It is well know that if $X_1,X_2,.., X_n$ are all independent and sampled from $Poisson(\lambda)$, then $\sum_{i=1}^n{X_i}\sim Poisson(n\lambda)$.

I have a following situation which is similar to the one here related discussion, however, I can not comment so I can not ask for a clarification on a specific term used in the discussion. The term is exposure. The situation is the following: I have an ab-test setup with two groups of users. Each about 3000 users. Then, in each group 45 and 53 cases are converted (the 0/1 event happened). I am treating these as rare events. Then we have poisson distribution for each group with $\hat{\lambda}=\frac{events}{n}$, where $n$ is number of , say, users per fixed time unit. Obviously, both lambdas will be small: (45/3000) and (53/3000), respectively. In the related discussion as well as in rate ratio test poisson this $n$ is called exposure. And now the question, if I follow the property of sum of poissons, I will get $\hat{\lambda n}=\frac{events}{n}\times n=43$ for group 1 and similarly for the other group. In this case a poisson with 43 events on average per unit of time is not rare anymore, but approximately normal. But in reality I know that 43 out of waiting 3000 "user-times" is rare. Isn't it a contradiction , is it? This exposure makes my individual rare count are now summed to a distribution with large lambda, not suitable for rare counts data.

Sorry for a bit of a messy question. May be I am missing some lambda scaling or other things.


closed as unclear what you're asking by Tim Sep 11 at 15:03

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  • $\begingroup$ I can't comprehend your question $\endgroup$ – Aksakal Sep 11 at 14:54