I just multiplied it by b, giving me -(a-1)/2p. Is this correct?

    Also, I have no idea how to do part b), if anyone could help me with it I'd appreciate it.

    Thanks in advance.
  2. jcsd
  3. Aug 12, 2009 #2
    Welcome to PF, Dark Wanderer. I would study the derivation of the mean and variance for geometric distribution, then just modify that derivation for your example.

    You probably will need to use two series closely related to the geometric series, if you use my approach.

    Your mean can't be right, because the distribution is symmetric about 0.
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