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Math Help - Linear combination of Poisson and approximation

  1. #1
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    Linear combination of Poisson and approximation

    There are 60 infections in village A per month and 48 infections in village B per month.
    Let A be no of infections in village A per month and B be no of infections in village B per month. Assume occurrence is independent and random.

    So
    Method 1 (Working method):

    A~Po (60) and B~Po (48)

    Using a suitable approximation, find the probability that in 1 month, the no of infections in B exceeds no of infections in A.

    Since λ>10, A~ N(60,60) and B~N(48,48) approximately

    B-A~(-12, 108)

    P(B>A)=P(B-A>0)=0.115

    This I can understand but when I tried another method, it didnt work.

    Method 2: Through linear combination of poisson (???Cannot get it to work???)


    A-B ~ Po(12)

    Since λ>10, A-B~N(12,12)

    P(A<B)=P(A-B<0)=2.66x10-4
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  2. #2
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    Quote Originally Posted by qazxsw11111 View Post
    There are 60 infections in village A per month and 48 infections in village B per month.
    Let A be no of infections in village A per month and B be no of infections in village B per month. Assume occurrence is independent and random.

    So
    Method 1 (Working method):

    A~Po (60) and B~Po (48)

    Using a suitable approximation, find the probability that in 1 month, the no of infections in B exceeds no of infections in A.

    Since λ>10, A~ N(60,60) and B~N(48,48) approximately

    B-A~(-12, 108)

    P(B>A)=P(B-A>0)=0.115

    This I can understand but when I tried another method, it didnt work.

    Method 2: Through linear combination of poisson (???Cannot get it to work???)


    A-B ~ Po(12)

    Since λ>10, A-B~N(12,12)

    P(A<B)=P(A-B<0)=2.66x10-4
    I haven't checked your numbers, but I think there is an error in assuming that if A has a Poisson(a) distribution and B has a Poisson(b) distribution, then A-B has a Poisson(a-b) distribution. This is not true, despite the fact that if C has a Poisson(a-b) distribution then B+C has a Poisson(a) distribution.

    For example, suppose X Has a Poisson(2) distribution and Y has a Poisson(1) distribution. Can X-Y have a Poisson(1) distribution? No, because X-Y can take on negative as well as positive values. A variable with a Poisson distribution takes on only non-negative values.
    Last edited by awkward; November 8th 2009 at 05:06 AM. Reason: Improved example
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  3. #3
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    Awkward is correct in stating that the difference between Poisson's is not a Poisson. Even if the coefficients of the linear combination are strictly positive, you don't in general get a Poisson back since clearly you can end up changing the support (the places where positive probability lives) even for positive coefficients. For example, X ~ Poisson(a), Y ~ Poisson(b), 3X + Y is not Poisson. To be honest, I'm not exactly sure what they want you to do with this. The difference between a pair of Poisson's DOES follow a named distribution, but it isn't one you typically encounter in undergrad work. Maybe they want you to use a normal approximation, since A and B are approximately normal given their parameters, and you can easily calculate the mean and variance of A - B. If we let X = A - B, X ~ N(12, 108) may be what they are looking for, but that's just what you did with your first method, so I don't know.
    Last edited by theodds; November 8th 2009 at 08:13 AM.
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