I need to use log-log regression and because I have lots of zero values I tried to add a very small constant c=8E-12 to x and it works pretty good. Xs are very small probabilities.
lnY= a + b ln (x+c)
But how do I interpret that model? Thanks
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1$\begingroup$ Why don't you tell us something about the original model and data, why "you need" to use log-log regression, and how the results will be used? $\endgroup$– Mark L. StoneCommented Apr 25, 2016 at 21:06
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$\begingroup$ I´m trying to create model for my bachelor thesis. I use gravity model to explain influence of religion on international trade and log-log regression is the very simple solution. $\endgroup$– DavidCommented Apr 25, 2016 at 21:13
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$\begingroup$ Also closely related: stats.stackexchange.com/questions/30728, stats.stackexchange.com/questions/18480, stats.stackexchange.com/questions/20397, and stats.stackexchange.com/questions/18694. $\endgroup$– whuber ♦Commented Apr 25, 2016 at 21:13
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$\begingroup$ Oh, thanks. I checked all of them, but now I see I missed few. $\endgroup$– DavidCommented Apr 25, 2016 at 21:19
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$\begingroup$ "Generally, using log(1+y)log(1+y) and then interpreting the estimates as if the variable were log(y)log(y) is acceptable when the data contain relatively few zeros" says Wooldridge. But what if I have lots of zero values? $\endgroup$– DavidCommented Apr 25, 2016 at 21:21
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1 Answer
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You are asserting (not testing) that x cannot actually be zero but rather that you simply cannot measure (or control) it below a certain level.
If it is a dose, for instance, you are asserting that there is always some of the compound in the background environment. If it is light, you are asserting that there is some background level of photons, etc.
The problem is, your $c$ is a guess and may bias the results even if your assertion is true. You may want to try several values.
Your model is related to $Y = e^a (x+c)^b \times \epsilon$.
If you intend to fit $Y = e^a (x+c)^b + \epsilon$ then how you fit the model matters. You maybe could estimate $c$ as a parameter but this could be problematic and may not be worth the bother.
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$\begingroup$ Thank you very much for your fast answer. I am just asking how to interpret that, not how to make model the right way. $\endgroup$– DavidCommented Apr 25, 2016 at 21:17
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$\begingroup$ I thought my first two paragraphs answered the interpretation question. I guess not. $\endgroup$ Commented May 3, 2016 at 22:01