Using Poisson Approximation to Compare Infection Rates in Village A and B

[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

Why the difference? Why does the second method not work?

 

[bpet]

A-B ~ Po(12)

Could you explain the reasoning for this step please?

 

[qazxsw11111]

I thought Poisson can perform linear combinations. I know A+B~Po(60+48) but A-B? I assumed minus is possible.

 

[bpet]

I know A+B~Po(60+48) but A-B?

Ok. If A-B were Poisson, what would be the frequency, mean and variance values?

 

[zli034]

λ1+λ2 in poisson formula is e^-(λ1+λ2)=e^-λ1*e^-λ2. but if you use λ1-λ2 we have e^-(λ1-λ2)=e^-λ1*e^λ2=e^λ2/e^λ1

 

[bpet]

I thought Poisson can perform linear combinations.

Linear combinations of Poisson random variables are actually Compound Poisson - not pure Poisson but an interesting topic in their own right.

P(A-B<0)=2.66x10^-4

On second thought, if A-B were Poisson then A-B can _never_ be negative, i.e. the probability would be exactly zero - but since A and B are independent Poisson there's always some chance. If you have access to math or stats software then it would be useful to run some random simulations to check which answer is correct.

Have fun!