Large value of $X\beta$ in logistic regression?

Question

In logistic regression, the probability is obtained from $$ Pr = \frac{\exp(X\beta)}{1 + \exp(X\beta)} ~~~~ (1) $$ From the plot below, it is obvious that if $X\beta$ > 10, the probability approaches 1 In my case (a penalized logistic regression, details are omitted), my estimate for $\beta$ (and therefore $X\beta$) is very large, > 2000, when plugging the equation (1), R gives out Inf.

(Updated details based on @whuber comments: I actually also have some negative values, say, <-2000, so the alternative Equation (2) is not working) $$ Pr = \frac{1}{1+\exp(-X\beta)} ~~~~~~~(2) $$

I am looking for some suggestions on how to deal with this situation, where the output is $Pr$.

Possible two approaches I can think of are:

• Use the if-condition: If $X\beta$ > 10, then Pr = 1; to compute the probability and skip the equation (1).
• Check my code on the estimation of $\beta$. (though I do not think my code has bugs)
• (based on @whuber's comments) If $X\beta$ is positive, use Equation (2); if $X\beta$ is negative, use Equation (1).
• (based on @BenBolker's comments) Use the R function plogis() to compute the $Pr$, which is the logistic of $X\beta$. The code is plogis(c(-2000,2000), lower.tail=TRUE). A related article is here.

Answer 1

If you use the built-in plogis() function (which returns the cumulative distribution function of the logistic distribution, which by definition is the same as the logistic/inverse-logit function) it will take care of your computational issues:

plogis(c(-2000,2000))
## 0 1

Even better, you can get a more precise answer by asking R to provide $\ln(\textrm{logistic}(X\beta))$ directly:

plogis(-2000,log.p=TRUE)
## -2000

In this case, because the logistic is approximately equal to the exponential function at its lower extent, $\ln(\textrm{logistic}(x)) \approx \ln(\exp(x)) = x$. If you want to convert this to base 10, $-2000/\ln(10)=-868.589$ tells you that $p \approx 10^{-868}$.

Since the logistic function is symmetric, the upper end works similarly: to compute $\ln(1-\textrm{logistic}(x))$, use

plogis(2000,log.p=TRUE,lower.tail=FALSE)
## -2000

So (as we could have figured out from symmetry), $\textrm{logistic}(2000) \approx 1-10^{-868}$.