# Equation for a logit link function for a series of events

I have modeled some data using generalized linear modeling with a binomial distribution and logit link function. However, my data is not dichotomous, it is actually a series of events. I have fixed the number of trials at 20 (because there were 20 trials) so it is not a binary outcome (i.e., logistic regression). I believe the logit link function looks like this: exp(X * beta)/[1 + exp(X * beta)] or In(natural log) (probability of the event happening/ 1 - the probability of the event not happening). I am trying to do something on excel so I need the actual equation for the link function I used but because this is not a traditional logit I am a bit confused. Can someone tell me based on this what the equation would be in a form I could use? In other words, what does the logit link function equation look like when instead of a binary DV it is viewed as some number of events, for a fixed amount of trials (20 in my case)? Thank you!

I assume that you run for each person 20 independent trials, and that you get a dataset that records for each person the number of successes ($k_i$) and some explanatory variables ($x_i$).

Since you want to model this as a binomial process with 20 trials the expected number of success ($E(k_i)=\mu_i$) is 20 times the probability of succeeding in a single trial. Since you want to model the probability of succeeding in single trial with a logit link function the expected outcome is:

$$E(k_i) = \mu_i = 20 \frac{ \exp(x_i\beta)}{1+\exp(x_i\beta)}$$

Alternatively, you can rewrite this as:

$$\ln\left(\frac{\mu_i}{20-\mu_i} \right) = x_i\beta$$

I looked this up in (Hardin and Hilbe 2001, p. 121), but I suspect you can find these formulas in most textbooks on Generalized Linear Models.

James W. Hardin and Joseph M. Hilbe (2001) Generalized Linear Models and Extensions, second edition. College Station, TX: Stata Press.

• I was thinking it was something along the lines of the second equation but I could not find a definitive source. Thank you so much! – user20334 Apr 6 '13 at 17:17