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I've just started studying Poisson regression and came across the two models:

$$ \begin{align*} \log{\mathbb{E}(count)} &= \beta_0 + \beta_1x_1 + \beta_2x_2 + \log(T) \\\\ \log{\mathbb{E}(count)} &= \beta_0 + \beta_1x_1 + \beta_2x_2 + \beta_T\log(T) \end{align*} $$ where T is time/exposure.

I'll start with the interpretation of a coefficient. Let's take $\beta_1$ for example. I'll do the interpretation both with respect to the count and the rate.

Offset model

  • When $x_1$ increases by one unit and $x_2$ stays constant, then the log count increases by $\beta_1$. Or, equivalently, the count is multiplied by $e^{\beta_1}$.
  • When $x_1$ increases by one unit and $x_2$ stays constant, then the log rate increases by $\beta_1$. Or, equivalently, the log rate is multiplied by $e^{\beta_1}$.

The interpretation is identical for both the mean and the rate of the Poisson distribution we are modeling.

Model with time as covariate

  • When $x_1$ increases by one unit, $x_2$ stays constant and $T$ stays constant, then the log count increases by $\beta_1$. Or, equivalently, the count is multiplied by $e^{\beta_1}$.
  • When $x_1$ increases by one unit, $x_2$ stays constant and $T$ stays constant, then the log rate increases by $\beta_1$. Or, equivalently, the log rate is multiplied by $e^{\beta_1}$.

The second interpretation is not so straightforward in this case but can be easily derived from the first. Keeping in mind that $rate = \frac{count}{T}$ and, based on the first interpretation, $T$ stays constant and count is multiplied by $e^{\beta_1}$, then the new rate is $rate_{new} = \frac{count_{new}}{T_{new}} = \frac{e^{\beta_1} count_{old}}{T_{old}} = e^{\beta_1} rate_{old}$

My points

  • The only difference between the two model interpretations is the bolded text.
  • The second model, not being restricted on the coefficient of $T$ being $1$, will be a better model.
  • Choosing between the two is sometimes a matter of whether you want to model rates or counts. However, you can get the rate by dividing count with $T$. So, using the second model is not an issue in that aspect.

Question

I get that the offset is used for easier interpretations. But upon exploring the above, the difference is not that great. Given that the second model will always give better results (I think?) why not always use the covariate time model? Is there another advantage to using offsets that I'm missing?

Note

People keep suggesting this question will help me In a Poisson model, what is the difference between using time as a covariate or an offset?. I read it before I made my question. I understand the difference between the two models. My question is different, I argue that offset model is inferior always. And even if my question is answered in the comments of that question (which is not, or at least not sufficiently for me to understand it) it would be helpful for other people to have this different question as a separate question so they can find it more easily and not search in comments of other questions.

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  • $\begingroup$ This question has already been discussed here. See this thread and this one $\endgroup$
    – Peter Flom
    Dec 21, 2023 at 21:16
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    $\begingroup$ If you use an offset, then that term is not multiplied by a parameter. If you use a covariate, it is. Sometimes you don't want it to be multiplied. Examples are given in threads. $\endgroup$
    – Peter Flom
    Dec 21, 2023 at 21:22
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    $\begingroup$ I think there's a flaw in your argument. The second model should be written as $\log{E(count)} = \alpha_0 + \alpha_1 x_1 + \alpha_2x_2 + \alpha_T\log(T)$ as the $\beta_i$'s are not equivalent in the two models (unless in the unlikely event that the estimate of $\alpha_T$ happens to be exactly 1). $\endgroup$
    – JimB
    Dec 21, 2023 at 23:01
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    $\begingroup$ However, if one is considering having an offset with $T$ (i.e., using $\log T$ as the offset in glm in R, for example) AND having $\log T$ as a covariate, then by all means just include $\log T$ in the model and don't use the offset option. You could include $\log T$ AND the offset $\log T$ and end up with an identical model just with the $\log T$ coefficient differing by exactly 1 in the two models. $\endgroup$
    – JimB
    Dec 21, 2023 at 23:27
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    $\begingroup$ This question was in fact answered where Peter From linked; sometimes you want to preserve the unit of your denominator (e.g. time is always in the constant weeks, not some multiplicative factor). For that you use an offset, not a covariate. $\endgroup$
    – PBulls
    Dec 22, 2023 at 5:52

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