I noticed some discrepancies between the expected output and actual output for binomial glm in R using customized weighted effect contrast coding.

The purpose of weighted effect coding is to use the overall sample mean as the reference group. All data is taken from the source package wec as shown below.



#Convert wage to a binary variable
PUMS <- as_tibble(PUMS) %>%
  mutate(wage = if_else(wage>50000, 1, 0))

#Switch to weighted effect contrasts
contrasts(PUMS$race) <- contr.wec(PUMS$race, "Black")

#Overall binomial percent
overall <- PUMS %>%
  summarise(Percent = sum(wage)/n())
overall_odds <- overall[[1,1]]/(1-overall[[1,1]])

#Calculated perfectly when treating the 0-1 binomial variable as linear
#The (intercept matches overall[[1, 1]])
tidy(lm(wage ~ race, data = PUMS))

#There are some odd slight discrepancies between the Odds_Ratio columns
tidy(glm(wage ~ race, family = binomial, data = PUMS)) %>%
  mutate(Odds_Ratio = exp(estimate))

PUMS %>%
  group_by(race) %>%
  summarise(Percent = sum(wage)/n()) %>%
  mutate(Odds_Ratio = (Percent/(1 - Percent))/overall_odds)

As you can see if you run the code, the output from the last two chunks is slightly off for some reason. For example, the odds ratio for the Asian group should be 1.59 but it appears as 1.61 in the glm model.

However, if the binomial outcome variable is treated as continuous a simple lm call predicts the odds ratios perfectly (the intercept matches overall[[1, 1]]).

Is there some mechanistic reason or rounding error that causes binomial regression to behave this way, and can I adjust settings to fix it? I would really appreciate help in this as it is critical for my research.


If anybody ever encounters this issue again, here is a modification of Ben Bolker's answer to get the weighted overall odds and expected coefficients for multiple groups, not just the unweighted odds in regular two-variable effect coding.


dd <- data.frame(x = factor(rep(c("a", "b", "c"), times = c(153, 46, 51))), 
                 y = rbinom(250, 1, 0.5))

contrasts(dd$x) <- wec::contr.wec(dd$x, omitted = "b")

dds <- (dd |>
          dplyr::group_by(x) |>
          dplyr::summarise(pct = mean(y), odds = pct/(1-pct), count = n()))

overall_odds <- exp(weighted.mean(log(dds$odds), dds$count))

dds |> dplyr::mutate(estimate = odds/overall_odds) |>
  relocate(estimate, .before = pct)

tidy(glm(y ~ x, family = binomial, dd), exponentiate = TRUE)
  • $\begingroup$ Welcome to the site. Please consider using base R, & commenting it extensively, when illustrating posts here with R code. Not everyone who will come to this page will be familiar with R, & not all of those will be able to read tidy-code. This is a Q&A site for statistics, not R. $\endgroup$ Jul 18, 2022 at 0:08
  • $\begingroup$ I haven't thought about this carefully at all, but I would suspect some kind of nonlinear averaging/Jensen's inequality thing going on ... $\endgroup$
    – Ben Bolker
    Jul 18, 2022 at 0:12

1 Answer 1


I made up a simpler, more extreme example that strongly hints that the problem is nonlinear averaging, not some rounding issue.

dd <- data.frame(x = factor(rep(c("a", "b"), each = 100)), 
                 y = rep(c(1,0,1,0), times = c(95, 5, 10, 90)))
contrasts(dd$x) <- wec::contr.wec(dd$x, omitted = "b")
overall_odds <- mean(dd$y)/(1-mean(dd$y))
dd |> dplyr::group_by(x) |> dplyr::summarise(pct = mean(y), 
                                             oddsratio = pct/(1-pct)/overall_odds)
##   x       pct   oddsratio
## 1 a      0.95 17.2  
## 2 b      0.1   0.101
tidy(glm(y ~ x, family = binomial, dd), exponentiate = TRUE)
## (exponentiate = TRUE returns estimates as odds ratios 
##    rather than log-odds diffs)
##   term        estimate std.error statistic  p.value
## 1 (Intercept)     1.45     0.284      1.32 1.88e- 1
## 2 xa             13.1      0.284      9.07 1.23e-19

A little bit more digging in te Grotenhuis et al 2017 explains what's going on:

Likewise, if a researcher wishes to investigate obesity (BMI > 30) and, therefore, uses a dichotomy of BMI in a logistic regression analysis, then the mean to be tested against is the average of all log odds.

(emphasis added).

So the value that you compare against should not be the overall odds for the data set ($p/(1-p)$), but the geometric mean of the odds for each group (i.e. $\exp(\textrm{mean}(\log(p_i/(1-p_i)))$).

Continuing the example above:

dds <- (dd |> dplyr::group_by(x)
    |> dplyr::summarise(pct = mean(y),
                        odds = pct/(1-pct)))
overall_odds <- exp(mean(log(dds$odds)))
dds <- (dds |> dplyr::mutate(odds_adj = odds/overall_odds))

Now the odds_adj column matches the values calculated by glm():

  x       pct   odds odds_adj
1 a      0.95 19.0    13.1   
2 b      0.1   0.111   0.0765

te Grotenhuis, Manfred, Ben Pelzer, Rob Eisinga, Rense Nieuwenhuis, Alexander Schmidt-Catran, and Ruben Konig. “When Size Matters: Advantages of Weighted Effect Coding in Observational Studies.” International Journal of Public Health 62, no. 1 (January 1, 2017): 163–67. https://doi.org/10.1007/s00038-016-0901-1.

  • $\begingroup$ Thanks so much for the quick reply! Is nonlinear averaging implemented somewhere in the glm source code, but only for binomial regression and not for linear regression? Why doesn't this affect all regression contrast coding schemes? svn.r-project.org/R/trunk/src/library/stats/R/glm.R $\endgroup$
    – dcsuka
    Jul 18, 2022 at 0:38
  • $\begingroup$ This kind of problem probably does affect all regression contrast coding schemes. Can you show that it doesn't? (Nonlinear averaging doesn't affect linear regression, or any GLM with an linear link function (typically the identity link).) $\endgroup$
    – Ben Bolker
    Jul 18, 2022 at 0:41
  • $\begingroup$ Massive thank you, Professor Bolker! This has been gnawing at my mind for about a month and I doubt I could have ever solved this on my own. Just one more quick question, and in 2 days I'll set up a bounty. Is it alright to publish contrast coding like this that refers to the mean, even though it is slightly inaccurate due to its logistic nature? Is it necessary to specify this in the manuscript? If I have 100 categorical variables with vastly different sample sizes and no clear reference group or order, should I select a different contrast system? Thank you so much. $\endgroup$
    – dcsuka
    Jul 18, 2022 at 2:21
  • $\begingroup$ *100 levels of a categorical variable $\endgroup$
    – dcsuka
    Jul 18, 2022 at 2:25
  • 1
    $\begingroup$ The answer to your last questions depends a lot on the expectations of your field. The weighted contrast defined by te Grotenhuis et al. seems technically fine (I wouldn't say it's "inaccurate"); it would be up to you to describe it clearly enough so that reviewers in your field understood it. You could use something like the emmeans package (see the section on "bias correction") to report your results. $\endgroup$
    – Ben Bolker
    Jul 18, 2022 at 13:11

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