Let's say value is TB treatment coverage in %.

What is the difference between:

  • linear regression in logit space :
logit       <- function(x){qlogis(log(x), log.p = TRUE)}
invlogit    <- function(x){exp(plogis(x, log.p = TRUE))}
fit <- predict(lm(logit(value/100) ~ year,data = df), newdata = data.frame(year = 2000:2025), se = T)
pred_val  <- as.matrix(fit$fit)[,1] 
data.frame(year = 2000:2025, pred_val = 100*invlogit(pred_val))
  • and glm using family = quasibnomial :
fit <- predict(glm(value/100 ~ year, family=quasibinomial),newdata=data.frame(year= 2000:2025),type="response",se = T)
data.frame(year=2000:2025, pred_val=fit$fit)

What makes the output be different?

Thank you

  • 1
    $\begingroup$ stats.stackexchange.com/questions/47840/… $\endgroup$
    – Ben Bolker
    Commented Sep 7, 2020 at 23:12
  • $\begingroup$ While the question at Ben's link discusses a comparison between a different pair of models, the underlying points made there apply here as well; it's definitely worth checking the answers and comments under them to follow the discussion, because it's relevant to your question. $\endgroup$
    – Glen_b
    Commented Sep 7, 2020 at 23:39
  • $\begingroup$ There's also stats.stackexchange.com/questions/357850/… - again not identical but the same issue (however, beta regression and quasi-binomial GLMs have the same variance function and typically both use the logit link, so they're quite similar) $\endgroup$
    – Glen_b
    Commented Sep 8, 2020 at 1:13

1 Answer 1


The comparison only makes sense when you can logit-transform observed proportions, which means you can't do it on 0-1 data, you must already have a set of proportions $p_i=X_i/n_i$ where none of the $X_i$'s are $0$ or $n_i$.

At heart, the distributional assumptions differ, so the likelihood function that is maximized is different; naturally if you optimize different functions of the data then the resulting parameter estimates (etc) will differ.

However, let's look in more basic terms by just considering the first couple of moments. When you apply a nonlinear transformation you impact the mean (introduce a bias relative to the original scale) and the variance (which affects relative weight on data). However, I'm not going to go through the algebra, and just show what happens when we look at the impact on transforming a binomial proportion to the logit scale.

The variance-stabilizing transform for the binomial is the arcsin-square root ($\sin^{-1}(\sqrt{p})$), which is different from the logit; relatively speaking* the logit pulls the most extreme values more strongly than the variance-stabilizing transformation does, which would be expected to make the variances get bigger near the ends.

If we generate a lot of binomial data across a range of values for the proportion that actually has a logistic expected value, and then logit-transform it, you can see the variance is not constant on the logit scale, even though your linear model fit treats it as if it were:

Plot of logit of binomial proportions against linear predictor

The spread is indeed wider near the ends than it is in the middle, as we anticipated. As a result the linear model on the logit scale puts (relatively) too much weight on the extreme values, which are noisier and so should be downweighted.

We also see in the plot that near the extremes, the transformation isn't close to symmetric - when the population proportion is near to either end, a (by chance) more-extreme-than-expected proportion gets pulled more strongly out (more extreme).

Note that $E(t(Y)) \neq t(E(Y))$ so when you apply a logit transform and fit a linear regression you introduce a (typically small) bias term; if you compare them on the original (proportion) scale the bias is small near the middle but becomes progressively larger. So for example, coefficients will be expected to be pushed away from 0 on average (assuming the cases of p=0 and p=1 are restricted to never occur, even though they're certainly possible).

If your proportions are all in the middle half, you would typically see only very small effects, but as proportions get more extreme, these effects all become more noticeable.

* in particular, compare a plot of $4\sin^{-1}(\sqrt{p})-\pi$ with $\operatorname{logit}(p)$

  • $\begingroup$ As @Glen_b pointed out, lm creates a model for E[g(Y)], and glm creates a model for g(E[Y]). In the glm case, g is the link function. $\endgroup$
    – igorkf
    Commented Sep 8, 2020 at 2:48
  • $\begingroup$ Hmm, I should probably be more explicit about it in my answer. $\endgroup$
    – Glen_b
    Commented Sep 8, 2020 at 5:32

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.