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Re: st: interpreting cofficient on interaction of two logged variables


From   David Hoaglin <[email protected]>
To   [email protected]
Subject   Re: st: interpreting cofficient on interaction of two logged variables
Date   Tue, 21 Aug 2012 16:24:56 -0400

Dear Anna,

Others have commented on the mathematics of the log function.

I assume that the interaction predictor is in that model for a good
reason, either from the theory underlying the model or because you (or
others) have found that it makes a useful contribution in actual sets
of data.  The justification for including the interaction term should
come from one of these sources.

After you have decided to transform x1 and x2 by taking their
logarithms (for any of a variety of reasons), they are simply
predictors in the model (that happen to be measured in the log scale).
 Their interaction represents (or summarizes) a nonlinearity in the
relation of log(y) to log(x1) and log(x2).  The slope of log(y)
against log(x1) varies with the value of log(x2), and similarly for
the slope of log(y) against log(x2).

Viewed in another way, the model describes a particular type of
quadratic surface in the variables log(x1) and log(x2).  By replacing
log(x1) and log(x2) by suitable linear combinations of those two
variables, it should be possible to eliminate the interaction term, in
trade for separate quadratic terms in log(x1) and log(x2), though that
may not improve the interpretability of the model.

David Hoaglin

On Tue, Aug 21, 2012 at 8:40 AM, D'Souza, Anna - ERS
<[email protected]> wrote:
> Hello,
>
> I am interested in estimating the following model: log y = b0 + b1 log x1 + b2 log x2 + b3 (log x1 * log x2) + e
>
> Question 1: Does anyone have a reference that describes the interaction of logged variables? I am looking for a justification as to why to use "log x1 * log x2" vs. log (x1 * x2).
>
> Question 2: What is the best way to calculate the (i) average marginal effect on x1, and (ii) the marginal effect of x1 evaluated at the mean of x2 in STATA?
>
> Thank you,
>
> Anna
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