I have a semi-log regression model, with two continuous predictors, two categorical predictors (0 or 1 dummy variables) and a non-zero intercept. The response variable is log10 transformed, none of the predictors are transformed. I'm trying to generate the appropriate antilog of the model so as to be able to make unbiased predictions of the response variable. My reading so far has shown me two things quite clearly:

1) It's not sufficient simply to exponentiate both sides, as this will cause the antilogged model to underestimate observations quite badly (by approximately a factor of 2 in the case of my model).

2) Dummy variables are a special case and have to be antilogged in a different manner from continuous variables.

At this point, however, I've started to draw a bit of a blank. There are lots of different methods suggested in the literature for different subjects (primarily econometrics and ecology), but I can't find anything that deals with both of the above points simultaneously.

Could anyone please suggest the most appropriate way for me to go about back-transforming my model so as to minimise its bias, or suggest a paper that deals with such.

  • $\begingroup$ What makes you think dummies have to be treated differently? $\endgroup$
    – Glen_b
    Commented Feb 4, 2015 at 13:06
  • $\begingroup$ Mainly reading some of Dave Giles' work, some of which is summaried here: davegiles.blogspot.co.uk/2011/03/dummies-for-dummies.html $\endgroup$
    – Rucky
    Commented Feb 4, 2015 at 13:08
  • $\begingroup$ I fear you have misunderstood the point there. The discussion does not suggest that when calculating a fitted mean or a prediction back on the original scale that the dummies are treated differently. It's talking about interpreting the impact of a variable (i.e. the effect of a unit change in the variable). $\endgroup$
    – Glen_b
    Commented Feb 4, 2015 at 13:10
  • $\begingroup$ I don't doubt for a moment that I've misunderstood. Could you elaborate? I'm an ecologist, not a statistician, so any help would be gratefully received. $\endgroup$
    – Rucky
    Commented Feb 4, 2015 at 13:12
  • $\begingroup$ See the edit above for elaboration. $\endgroup$
    – Glen_b
    Commented Feb 4, 2015 at 13:24

1 Answer 1


Here I'm working with logs to base $e$.

If you are prepared to assume normality on the log scale and your sample size is so large you're prepared to treat the sample variance as a fixed quantity, and you want to unbias for the mean then you can use the mean of a lognormal ($\exp(\mu+\frac{1}{2}\sigma^2)=\exp(\mu)\exp(\frac{1}{2}\sigma^2)$). So you multiply by $\exp(\frac{1}{2}\sigma^2)$, where since the variance was assumed fixed, would be estimated by $\exp(\frac{1}{2}\hat{\sigma^2})$. [If you try to incorporate the uncertainty in $\hat{\sigma^2}$, the resulting distribution is log-t not lognormal ... and the mean -- that goes off to $\infty$]

(However, you need to take some care over what it is you're predicting, as it will affect the variance you use.)

Alternatively, you can use a Taylor expansion to get an approximate adjustment to the exponentiated forecast.

(If the standard error of the estimate is very small, it won't make much difference.)

Either way should improve things. This calculation applies the same way to both dummies and to continuous IVs.


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