Offset in Poisson regression

Question

Consider an offset term in a Poisson regression:

$$\log \mu_x = \log t_x+ \beta_0 + \beta_{1} x$$

To interpret $\beta_0$, would you need to consider $(\beta_0+ \log t_x)$? Because what if $t_x \neq 1$? Is the interpretation of $\beta_0$ the mean number of events when $t_x = 1$ and $x=0$? Suppose $t_1 = 2, t_2= 3$ and $t_3 = 4$. Then there is no $x$ such that $t_x = 1$. Also $x \neq 0$.

Also is the interpretation of $\beta_1$ the following: The mean number of events comparing $x+1$ and $x$ for a fixed $t_x$?

You will also want to read @ocram's answer here: [When to use an offset in Poisson regression?](http://stats.stackexchange.com/q/11182/7290) — gung - Reinstate Monica, Jun 01 '14 at 19:30

score 3 · Answer 1 · answered Jun 01 '14 at 18:16

An alternative formulation of your model is $$ \log\left(\frac{\mu_x}{t_x}\right) = \beta_0 + \beta_1 x $$ Then, we can see that $$ \beta_0 = \log\left(\frac{\mu_0}{t_0}\right) $$ and $$ \beta_1 = \log\left(\cfrac{\mu_{x+1}}{t_{x+1}}\right) - \log\left(\cfrac{\mu_{x}}{t_{x}}\right) = \log \left( \cfrac{\,\,\,\,\cfrac{\mu_{x+1}}{t_{x+1}}\,\,\,\,}{\cfrac{\mu_{x}}{t_{x}}} \right) $$ In words,

$\beta_0$ is the log rate at $x=0$;
$\beta_1$ is a log rate ratio ($x+1$ versus $x$)

+1, but I wonder if your equation for $\beta_0$ could be clarified... — gung - Reinstate Monica, Jun 01 '14 at 18:19

Offset in Poisson regression

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