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?

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.

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.

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?

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?