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This question already has an answer here:

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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marked as duplicate by whuber Apr 25 '16 at 21:14

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ 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. Stone Apr 25 '16 at 21:06
  • $\begingroup$ 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. $\endgroup$ – David Apr 25 '16 at 21:13
  • $\begingroup$ Oh, thanks. I checked all of them, but now I see I missed few. $\endgroup$ – David Apr 25 '16 at 21:19
  • $\begingroup$ "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? $\endgroup$ – David Apr 25 '16 at 21:21
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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. $\endgroup$ – David Apr 25 '16 at 21:17
  • $\begingroup$ I thought my first two paragraphs answered the interpretation question. I guess not. $\endgroup$ – StatNoodle May 3 '16 at 22:01

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