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I have been given a data set that contains the number of awards earned by students at one high school where predictors of the number of awards earned include the type of program in which the student was enrolled and the score on their final exam in maths.

I was wondering if anyone could tell me why a linear regression model may be unsuitable in this instance and why it would be better to use a Poisson regression? Thanks.

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4 Answers 4

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Three points about Poisson vs Normal regression, all concerning model specification:

Effect of changes in predictors

With a continuous predictor like math test score Poisson regression (with the usual log link) implies that a unit change in the predictor leads to a percentage change in the number of awards, i.e. 10 more points on the math test is associated with e.g. 25 percent more awards. This depends on the number of awards the student is already predicted to have. In contrast, Normal regression associates 10 more points with a fixed amount, say 3 more awards under all circumstances. You should be happy with that assumption before using the model that makes it. (fwiw I think it is very reasonable, modulo the next point.)

Dealing with students with no awards

Unless there are really many awards spread over lots of students then your award counts will mostly be be rather low. In fact I would predict zero-inflation, i.e. most students don't get any award, so lots of zeros, and some good students get quite a few awards. This messes with the assumptions of the Poisson model and is at least as bad for the Normal model.

If you have a decent amount of data a 'zero-inflated' or 'hurdle' model would then be natural. This is two models tied together: one to predict whether the student gets any awards, and another to predict how many she gets if she gets any at all (usually some form of Poisson model). I would expect all the action to be in the first model.

Award exclusivity

Finally, a small point about awards. If awards are exclusive, i.e. if one student gets the award then no other students can get the award, then your outcomes are coupled; one count for student a pushes down the possible count of every other one. Whether this is worth worrying about depends on the awards structure and the size of the student population. I'd ignore it at a first pass.

In conclusion, Poisson comfortably dominates Normal except for very large counts, but check the assumptions of the Poisson before leaning on it to heavily for inference, and be prepared to move to a mildly more complex model class if necessary.

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Poisson regression would be more suitible in this case because your response is the count of something.

Putting things simply, we model that the distribution of number of awards for an individual student comes from a poisson distribution, and that each student has their own $\lambda$ poisson parameter. The Poisson regression then relates this parameter to the explanatory variables, rather than the count.

The reason this is better than normal linear regression is to do with the errors. If our model is correct, and each student has their own $\lambda$, then for a given $\lambda$ we would expect a poisson distribution of counts around it - i.e. an asymmetric distribution. This means unusually high values are not as surprising as unusually low.

Normal linear regression assumes normal errors around the mean, and hence equally weights them. This says that if a student has an expected number of awards of 1, it is just as likely for them to receive -2 awards as for them to receive 3 awards: this is clearly nonsense and what poisson is built to address.

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Ordinary least-squares regression of awards on predictors will yield consistent parameter estimates as long as the conditional mean of awards is linear in the predictors. But this is often inadequate since it allows the predicted number of awards to be negative (even for "reasonable" values of predictors), which makes no sense. Folks will often try to remedy this by taking the natural log of awards and using OLS. But this fails since some students receive no awards, so then you have to use something like $\ln(awards+0.5)$, but this creates its own problems since you presumably care about awards, and the re-transformation is non-trivial.

Also, as the expected number of awards becomes very large, OLS should perform better for reasons outlined by @Corone. In Lake Wobegon, OLS is the way to go.

If the expected number is low, with lots of zeros, I would use the Poisson with robust standard errors over the negative binomial model. NB regression makes a strong assumptions about the variance that appear in the first-order conditions that produce the coefficients. If these assumptions are not satisfied, the coefficients themselves could be contaminated. That is not the case with the Poisson.

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@corone raises good points, but note that the Poisson is only really asymmetric when $\lambda$ is small. Even for $\lambda$ = 10, it is pretty symmetric e..g.

set.seed(12345)
pois10 <- rpois(1000, 10)
plot(density(pois10))
library(moments)
skewness(pois10)

shows a skewness of 0.31, which is pretty close to 0.

I also like @conjugateprior 's points. In my experience, it is rare for Poisson regression to fit well; I usually wind up using either a negative binomial or a zero-inflated model.

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