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I have a lin-log regression model like

$$Y = b_0 + b_1 \log(x_1 + 1) + e.$$

The distribution of $x_1$ is very skewed, thus I use the natural logarithm to get a more Gaussian like distribution. Because 3 out of 100 values have zero as entry I add a constant c, in my case plus 1, to avoid -Inf.

The resulting estimation of $b_1$ is about -0.14.

Without the constant the interpretation is clear: a 1% change in $x$ results in a $0.01\cdot b_1$ change in $y$.

I struggle with the constant. How can I account for it in my interpretation? If I change the value of c I get, of course, other estimates. I have chosen + 1 because this results in positive log values (the values of $x_1$ are originally positive too).

Or should I add a small value just to the three 0s?

Many thanks in advance and, please, a non mathematical answer ;-) Marco

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    $\begingroup$ since it is not log-log regression, without constant you have not the elasticity interpretation as you wrote. Why you believe that log transform is needed? Is your $x_1$ log-normally distributed? Is it a time series or just cross-sectional data you work with? What is the nature of your data? $\endgroup$ – Dmitrij Celov Jun 15 '11 at 14:57
  • $\begingroup$ Thank you for your response and sorry for my unclear thread. My database are environmental cross-sectional data. I thought that choosing log() is a better choice compared to the raw data which show a skewed distribution. Using the log I get a nearly normal distribution and the model fits the data better. I hope that this is the right interpretation of my coefficient: (-0.14/100) (%x) Thus e.g. if x is 10% higher y is predicted to be 0.014 units lower. Hope that is correct now, but still not considering the added constant. $\endgroup$ – Marco Jun 15 '11 at 18:19
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    $\begingroup$ There is no implicit need in linear regression wrt the distribution of your covariates. So it is not necessary to do the transform: your x can be as skewly distributed as it likes. If the fit gets better by doing this transformation (as opposed to without any transformation): lucky you. However, you would need x+1 to increase with 10% for you kind of interpretation! Maybe you were too lucky and have overfit with the transformation? Is the difference really noteworthy? $\endgroup$ – Nick Sabbe Jun 16 '11 at 6:45
  • $\begingroup$ As @Nick wrote there is no need for your covariate to look like normal, regression is roughly something about multivariate distribution of $Y$ and $x_1$ so if both of them are skewed as taking a pair of variables ($Y$,$x_1$) it is still possible to do regression. What your concern is about is if a linear model relevant? It would be easier to understand your transformations if we know what is $Y$ and $x_1$. $\endgroup$ – Dmitrij Celov Jun 16 '11 at 7:43
  • $\begingroup$ @ all, thank you very much for clarification, I wrongly assumed that the independent variable(s) must be normally distributed as well. the model assumptions are fulfilled without transformations. so I think i will use this model. btw, my Y are mortality ratios and x are a density index of medical practitioner per 10000 inhabitants. $\endgroup$ – Marco Jun 16 '11 at 8:27

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