Does anyone have a derivation of how an offset works in binary models like probit and logit?

In my problem, the follow-up window can vary in length. Suppose patients get a prophylactic shot as treatment. The shot happens at different times, so if the outcome is a binary indicator of whether any flare-ups happened you need to adjust for the fact that some people have more time to exhibit symptoms. It seems that the probability of a flare-up is proportional to the length of the follow-up period. It's not clear to me mathematically how a binary model with an offset captures this intuition (unlike with the Poisson).

The offset is a standard option in both Stata (p.1666) and R, and I can easily see it for a Poisson, but the binary case is a bit opaque.

For example, if we have \begin{equation} \frac{E[y \vert x]}{Z}=\exp\{x'\beta\}, \end{equation} this is algebraically equivalent to a model where \begin{equation}E[y \vert x]=\exp\{x'\beta+\log{Z}\}, \end{equation} which is the standard model with the coefficient on $\log Z$ constrained to $1$. This is called a logarithmic offset. I am having trouble figuring out how this works if we replace $\exp\{\}$ with $\Phi()$ or $\Lambda()$.

Update #1:

The logit case was explained below.

Update #2:

Here's an explanation of what seems to be the main use of offsets for the non-poisson models like probit. The offset can be used to conduct likelihood ratio tests on index functions coefficients. First you estimate the unconstrained model and store the estimates. Say you want to test the hypothesis that $\beta_x=2$. Then you create the variable $z=2 \cdot x$, fit the model dropping $x$ and using $z$ as an non-logarithmic offset. This is the constrained model. The LR tests compares the two, and is an alternative to the usual Wald test.


2 Answers 2


You can always include an offset in any GLM: it's just a predictor variable whose coefficient is fixed at 1. Poisson regression just happens to be a very common use case.

Note that in a binomial model, the analogue to log-exposure as an offset is just the binomial denominator, so there's usually no need to specify it explicitly. Just as you can model a Poisson RV as a count with log-exposure as an offset, or as a ratio with exposure as a weight, you can similarly model a binomial RV as counts of successes and failures, or as a frequency with trials as a weight.

In a logistic regression, you would interpret a $\log Z$ offset in terms of the odds ratios: a proportional change in $Z$ results in a given proportional change in $p/(1-p)$.

$$\begin{equation}\begin{split} \log (p/(1-p)) &= \beta' \mathrm{X} + \log Z \\ p/(1-p) &= Z \exp(\beta' \mathrm{X}) \end{split}\end{equation}$$

But this doesn't have any particular significance like log-exposure does in a Poisson regression. That said, if your binomial probability is small enough, a logistic model will approach a Poisson model with log link (since the denominator on the LHS approaches 1) and the offset can be treated as a log-exposure term.

(The problem described in your linked R question was rather idiosyncratic.)

  • $\begingroup$ The weighting part was missing from my understanding of the equivalence of the two. That was very helpful. I am still a bit confused how one can turn something like $\Pr(Y=1 \vert X)=\Phi(x'\beta+\ln(t))$ into a statement about the probability of a flare-up being proportional to the length of the follow-up period $t$, though I can see how it is increasing in $t$. $\endgroup$
    – dimitriy
    Commented Aug 9, 2013 at 0:44
  • $\begingroup$ It's not the probability, but the odds ratio. Hopefully the edit makes it clearer. $\endgroup$
    – Hong Ooi
    Commented Aug 9, 2013 at 9:21
  • $\begingroup$ Expressing the problem in terms of the odds ratio makes it very clear. What about the probit? $\endgroup$
    – dimitriy
    Commented Aug 9, 2013 at 15:54
  • $\begingroup$ I would not expect this to work for probit, or at least to have a clean interpretation, since $\Phi(\cdot)$ is not a canonical link, and a binary dependent variable with probit does not fall into the exponential family. $\endgroup$
    – StasK
    Commented Aug 12, 2013 at 21:24
  • $\begingroup$ @StasK That seems right, but then why do these options exist in Stata and R? What do they accomplish? $\endgroup$
    – dimitriy
    Commented Aug 13, 2013 at 1:20

Recasting this as a time-to-event problem, wouldn't a logistic model with a ln(time) offset effectively commit you to a parametric survival function that may or may not fit the data well?


$p = [Z*\exp(\text{xbeta})]/[1+Z*\exp(\text{xbeta})]$

Predicted survival at time $Z = 1-[Z*\exp(\text{xbeta})]/[1+Z*\exp(\text{xbeta})]$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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