How to back-transform inverse transformation in emmeans when using contrast function

Question

I am trying to calculate pairwise comparisons using the {emmeans} package after fitting a linear model with an inverse-transformed response. Here is the data and fitted model. The data comes from the {faraway} package.

> data("rats", package = "faraway")
> m <- lm(1/time ~ treat + poison, data = rats)

After fitting the model I can get estimated marginal means that are back-transformed using the following code:

> emmeans(m3, specs = "treat", type = "response", tran = "inverse") 
 treat response     SE df lower.CL upper.CL
 A        0.284 0.0115 42    0.263    0.309
 B        0.537 0.0411 42    0.465    0.635
 C        0.339 0.0164 42    0.309    0.376
 D        0.463 0.0305 42    0.408    0.534

Results are averaged over the levels of: poison 
Confidence level used: 0.95 
Intervals are back-transformed from the inverse scale 

However, when I then try to calculate pairwise comparisons using the contrast() function, the results are not back-transformed.

> emmeans(m3, specs = "treat", type = "response", tran = "inverse") |> 
+   contrast("pairwise")
 contrast estimate    SE df t.ratio p.value
 A - B       1.657 0.201 42   8.233  <.0001
 A - C       0.572 0.201 42   2.842  0.0335
 A - D       1.358 0.201 42   6.747  <.0001
 B - C      -1.085 0.201 42  -5.391  <.0001
 B - D      -0.299 0.201 42  -1.485  0.4551
 C - D       0.786 0.201 42   3.905  0.0018

Results are averaged over the levels of: poison 
Note: contrasts are still on the inverse scale 
P value adjustment: tukey method for comparing a family of 4 estimates 

I have tried moving the type = "response", tran = "inverse" arguments to the contrast() function but that produces unusual results:

> emmeans(m3, specs = "treat") |> 
+   contrast("pairwise", type = "response", tran = "inverse")
 contrast response  SE df null t.ratio p.value
 A - B           1   0 42  Inf   8.233  <.0001
 A - C           2   1 42  Inf   2.842  0.0335
 A - D           1   0 42  Inf   6.747  <.0001
 B - C         Inf Inf 42  Inf  -5.391  <.0001
 B - D         Inf Inf 42  Inf  -1.485  0.4551
 C - D           1   0 42  Inf   3.905  0.0018

I have also tried the following but still get the same result (comparisons still on inverse scale).

tran <- make.tran("inverse")
m2 <- with(tran, lm(linkfun(time) ~ treat + poison, data = rats))
emmeans(m2, specs = "treat", type = "response") |> 
  contrast("pairwise")

My question: Is there a way to get contrasts that are back-transformed from the inverse scale?

Answer 1

This is well-documented and is a matter of deciding what you want to be talking about. If you do

emm <- emmeans(..., type = "response")

then the means in emm are still on the transformed scale, but back-transformed to the response scale. When we do

contrast(emm, ...)

the contrasts are still computed on the transformed scale. If type = "response" is in force, then those results are back-transformed, but only if there is a valid way to do that. For example, if the transformation is the log, then a comparison of two means is of the form log(a) - log(b) which is equal to log(a/b), hence back-transforming it will yield an estimate of a/b.

However, with the inverse scale, because there is no way to back-transform 1/a - 1/b, so the results are reported on the inverse scale, with a message.

If really what you wanted was an estimate of mean(1/a) - mean(1/b), that can be done but we don't do mind-reading, you have to ask for it. That is accomplished via the regrid() function, which re-casts all the estimates to a new scale:

remm <- regrid(emm)    # back-transforms to the response scale by default
contrast(remm, ...)

By the same token, if you really wanted to estimate the ratios of the response means, you could do

contrast(regrid(emm, transform = "log"), ...)

See the help page for regrid() for more information. You can also find a detailed discussion of how transformations are dealt with in the "transformations" vignette.