5
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.