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Proving memoryless property.

  1. Oct 28, 2009 #1
    1. The problem statement, all variables and given/known data

    Let X have a geometric distribution. Show that

    P(x>or= k+j|x>or=k) = P(X>or=j)

    where k and j are nonnegative integers.

    3. The attempt at a solution

    P(x>or= k+j|x>or=k) = P(x>or= k+j intersect x>or=k)/P(x>or=k) = P(x>or=k+j)/P(x>or=k) for j>or= 0 = [1-P(x=k+j)]/[1-P(x=k)] = [1 - p(1-p)^(k+j)]/[1-p(1-p)^k] = [1-p(1-p)^k(1-p)^j]/[1-p(1-p)^k] = (1-p)^j which is equal to P(x>or=j) if it had another p so that it was p(1-p)^j....so where am I losing this p in my proof?
  2. jcsd
  3. Oct 28, 2009 #2
    That's an error. It should be = [1-P(x<k+j)]/[1-P(x<k)]. However, don't do it this way.

    That's an algebra error. Are you saying (a+bcd)/(a+bc) would be equal to d? It isn't.

    This is right, and it is the key. Since you know this, you should go back and use this same idea in the expression P(x>or=k+j)/P(x>or=k).
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