# Diagnostics for quasipoisson glm for continuous data

I'm a little confused about how to use the quasipoisson family in the glm function. It was recommended by someone that I use it for my analysis, even though the data are continuous - and as such, I don't get the same warning messages (regarding the fact that my data are non-discrete) as I do when I use it for the Poisson family. Why is this? Is the quasipoisson family actually compatible with continuous data? I thought Gamma might be more suitable for a continuous dataset, but about 50% of my values are zeroes (this was the original reason quasipoisson was recommended).

I don't quite understand all of the information I've been given about lm's/glm's in general - do they assume anything about the distribution of y (the response variable), or just the distribution of the residuals, when a model has already come up with fitted values?

As I said earlier, I was given the typical advice of 'if the assumptions for the Poisson family aren't met, use quasipoisson.' ...but how can I test if the assumptions of the quasipoisson distribution are met (or Gamma, if I just add one to everything)?

I can't use a Kolmogorov-Smirnov test, as the function does not like repeated values (as I said, about half of the y-values are zero).

Would you suggest I just give up and use nonparametric stats? My project is already overdue and I have lots of other work to do!

I'm ok with sharing my data if that helps

From what you stated, I think you may have been misled.

In a standard Poisson GLM, you are assuming that your data follow

$Y_i \sim Pois(\lambda_i)$

where $\lambda_i = \exp(X^T \beta)$

$X_i$ are the covariates for subject $i$ and $\beta$ are the coefficients you are interested in estimating.

This builds many assumptions into your model. One (and only one of many), is that $Var(Y_i) = E(Y_i)$. This assumption is relaxed in a quasi-Poisson model; it is rather assumed that $Var(Y_i) = \alpha E(Y_i)$, where $\alpha$ is a parameter to be estimated. This allows for over dispersion when $\alpha > 1$, i.e. more variance than allowed by the exact Poisson case. It's important to note that you're still assuming the variance is proportional to the mean (but of course that's less restrictive than being equal to the mean).

This does not directly answer your issue of being zero-inflated, although it is probably a slight improvement compared with the standard Poisson GLM. As for what I think you should do, it is hard to say without knowing more about your data. If it's the case that you're just comparing response variables in different groups (as opposed to across different different levels of a continuous variable), then I would certainly endorse simple non-parametric methods rather than trying to check all the assumptions of different potential parametric models.

• My study is trying to determine if there is more common ivy biomass on some species of tree than others, but I want to factor in other things, such as location and tree circumference, because these are confounding variables (ie if a certain species of tree is more common at one site, which happens to have higher ivy abundance on average, a non-parametric test could easily fool you into thinking that this is a direct affect of the species of tree). What are the other assumptions and is there any way to test that they're being met, or do you need an a priori reason to use a specific glm family?
– Viv
Nov 21, 2015 at 20:29
• I got a lot of zeroes, because so many trees had no ivy at all, but I think removing zeroes would remove meaning... what if 2 trees of a species had the highest ivy biomass in the whole dataset, but absolutely none on the rest? How numerous are the assumptions? My supervisor for the project (I'm an undergrad) showed me some normal qq plots, which of course show that my data aren't normal, but I can't confirm that they ARE something else...
– Viv
Nov 21, 2015 at 20:32
• What percentage of your data has 0 biomass? Personally, I might be inclined to break in up into two questions: first, 0,1 response for whether the tree had any biomass (adjusting for covariates with logistic regression), and then perhaps looking conditional on them having biomass, does log(biomass) change given our covariates (i.e. tree type, size, etc.). This could be done by log transforming the biomass and then using linear regression on only the observations with positive biomass. Nov 21, 2015 at 20:36
• One idea I had was to form the most complex model I could think of - an overfitting model - and find out the residual deviance for the model for each distribution, then take the distribution that gave the lowest score, and use AIC or BIC to cut the model down. Do you think that would work?
– Viv
Nov 21, 2015 at 20:37
• @Viv: Personally, I do not like the overfitting and then reducing model you just mentioned. The reason for this is that these results are extremely unlikely to be reproducible; if you reran the experiment and applied the same procedure, you are very likely to get a totally different parametric model, set of covariates, etc. It also greatly introduces bias and inflates type I error rates. Most statisticians do not like these types of model building strategies for inference (but sadly they still do appear in applied literature). Nov 21, 2015 at 20:41