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In this paper titled "CHOOSING AMONG GENERALIZED LINEAR MODELS APPLIED TO MEDICAL DATA" the authors write:

In a generalized linear model, the mean is transformed, by the link function, instead of transforming the response itself. The two methods of transformation can lead to quite different results; for example, the mean of log-transformed responses is not the same as the logarithm of the mean response. In general, the former cannot easily be transformed to a mean response. Thus, transforming the mean often allows the results to be more easily interpreted, especially in that mean parameters remain on the same scale as the measured responses.

It appears they advise the fitting of a generalized linear model (GLM) with log link instead of a linear model (LM) with log-transformed response. I do not grasp the advantages of this approach, and it seems quite unusual to me.

My response variable looks log-normally distributed. I get similar results in terms of the coefficients and their standard errors with either approach.

Still I wonder: If a variable has a log-normal distribution, isn't the mean of the log-transformed variable preferable over the log of the mean untransformed variable, as the mean is the natural summary of a normal distribution, and the log-transformed variable is normally distributed, whereas the variable itself is not?

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    $\begingroup$ I agree with your clue if we have a log-normal distributed variable. However, the mean need to be "back transformed" to get an easily understandable statistic based on the original scale of data. This might explain the article's conclusion. Also, after log-transformation, we might not get a normally distributed variable and in this case, I do not know which approach would be better. $\endgroup$ – soufanom Jan 17 '13 at 4:46
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Although it may appear that the mean of the log-transformed variables is preferable (since this is how log-normal is typically parameterised), from a practical point of view, the log of the mean is typically much more useful.

This is particularly true when your model is not exactly correct, and to quote George Box: "All models are wrong, some are useful"

Suppose some quantity is log normally distributed, blood pressure say (I'm not a medic!), and we have two populations, men and women. One might hypothesise that the average blood pressure is higher in women than in men. This exactly corresponds to asking whether log of average blood pressure is higher in women than in men. It is not the same as asking whether the average of log blood pressure is higher in women that man.

Don't get confused by the text book parameterisation of a distribution - it doesn't have any "real" meaning. The log-normal distribution is parameterised by the mean of the log ($\mu_{\ln}$) because of mathematical convenience, but equally we could choose to parameterise it by its actual mean and variance

$\mu = e^{\mu_{\ln} + \sigma_{\ln}^2/2}$

$\sigma^2 = (e^{\sigma^2_{\ln}} -1)e^{2 \mu_{\ln} + \sigma_{\ln}^2}$

Obviously, doing so makes the algebra horribly complicated, but it still works and means the same thing.

Looking at the above formula, we can see an important difference between transforming the variables and transforming the mean. The log of the mean, $\ln(\mu)$, increases as $\sigma^2_{\ln}$ increases, while the mean of the log, $\mu_{\ln}$ doesn't.

This means that women could, on average, have higher blood pressure that men, even though the mean paramater of the log normal distribution ($\mu_{\ln}$) is the same, simply because the variance parameter is larger. This fact would get missed by a test that used log(Blood Pressure).

So far, we have assumed that blood pressure genuinly is log-normal. If the true distributions are not quite log normal, then transforming the data will (typically) make things even worse than above - since we won't quite know what our "mean" parameter actually means. I.e. we won't know those two equations for mean and variance I gave above are correct. Using those to transform back and forth will then introduce additional errors.

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  • $\begingroup$ I found this very helpful: christoph-scherber.de/content/PDF%20Files/… $\endgroup$ – Aditya Jul 9 '17 at 7:10
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    $\begingroup$ Corone, I highlighted two important sentences in your answer. I hope you don't mind. Please roll back if you disagree. $\endgroup$ – Stefan Apr 13 '18 at 15:29
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If the response veritable is not symmetric (not distributed as normal) but log transformed response is normal then linear regression on transformed response be used and the exponent coefficient gives us the ration of geometric mean.

If the response veritable is symmetric (distributed as normal) but relation between explanatory (X) and response is not linear but log expected value is linear function of X then GLM with log link be used and exponent coefficient gives us the ratio of arithmetic mean

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  • $\begingroup$ This answer is not clear. Did you mean "variable" rather than "veritable"? $\endgroup$ – Michael Chernick Jul 23 '18 at 4:46
  • $\begingroup$ This is a fragment of an answer. You need to make it clear how this relates to the question and what the answer to the question actually is based on this piece of insight. $\endgroup$ – ReneBt Jul 23 '18 at 8:29
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Here are my two cents from an advanced data analysis course I took while studying biostatistics (although I don't have any references other than my professor's notes):

It boils down to whether or not you need to address linearity and heteroscedasticity (unequal variances) in your data, or just linearity.

She notes that transforming the data affects both the linearity and variance assumptions of a model. For example, if your residuals exhibit issues with both, you could consider transforming the data, which potentially could fix both. The transformation transforms the errors and thus their variance.

In contrast, using the link function only affects the linearity assumption, not the variance. The log is taken of the mean (expected value), and thus the variance of the residuals is not affected.

In summary, if you don't have an issue with non-constant variance, she suggests using the link function over transformation, because you don't want to change your variance in that case (you're already meeting the assumption).

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    $\begingroup$ While the link function only impacts the mean, the link function is only one part of a GLM. Your comments work for a Gaussian glm with log link. A gamma GLM with log link will have the same variance-function assumption (variance proportional to mean squared) as taking logs and fitting a constant variance on that log scale. Other families within the GLM framework will have other variance functions. Unfortunately the table on the wikipedia page for GLMs omits the variance functions for the distribution families it gives. $\endgroup$ – Glen_b Oct 15 '14 at 2:50
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    $\begingroup$ They do mention some examples here though. Here's the gamma $\endgroup$ – Glen_b Oct 15 '14 at 2:56

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