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Thread: Poisson Process

  1. #1
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    Poisson Process



    I got a question.
    Q) Let {X(t), t>=0} and {Y(t), t>=0} be independent Poisson processes with parameter (lambda)_1 and (lambda)_2 respectively. Define Z_1(t) = X(t) + Y(t) and Z_2(t) = X(t) - Y(t) and Z_3 = X(t) + k, k a positive integer. Determine which of the above are Poisson and find lambda.

    so I figured that Z_1(t) is limiting case to the Binomial distribution and the lambda is ((lambda)_1)/((lambda)_1 + (lambda)_2)

    and for Z_2(t), it is Skellam distribution and lambda is (lambda)_1 - (lambda)_2

    and for Z_3(t), even though you add positive integer k to poisson process, you get (lambda)_1 + k.

    I not sure that this is right...
    I got the idea from the internet (except for the last one).
    and for Z_1(t), there is no lambda?
    b/c Binomial distribution doesn't have mean.
    Please tell me what I did wrong.
    Thank you....

  2. #2
    Bhoot
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    Re: Poisson Process

    Are you sure about the first one. What is the distribution of sum of two independent Poisson random variables?
    In the long run, we're all dead.

  3. #3
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    Re: Poisson Process


    I actually found that the mean of binomial distribution is number of trial * success probability in each trial, so this is lambda...
    how about Z2 and Z3? are they okay?
    Last edited by jjangub; 11-03-2010 at 08:31 PM.

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