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I've been thinking about a regression model with dependent variable a ratio in the range [0-600+]. About 50%+ are 0 values though, 40%+ values [0-5], and a very small percentage of the values are higher than 5, so the picture is quite skewed.

I've been wondering what kind of transformation I need to do in order to linearize the model. Is the natural log transformation appropriate?

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    $\begingroup$ If you take the log of the dependent variable, then you lose all observations with zero, i.e., 50%+. I wouldn't do that. What is the ratio exactly? Maybe you can multiply with the denominator, thus making everything linear. $\endgroup$
    – Nameless
    Commented Jul 3, 2013 at 13:01
  • $\begingroup$ What are the numerator and denominator of the ratio? Are they counts? Why are they divided? $\endgroup$
    – Glen_b
    Commented Jul 3, 2013 at 22:54
  • $\begingroup$ I've been interested in the success rate of a crowdfunding campaigns, that why I am taking the ratio beteen the actual pledge and the goal. The ratios do not have the same denominator, thus multipling wont work :( $\endgroup$
    – user27588
    Commented Jul 4, 2013 at 14:47

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One answer is no transformation at all. Poisson regression in fact can give the best of both worlds, the ability to cope with observed zeros and a logarithmic link. The point is that the mean response is modelled as positive, so that it can be logged, and that can be consistent with some observed values being zero. You don't transform the response. In effect the software takes a logarithmic view of the data but returns predictions on the original scale. (This is one kind of generalized linear model.)

See http://blog.stata.com/2011/08/22/use-poisson-rather-than-regress-tell-a-friend/ for one encouraging discussion. By the way, it's a myth that Poisson regression requires counted data or even a marginal Poisson distribution for the response. The leading idea is simply that a non-negative response and a functional form $Y = \exp(X\beta)$ just as the leading idea of classical regression is $Y = X\beta$ and assumptions of Gaussian distributions are far less crucial, just nice if they are accurate.

Much, however, depends on those zeros and whether they are all comparable. For example, number of alcoholic drinks per day is a response that could be zero, but if the population mixed people who never drink alcohol at all and people who in some cases just drank no alcohol on some days, then there is heterogeneity that might require explicit modelling. The best way I can think of putting this is that the simplest situation is that the zeros you observe are all sampling zeros, so that in principle they could have been different. If some are structural zeros, i.e. zeros by necessity or definition, your situation is more complicated.

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  • $\begingroup$ Even if they're all sampling zeros, don't you think he might run into an excess zeros problem using Poisson with 50% or more zeros? $\endgroup$
    – Nameless
    Commented Jul 3, 2013 at 14:18
  • $\begingroup$ @Nameless Could well be. But the mean being very low and near zero is consistent with Poisson regression. Also, I would always try Poisson first before trying something more exotic such as a zero-inflated Poisson. $\endgroup$
    – Nick Cox
    Commented Jul 3, 2013 at 14:21
  • $\begingroup$ The zeroes are sampling zeroes, however the sample is random and I do not expect to be that different from the whole population. I do not believe I will run into unobserved heterogeneity problem. I am still not sure how to proceed, probably I will read more about Poisson. Just to clarify...ln transformation is not an option, right? $\endgroup$
    – user27588
    Commented Jul 3, 2013 at 15:21
  • $\begingroup$ ln(0) is indeterminate, so a ln() transformation is, as already stated, emphatically not an option for you. You can find literature using ln(x + 1) as a work-around, on which there are threads in this forum. I advise against, but there are different views. $\endgroup$
    – Nick Cox
    Commented Jul 3, 2013 at 15:27
  • $\begingroup$ What you think about using Gamma GLM since the distibution is heavily skewed? I am mostly concerned about the academic support for the use of Poisson in similar situations. $\endgroup$
    – user27588
    Commented Jul 4, 2013 at 14:41

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