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Geometric Distribution, Poisson

  1. Aug 17, 2005 #1
    The problem is the following;

    N has a geometric distribution with Pr(N=0)>0. M has a Poisson distribution. You are given:

    E(N) = E(M); Var(N) = 2Var(M)

    Calculate Pr (M>1).

    From general knowledge we know that the expected value of a variable in a geometric distribution E(N) = q/p, and Var(N) = q/(p^2).
    Also; the expected value of a variable in a Poisson distribution E(M) = lambda and Var(M) also = lambda.

    I believe that the answer is 1 - pr(M=0) - pr(M=1) which is the equivalent of

    But this would require solving for lambda, a feat I have not yet accomplished.

    Any pointers?..

    Thanks in advance,

  2. jcsd
  3. Aug 18, 2005 #2


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    You should be able to solve for p and [itex]\lambda[/itex], from [itex]\lambda[/itex] = (1-p)/p, and 2[itex]\lambda[/itex] = (1-p)/(p^2).

    Note that Pr(N=0) = p > 0 so [itex]\lambda < +\infty[/itex].
    Last edited: Aug 18, 2005
  4. Aug 18, 2005 #3
    Thanks Enuma, I was able to solve for it.
  5. Aug 23, 2005 #4
    NM i saw the reply.
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