# Can I model incidence per 1000 people per month using poisson regression without an offset or using weights?

I am trying to model incidence rates (number of malaria cases per 1000 people per month) over time. I only have the data in the form of rates per 1000 people per month, i.e. I do not know the total population size each month. i.e. in January, there were 3.3 cases per 1000 individuals, in February, there were 5.6 cases per 1000 individuals etc. etc.

What I am unsure about is whether I can use a Poisson regression model to model the rates i.e 5.6 and 3.3, without using an offset, or weighted Poisson regression? I haven't done either of these things as the population would always be 1000 so I don't think using an offset or a weighted Poisson regression approach would achieve anything.

I am just not sure whether my approach is valid when the "count data" aren't the actual number of events, and indeed aren't count data as they are not integers.

You have data for $$n$$ regions/populations/months, each of size $$N_i$$ (which could vary from month to month.) The number of malaria cases is $$Y_i$$, but that is not given to you, neither is the $$N_i$$'s known to you. Apart from trying to obtain that missing information, you have to work with what is given, the rates $$\DeclareMathOperator{\E}{\mathbb{E}} R_i =1000\frac{Y_i}{N_i}.$$ We start with a Poisson regression model in terms of $$Y_i$$ $$\E [Y_i \mid x_i] = \lambda_i, \quad \log\lambda_i= \log N_i + x_i^T\beta$$ which we cannot use directly, since $$Y_i$$ is not known.
Rewriting in terms of $$R_i$$ we get $$\E [R_i \mid x_i] = 1000\frac{\lambda_i}{N_i}$$ with logarithm $$x_i^T\beta +\log 1000$$. This is a quasi-poisson model for $$R_i$$ with offset $$\log 1000$$, but since the offset is the same for all the observations it can just be absorbed in the intercept. Note that even with $$N_i$$ unknown we have no problem with expectation structure, since the expectation do not depend on $$N_i$$. But there will be one problem: The quasi-poisson model do use the assumption that the variance is proportional to the expectation. But that will no longer be true, because if the unknown $$N_i$$ are nonconstant, that will not hold here. See: Assume that $$Y_i \mid X_i=x_i$$ is Poisson, then we find that $$\DeclareMathOperator{\V}{\mathbb{V}} \V [R_i \mid x_i]=\frac{1000^2 \lambda_i}{N_i^2}$$. So while the expectation structure is right, the variance structure is not. So maybe use robust standard errors.
• (+1) This is probably a dumb question but why is the offset $+\log(1000)$ instead of $-\log(1000)$ in the logarithm, because the $1000$ is not in the denominator? – COOLSerdash Apr 21 at 19:15
• Because the data is rates per thousand, so the denominator is $N_i/1000$. – kjetil b halvorsen Apr 21 at 19:17