# Thread: Linear combination of Poisson and approximation

1. ## 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

2. Originally Posted by qazxsw11111
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.

3. 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.