While looking for a link function for my model I realized I cannot find a good fit for the distribution of my Y (see fig. below). It is the distribution of number of offspring for a given season, the mode is 3 and so poisson and negbinomial tend to underestimate the bulk of the data. I would be curious to know which distribution it follows!

(I am aware that the distribution of the residuals is what matters for the interpretation)

Mean=3.3 SD= 2 N=82

I ran models with poisson (and zero inflated), negbinomial(and zero inflated) and skew_normal. Other solutions?

enter image description here

EDIT: To show what it made me ponder about the distribution of my variable I add 2 diagnostic plots to show why I do not think the fitting (in this case from a zero inflated negative binomial) is very good and the model tends to overestimate the number of 1s and 2s. My guess is because of the low number of 1s and 2s in my Y variable, which made me realize it was a distribution I had never seen..and wondered if someone could name it.

I am trying to find a link function for my model. The residuals for the ones I tried are not normally distributed

error histograms fitting enter image description here

  • 5
    $\begingroup$ Can you recreate your histogram with breaks=seq(-0.5,9.5)? That would be more informative. $\endgroup$ Commented May 2, 2019 at 15:02
  • 3
    $\begingroup$ What problem are you trying to solve? How does fitting a density to this data help you solve that problem? $\endgroup$
    – Sycorax
    Commented May 2, 2019 at 15:43
  • 1
    $\begingroup$ Zero inflated negative binomial. $\endgroup$
    – usεr11852
    Commented May 2, 2019 at 17:21
  • 3
    $\begingroup$ Also to state the obvious: Avoid dividing by the max and using beta regression. There are counts not continuous proportions. :) For the record, I quickly tried countreg::hurdle with intercept only and I got an mean of ~ 3.3 so it is spot on. I cannot understand what is meant by "shifted towards 1s and 2s". $\endgroup$
    – usεr11852
    Commented May 2, 2019 at 17:53
  • 1
    $\begingroup$ multinomial distribution. $\endgroup$
    – user158565
    Commented May 3, 2019 at 3:01

1 Answer 1

  1. The distribution of the outcome variable is not really the point.

  2. The distribution of the residuals is more relevant.

  3. It is not clear how you have obtained the histograms of residuals, but as I see them, they do not depart drastically from normality.

  4. Even if the residuals are not plausibly normally distributed, this is not necessary a problem, particularly if the model is to be used for prediction rather than inference.

  5. Even if inference is the goal, non-normality of residuals can still not be a problem - normality of residuals is one of the least important assumptions/conditions of regression models.

  6. Without details of how these data arise, the design of the experiment, study, survey etc. and the research question(s) it is very hard to give specific advice.

  • $\begingroup$ Thanks for your answer. I am well aware that the residuals are what matters (see comments) and that overall the fitting is not too bad (but the fitting of my model is not the point of my question, I put the residual plots just to better illustrate how I started to reason about my question). My question is about the distribution of my variable I realized that it did not fit any distribution I know and I was wondering (for curiosity) if it had a name. Also, could you give me reference for your point 5? I would be interested to read more about it. $\endgroup$
    – have fun
    Commented May 2, 2019 at 21:22
  • $\begingroup$ As i said in my point 6, it is very hard to give any advice without more information. As for point 5, you could look up robust standard errors. Also check this. A lot will depend on the reasons for non-normality. $\endgroup$ Commented May 2, 2019 at 21:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.