3
$\begingroup$

My independent variables were highly skewed, so to normalise the distribution they were log transformed. Also since there were zeros in the data, I've added + 1 to transform the variables. This is what the model looks like (negative binomial regression):

Dependant_var ~ log(Independent_var_1 + 1) + log(Independent_var_2 + 1)

Coefficients: 

                            Est.       Std. Err.  z-value   sig.
log(Independent_var_1 + 1)  0.031907   0.004701   6.787 1.14e-11 ***
log(Independent_var_2 + 1) -0.019007   0.004735  -4.015 5.96e-05 ***

IRRs:

log(Independent_var_1 + 1)  1.0324219
log(Independent_var_2 + 1)  0.9811724

Now, I'm having problems understanding how to interpret the results. If the data were not log transformed, I would interpret this as follows:

If everything else is held constant, a one unit increase in Independent_var_1 would result in the decrease by 0.031 units of Dependent_var. And for IRRs – a one unit increase of Independent_var_1 will result in an expected increase of the Dependent_var by a factor of 1.032 (everything else constant).

However, I'm confused since I don't have "units" anymore, but log transformed vars. Thanks.

|cite|improve this question
$\endgroup$
  • 2
    $\begingroup$ Welcome to the site, @Peter. Can you clarify a few things? You included the tag for [negative-binomial] & included that in the title. The NB is a distribution of counts. Are your data counts? You state that your DV was skewed, but your example shows logs of your IVs. Which was skewed? The distribution (skewed or not) of your IVs doesn't matter, & the overall (ie, marginal) distribution of your DV doesn't matter, only the distribution of your residuals does & even then the NB is supposed to be skewed. Can you say more about your situation, your data & your goals? $\endgroup$ – gung♦ Dec 20 '14 at 1:10
  • $\begingroup$ Hi @gung. There was a typo – the independent variables are skewed and log transformed. The DV is count data and left as is (no log transformation). The IVs, however, follow almost a power-law distribution (without log transformation). Are log transformations unnecessary in this case? $\endgroup$ – Peter Dec 20 '14 at 1:48
  • 2
    $\begingroup$ Why would you attempt to make your IV's "normal"? There's no assumption that they should be. $\endgroup$ – Glen_b♦ Dec 20 '14 at 9:16
1
$\begingroup$

The interpretation of coefficients associated with log-transformed independent variables is straightforward. You now have log units, which depend on the choice of basis for the logarithm. For natural log as in your example, an $e$-fold change in (Independent_var_1 + 1) is associated with the indicated change in the dependent variable. It might be simpler for a reader to understand if you to use base-2 or base-10 logarithms, so that the regression coefficient represents a doubling or a 10-fold increase in (Independent_var_1 + 1).

You might want to consider the suggestion provided here for an alternate way of dealing with the 0-value problem in logarithmic transformations, which handles cases having 0 values of an independent variable separately from cases having positive values.

|cite|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.