It appears that in Poisson regression, using an identity link function means your betas will be rate differences, and using a log link function your (exponentiated) betas will be rate ratios.

You can use an offset when using a log link function because log(events/time) = log(events) - log(time) and you can then move log(time) to the predictors side of the equation to make it the offset. But when using an identity link function, the outcome is just "events/time" with no log, so you can't use logarithm math to turn the division into a subtraction and make it easily moveable to the other side.

At first I thought you could just make sure your rates have the same denominator before you fit the model, the same as you would do if calculating a rate difference by hand (eg. if you are comparing 2 events in 50 person-years vs. 7 events in 100 person-years, convert them to 4/100 and 7/100, and then for Poisson regression drop the "per hundred person-years" because it now cancels out) but then I realized that because the Poisson distribution only has a single parameter for center and variance, 4 events in 100 person-years would not have the same variance as 2 events in 50 person-years (I think -- if I'm wrong on this, please explain how/why!).

So is it just not possible to use an identity link function if you need an offset, or is there a way to do it? And if there is a way to do it, what does the equation look like?

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    $\begingroup$ you'll probably have to use a more general-purpose maximum likelihood machine (e.g. the bbmle package in R) $\endgroup$ – Ben Bolker Apr 26 '17 at 3:10

Try this: Suppose your has $Y_i$, $X_{i1}$, $X_{i2}$, and $N_i$. where $Y_i$ is # of events, $X_{i1}$ and $X_{i2}$ are covariates and $N_{i}$ is person-time, where $i$ indicates the subject. Then $Y_i$ ~ Poisson($\pi_iN_i$), where $\pi_i= \beta_0 + \beta_1X_{i1} + \beta_2X_{i2}$. (based on your requirement of identity link). So $Y_i$ ~ Poisson($\beta_0N_i + \beta_1X_{i1}N_i + \beta_2X_{i2}N_i$). So you need to generate two new variables $Z_{i1} = X_{i1}N_i$ and $Z_{i2} = X_{i2}N_i$. Then fit a Poisson regression model with identity link function on $N_i$, $Z_{i1}$ and $Z_{i2}$ WITHOUT intercept. In this new model, the regression coefficient for covariate $N_i$ is the original intercept, so no additional intercept is needed. Then you get what you wanted.

  • $\begingroup$ You should fix the typo, by editing your original post, not just commenting it! Plese do it. $\endgroup$ – kjetil b halvorsen Apr 26 '17 at 7:47
  • $\begingroup$ Thank you! Why without an intercept? $\endgroup$ – bluemouse Apr 27 '17 at 21:55
  • $\begingroup$ @bluemouse Just learned how to use Latex here, so I modified Answer and also one sentence was added to answer your question. $\endgroup$ – user158565 Apr 28 '17 at 2:24
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    $\begingroup$ @bluemouse the intercept-variable is a column of 1's. When you multiply through by offset you get the column of $N_i$ values, so that is playing the role of the intercept. an intercept term in there would correspond to a predictor of $1/N_i$ in your original model -- what would that be for? $\endgroup$ – Glen_b Apr 28 '17 at 3:05

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