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Context

This question uses R, but is about general statistical issues.

I'm analysing the effects of mortality factors (% mortality due to disease and parasitism) on moth population growth rate over time, where larval populations were sampled from 12 sites once a year for 8 years. The population growth rate data displays a clear but irregular cyclical trend over time.

The residuals from a simple generalised linear model (growth rate ~ %disease + %parasitism + year) displayed a similarly clear but irregular cyclical trend over time. Therefore, generalised least squares models of the same form were also fitted to the data with appropriate correlation structures to deal with the temporal autocorrelation, e.g. compound symmetry, autoregressive process order 1 and autoregressive moving average correlation structures.

Models all contained the same fixed effects, were compared using AIC, and were fitted by REML (to allow comparison of different correlation structures by AIC). I'm using the R package nlme and the gls function.

Question 1

The GLS models' residuals still display almost identical cyclical patterns when plotted against time. Will such patterns always remain, even in models that accurately account for the autocorrelation structure?

I have simulated some simplified but similar data in R below my second question, which shows the issue based on my current understanding of the methods needed to assess temporally autocorrelated patterns in model residuals, which I now know are wrong (see answer).

Question 2

I have fitted GLS models with all possible plausible correlation structures to my data, but none are actually substantially better fitting than the GLM without any correlation structure: just one GLS model is marginally better (AIC score = 1.8 lower), whilst all the rest have higher AIC values. However, this is only the case when all models are fitted by REML, not ML where GLS models are clearly much better, but I understand from stats books you must only use REML to compare models with different correlation structures and the same fixed effects for reasons I won't detail here.

Given the clearly temporally autocorrelated nature of the data, if no model is even moderately better than the simple GLM what is the most appropriate way to decide which model to use for inference, assuming I'm using an appropriate method (I eventually want to use AIC to compare different variable combinations)?

Q1 'simulation' exploring residual patterns in models with and without appropriate correlation structures

Generate simulated response variable with a cyclical effect of 'time', and a positive linear effect of 'x':

time <- 1:50
x <- sample(rep(1:25,each=2),50)
y <- rnorm(50,5,5) + (5 + 15*sin(2*pi*time/25)) + (x/1)

y should display a cyclical trend over 'time' with random variation:

plot(time,y)

And a positive linear relationship with 'x' with random variation:

plot(x,y)

Create a simple linear additive model of "y ~ time + x":

require(nlme)
m1 <- gls(y ~ time + x, method="REML")

The model displays clear cyclical patterns in the residuals when plotted against 'time', as would be expected:

plot(time, m1$residuals)

And what should be a nice, clear lack of any pattern or trend in the residuals when plotted against 'x':

plot(x, m1$residuals)

A simple model of "y ~ time + x" that includes an autoregressive correlation structure of order 1 should fit the data much better than the previous model because of the autocorrelation structure, when assessed using AIC:

m2 <- gls(y ~ time + x, correlation = corAR1(form=~time), method="REML")
AIC(m1,m2)

However, the model should still display almost identically 'temporally' autocorrelated residuals:

plot(time, m2$residuals)

Thank you very much for any advice.

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  • $\begingroup$ Your model doesn't properly capture the time dependence caused by the cycles (even for your simulated case), so your characterization of 'accurately account for' is not appropriate. The reason you still have pattern in your residuals is probably because of that. $\endgroup$ – Glen_b Dec 30 '13 at 1:48
  • $\begingroup$ I think you have it backwards. Estimates should be obtained using full maximum likelihood rather than REML. Choosing method="ML" is required for carrying out likelihood ratio tests and is necessary if you wish to use AIC to compare models with different predictors. REML provides better estimates of the variance components and standard errors than does ML. Having used method="ML" to compare different models it is sometimes recommended that the final model be refit using method="REML" and that the estimates and standard errors from the REML fit be used for the final inference. $\endgroup$ – theforestecologist Apr 11 '18 at 0:34
  • $\begingroup$ For further evidence, see the answers to this post: REML or ML to compare two mixed effects models with differing fixed effects, but with the same random effect? $\endgroup$ – theforestecologist Apr 11 '18 at 0:36
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Q1

You are doing two things wrong here. The first is a generally bad thing; don't in general delve into model objects and rip out components. Learn to use the extractor functions, in this case resid(). In this case you are getting something useful but if you had a different type of model object, such as a GLM from glm(), then mod$residuals would contain working residuals from the last IRLS iteration and are something you generally don't want!

The second thing you are doing wrong is something that has caught me out too. The residuals you extracted (and would also have extracted if you'd used resid()) are the raw or response residuals. Essentially this is the difference between the fitted values and the observed values of the response, taking into account the fixed effects terms only. These values will contain the same residual autocorrelation as that of m1 because the fixed effects (or if you prefer, the linear predictor) are the same in the two models (~ time + x).

To get residuals that include the correlation term you specified, you need the normalized residuals. You get these by doing:

resid(m1, type = "normalized")

This (and other types of residuals available) is described in ?residuals.gls:

type: an optional character string specifying the type of residuals
      to be used. If ‘"response"’, the "raw" residuals (observed -
      fitted) are used; else, if ‘"pearson"’, the standardized
      residuals (raw residuals divided by the corresponding
      standard errors) are used; else, if ‘"normalized"’, the
      normalized residuals (standardized residuals pre-multiplied
      by the inverse square-root factor of the estimated error
      correlation matrix) are used. Partial matching of arguments
      is used, so only the first character needs to be provided.
      Defaults to ‘"response"’.

By means of comparison, here are the ACFs of the raw (response) and the normalised residuals

layout(matrix(1:2))
acf(resid(m2))
acf(resid(m2, type = "normalized"))
layout(1)

enter image description here

To see why this is happening, and where the raw residuals don't include the correlation term, consider the model you fitted

$$y = \beta_0 + \beta_1 \mathrm{time} + \beta_2 \mathrm{x} + \varepsilon$$

where

$$ \varepsilon \sim N(0, \sigma^2 \Lambda) $$

and $\Lambda$ is a correlation matrix defined by an AR(1) with parameter $\hat{\rho}$ where the non-diagonal elements of the matrix are filled with values $\rho^{|d|}$, where $d$ is the positive integer separation in time units of pairs of residuals.

The raw residuals, the default returned by resid(m2) are from the linear predictor part only, hence from this bit

$$ \beta_0 + \beta_1 \mathrm{time} + \beta_2 \mathrm{x} $$

and hence they contain none of the information on the correlation term(s) included in $\Lambda$.

Q2

It seems you are trying to fit a non-linear trend with a linear function of time and account for lack of fit to the "trend" with an AR(1) (or other structures). If your data are anything like the example data you give here, I would fit a GAM to allow for a smooth function of the covariates. This model would be

$$y = \beta_0 + f_1(\mathrm{time}) + f_2(\mathrm{x}) + \varepsilon$$

and initially we'll assume the same distribution as for the GLS except that initially we'll assume that $\Lambda = \mathbf{I}$ (an identity matrix, so independent residuals). This model can be fitted using

library("mgcv")
m3 <- gam(y ~ s(time) + s(x), select = TRUE, method = "REML")

where select = TRUE applies some extra shrinkage to allow the model to remove either of the terms from the model.

This model gives

> summary(m3)

Family: gaussian 
Link function: identity 

Formula:
y ~ s(time) + s(x)

Parametric coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)  23.1532     0.7104   32.59   <2e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Approximate significance of smooth terms:
          edf Ref.df      F  p-value    
s(time) 8.041      9 26.364  < 2e-16 ***
s(x)    1.922      9  9.749 1.09e-14 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

and has smooth terms that look like this:

enter image description here

The residuals from this model are also better behaved (raw residuals)

acf(resid(m3))

enter image description here

Now a word of caution; there is an issue with smoothing time series in that the methods that decide how smooth or wiggly the functions are assumes that the data are independent. What this means in practical terms is that the smooth function of time (s(time)) could fit information that is really random autocorrelated error and not only the underlying trend. Hence you should be very careful when fitting smoothers to time series data.

There are a number of ways round this, but one way is to switch to fitting the model via gamm() which calls lme() internally and which allows you to use the correlation argument you used for the gls() model. Here is an example

mm1 <- gamm(y ~ s(time, k = 6, fx = TRUE) + s(x), select = TRUE,
            method = "REML")
mm2 <- gamm(y ~ s(time, k = 6, fx = TRUE) + s(x), select = TRUE,
            method = "REML", correlation = corAR1(form = ~ time))

Note that I have to fix the degrees of freedom for s(time) as there is an identifiability issue with these data. The model could be a wiggly s(time) and no AR(1) ($\rho = 0$) or a linear s(time) (1 degree of freedom) and a strong AR(1) ($\rho >> .5$). Hence I make an educated guess as to the complexity of the underlying trend. (Note I didn't spend much time on this dummy data, but you should look at the data and use your existing knowledge of the variability in time to determine an appropriate number of degrees of freedom for the spline.)

The model with the AR(1) does not represent a significant improvement over the model without the AR(1):

> anova(mm1$lme, mm2$lme)
        Model df      AIC      BIC    logLik   Test   L.Ratio p-value
mm1$lme     1  9 301.5986 317.4494 -141.7993                         
mm2$lme     2 10 303.4168 321.0288 -141.7084 1 vs 2 0.1817652  0.6699

If we look at the estimate for $\hat{\rho}} we see

> intervals(mm2$lme)
....

 Correlation structure:
         lower      est.     upper
Phi -0.2696671 0.0756494 0.4037265
attr(,"label")
[1] "Correlation structure:"

where Phi is what I called $\rho$. Hence, 0 is a possible value for $\rho$. The estimate is slightly larger than zero so will have negligible effect on the model fit and hence you might wish to leave it in the model if there is a strong a priori reason to assume residual autocorrelation.

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  • $\begingroup$ Thank you very much Gavin for that excellent, in-depth detailed answer. It looks like my data produces a qualitatively similar outcome with GAMs, in that there is either very little improvement or a worsening of fit (assessed via AIC/AICc) when comparing a GAM with and without the standard correlation structures. Do you/anyone know: if there are very clear, if irregular, cyclical trends in the data/residuals, would it then be most appropriate to stick with the most well fitting correlation structure rather than a model with none? Thanks again. $\endgroup$ – JupiterM104 Dec 31 '13 at 10:20
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    $\begingroup$ Coming in super late, but wanted to thank Gavin for this fantastic response. Helped me out a ton. $\endgroup$ – giraffehere Apr 13 '16 at 21:05

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