# Consequences of violating assumptions of nonlinear regression when comparing models and/or datasets

I have a question about the consequences of using non-linear regression when the data violate the assumptions of (1) homoscedasticity and (2) normal distribution. Specifically, I am wondering about how it affects model comparison and the comparison of two data sets with one model. That is, how does it affect the validity and/or strength of those comparisons?

To give the question a little context, there are many instances in agronomy (my particular field) where the data are always heteroscedastic and non-normal--especially when they approach some sort of natural limit. It is nevertheless very common to see nonlinear regressions fit through these kind of data and then see subsequent model and/or data set comparison.

Here is an imaginary example where weed biomass influences crop yield loss.

Issues of heteroscedasticity. The uncertainty is clearly and regularly related to the predictor value. As we increase weed biomass there is much more certainty in the crop yield loss values than there is at lower levels of weed biomass.

Issues of non-normality. Additionally, at the higher levels of weed biomass, where the crop yield loss approaches but cannot exceed 100%, the data is not distributed normally. Rather, it's something of a truncated normal distribution, where half of the distribution is missing (in this case, the half that would go above 100%).

So, what does this mean for further analysis in terms of model and/or data set comparison?

The former is often done in agronomy to determine the better model for prediction purposes (e.g., which model is better for predicting crop response?). The latter is often done to evaluate the effect of certain treatments on a relationship (e.g., does fertilization affect the relationship between weed biomass and crop yield loss?).

Can we proceed with these kinds of comparisons despite the data inherently and always violating two of the major assumptions of nonlinear regression? If not, why not and what might be an alternative approach to answering the aforementioned kind of research questions?

Thanks so much,

Angela

Just for reference, here is the code used to generate the figure.

### Simulated data set (idealized) ###
set.seed(123)
density<-as.numeric(seq(0,1000,10)) # hypothetical range of predictor variables
error <- rnorm(n = length(density), mean = 0, sd = 20) # normally distributed errors
A = 100 # asymptote parameter = 100%
I = 1 # slope parameter = 1
YL = I*density/(1+(I/A)*density) + error # yield loss + random error
plot(density, YL) # idealized but unrealistic data for this kind of study

### Modified data set (more realistic) ###
# That is ...
# (1) a maximum yield loss of 100% and
# (2) decreased variability at higher density values
temp.logical <- density >= 500 & YL <= 95
new.error <- rnorm(n = length(temp.logical[temp.logical==TRUE]), mean = 0, sd = 5)
realistic.YL<-ifelse(density >= 500 & YL <= 100, 100 - abs(new.error),
ifelse(YL >= 100, 100, YL))
plot(density, realistic.YL,
xlab = expression(Weed ~ biomass ~ (g ~ m^{-2})),
ylab = "Crop yield loss (%)") # more realistic dataset

### Fitting a rectangular hyperbola to the modified data set ###
mod.1 <- nls2(realistic.YL ~ I*density/(1+(I/A)*density),
start = list (I= 1, A=100),
trace = T)
summary(mod.1)
I<-summary(mod.1)$coefficients[1,1] A<-summary(mod.1)$coefficients[2,1]
pred.YL = I*density/(1+(I/A)*density)
lines(pred.YL~dens)

• The normality assumption applies to the distribution of residuals not to the data distribution – Alexis Apr 15 '15 at 17:20
• Oh goodness, thank you. I will edit the post to confine my concern to the heteroscedasticity issue. – Angela Apr 15 '15 at 17:34
• Wait, though. From Motulsky and Christopoulos (2003), under the assumptions for nonlinear regression on p. 30 - "The variability of Y values at any particular X value follows a known distribution. Almost always, this is assumed to be a Gaussian (normal) bell-shaped distribution." That refers to the data distribution, doesn't it? – Angela Apr 15 '15 at 17:49
• Angela, no: that refers to the variability of $Y$ about the regression curve, not to the values of $Y$ independent of $Y$ as a function of $X$. – Alexis Apr 15 '15 at 18:43
• I think I see what you're saying. In which case, in my example, I would need to demonstrate the issue of non-normality of Y at each particular value of X by having multiple observations at each X value. Since that would be unrealistic in this particular example though (having precisely the same amount of weed biomass in different plots, that is), I thought perhaps one could infer that the data would not be normal about the regression curve given the data eventually collecting at the upper limit of 100%. Would that be a fair thing to conclude? Thanks for the assistance/clarification! – Angela Apr 16 '15 at 15:00

## 1 Answer

Normality assumption is not necessary for nonlinear regression. It is often used because it's convenient. However, if it's clearly violated then I wouldn't use it. The same with heteroscedasticity.

In your example the dependent variable seems to be confined between 0 and 100%. You could still use normal distributions and homoscedasticity if the data were "far" from the bounds. However, you show the sample where data spans all range, with substantial portion clustered by the borders. In this case neither homoscedasticity nor normality seems like reasonable assumptions.

• Thanks for the response! Besides wondering whether or not I should engage in this kind of analysis though, I am also curious as to how these violations of assumptions should affect our interpretation of the results of others. That is, if a researcher concludes that model A is better than model B or treatment X has an effect on Y on the basis of models that violate these assumptions, are their conclusions still valid or is the strength of their evidence changed? – Angela Apr 16 '15 at 15:16
• The distributional assumptions impact the confidence bands first. However, they also may lead to bias in the estimates when using something like MLE. So, if the assumptions are clearly badly violated then at the very least it would be hard to make statements about relationships between independent and dependent variables, for instance. – Aksakal Apr 16 '15 at 16:08
• That makes sense. So it would be really quite inappropriate to use the confidence bands on a model fitted under these circumstances. I wonder though if it might be valid to still use the residual sum of squares for interpretation. For example, it seems that it would still allow us to tell which model is more predictive? Appreciate your help! – Angela Apr 17 '15 at 14:17
• Yes, sums of squares will work, of course. The trouble is with interpreting them. Less is better, but you'll have to figure out where the differences are significant to judge the goodness of models. On any given sample you'll have random sums of squares, so if the differences between two models are "small" it's hard to tell whether it's random or the model is better than the other. When you know the distributions you'll have some statistical tools to help you separate. In your case using normality assumption these tools will be useless. – Aksakal Apr 17 '15 at 14:29
• Forgive me my ignorance, but does the failure to meet the normality assumption also render F-tests useless in comparing models? In looking at the calculation for F-values, I see that it only consists of residual sums of squares and degrees of freedom--which makes it seem (at least on the surface) that it doesn't rely on the normality assumption. If it does though and F-tests are inappropriate, that would be a good thing to know. – Angela Apr 17 '15 at 15:07