A Poisson regression is a GLM with a log-link function.

An alternative way to model non-normally distributed count data is to preprocess by taking the log (or rather, log(1+count) to handle 0's). If you do a least-squares regression on log-count responses, is that related to a Poisson regression? Can it handle similar phenomena?

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    $\begingroup$ How do you plan on taking logarithms of any counts that are zero? $\endgroup$ – whuber Mar 19 '11 at 18:47
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    $\begingroup$ Definitely not equivalent. An easy way to see this is to look at what would happen if you observed zero counts. (Comment created before seeing @whuber's comment. Apparently this page didn't refresh appropriately on my browser.) $\endgroup$ – cardinal Mar 19 '11 at 18:52
  • $\begingroup$ OK, I obviously should say, log(1+count). Obviously not equivalent, but wondering if there was a relationship, or if they can handle similar phenomena. $\endgroup$ – Brendan OConnor Mar 21 '11 at 0:45
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    $\begingroup$ There is useful discussion of this issue here: blog.stata.com/2011/08/22/… $\endgroup$ – Michael Bishop Dec 9 '11 at 23:55

On the one hand, in a Poisson regression, the left-hand side of the model equation is the logarithm of the expected count: $\log(E[Y|x])$.

On the other hand, in a "standard" linear model, the left-hand side is the expected value of the normal response variable: $E[Y|x]$. In particular, the link function is the identity function.

Now, let us say $Y$ is a Poisson variable and that you intend to normalise it by taking the log: $Y' = \log(Y)$. Because $Y'$ is supposed to be normal you plan to fit the standard linear model for which the left-hand side is $E[Y'|x] = E[\log(Y)|x]$. But, in general, $E[\log(Y) | x] \neq \log(E[Y|x])$. As a consequence, these two modelling approaches are different.

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    $\begingroup$ Actually, $\mathbb{E}(\log(Y) | X) \neq \log(\mathbb{E}(Y | X))$ ever unless $\mathbb{P}(Y = f(X) | X ) = 1$ for some $\sigma(X)$-measurable function $f$, i.e., $Y$ is fully determined by $X$. $\endgroup$ – cardinal Mar 19 '11 at 18:50
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    $\begingroup$ @cardinal. Very well put. $\endgroup$ – suncoolsu Mar 20 '11 at 12:34

I see two important differences.

First, the predicted values (on the original scale) behave different; in loglinear least-squares they represent conditional geometric means; in the log-poisson model the represent conditional means. Since data in this type of analysis are often skewed right, the conditional geometric mean will underestimate the conditional mean.

A second difference is the implied distribution : lognormal versus poisson. This relates to the heteroskedasticity structure of the residuals : residual variance proportional to the squared expected values (lognormal) versus residual variance proportional to the expected value (Poisson).


One obvious difference is that Poisson regression will yield integers as point predictions whereas log-count linear regression can yield non-integers.

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    $\begingroup$ How does that work? Doesn't the GLM estimate expectations, which are not necessarily integral? $\endgroup$ – whuber Mar 20 '11 at 17:50
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    $\begingroup$ This is untrue. Mechanically, poisson regressions are perfectly able to handle non-integers. The standard errors won't be poisson distributed, but you can just use robust standard errors instead. $\endgroup$ – Matthew Aug 30 '18 at 16:19

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