# [R-sig-ME] Assessing linearity

Andrew Dolman andydolman at gmail.com
Tue Oct 26 15:18:01 CEST 2010

```Hi Mike,

I don't see why you would want to fit such a complicated trend to your
data. If you need anything more than a cubic then just treat grade as
a factor. I would start by fitting a model with natural splines ns(),
not bs() with 1 and 2 degrees of freedom and then see whether your 2
degree model fits better - this would be a reasonable test for
non-linearity in grade. If you expect the effect of grade to jump
around all over the place then treating it as a factor if probably
more appropriate. Even a cubic is probably overkill, do you expect
performance to ever decline with grade?

andydolman at gmail.com

On 25 October 2010 20:15, Mike Lawrence <Mike.Lawrence at dal.ca> wrote:
> Thanks Andrew. Jarrod mentioned using splines as well, and I do think
> they sound promising since I indeed do not have strong a priori
> reasons to expect polynomial shaped trends.
>
> I'm uncertain, however, how to choose values for the "degree" and "df"
> arguments to the bs. I see that df has to be equal-to-or greater than
> degree, and less than the number of unique values in grade minus 2
> (minus one makes it equivalent to factorizing grade, which I'm doing
> already). I also see that degree=1 and df=1 is equivalent to the
> linear case, which I'm already estimating in the original lmer fit.
>
> However, that still leaves a large parameter space (df at degree:
> 2 at 1;3 at 1,4 at 1, 2 at 2, 3 at 2; 4 at 2; 3 at 3; 4 at 3; 4 at 4), and I'm not sure it's
> appropriate to compare the linear case to fits from every point in
> this space. Thoughts?
>
> Mike
>
>
> On Mon, Oct 25, 2010 at 12:59 PM, Andrew Dolman <andydolman at gmail.com> wrote:
>> Hi Mike,
>>
>> Couple of things.
>>
>> poly() is used like this
>>
>> a <- rnorm(100)
>> b <- rnorm(100)
>>
>> m2 <- lm(a~poly(b,2))
>> summary(m2)
>>
>> To compare a model with a linear predictor and a quadratic polynomial
>> you can do this:
>> m0 <- lm(a~1)
>> m1 <- lm(a~poly(b,1))
>> m2 <- lm(a~poly(b,2))
>> anova(m1,m2)
>>
>> In the case of likelihood fitted models you'll get a likelihood ratio test.
>>
>>
>> You wouldn't normally fit a quadratic term instead of the linear term,
>> rather in addition, so m2 has 1 extra parameter. You can compare
>> multiple models at once with anova(m0,m1,m2), or you can go the AIC
>> route and calculate deltaAIC for your set of candidate models.
>>
>> Unless you expect the non-linearity to be polynomial shaped you might
>> be better off fitting splines with varying number of knots/df.
>>
>>
>>
>> andydolman at gmail.com
>>
>>
>>
>> On 24 October 2010 18:43, Mike Lawrence <Mike.Lawrence at dal.ca> wrote:
>>> Ah, this looks promising! So how does this sound:
>>>
>>> I typically assess the evidence for a relationship between the
>>> predictor and response variables by comparing the AIC values for a
>>> model including the predictor to a model without it. In the case of
>>>
>>> fit_null = lmer(
>>>    formula = response ~ (1|individual)
>>>    data = my_data
>>> )
>>> fit_null_AIC = AIC(fit_null)
>>>
>>> fit_alt = lmer(
>>>    formula = response ~ (1|individual) + grade_as_numeric
>>>    data = my_data
>>> )
>>> fit_alt_AIC = AIC(fit_alt)
>>>
>>> grade_loglikratio = fit_null_AIC - fit_alt_AIC
>>>
>>> Now, if I wanted to check whether there is a quadratic component to
>>> the grade effect, I'd first compute an analogous likelihood ratio for
>>> the quadratic fit compared to the null:
>>>    formula = response ~ (1|individual) + poly(grade_as_numeric)^2
>>>    data = my_data
>>> )
>>>
>>> Then compute a final log likelihood ratio between the improvement over
>>> the null caused by grade versus the improvement over the null caused
>>>
>>>
>>> I could repeat this for higher-order polynomials of grade, each
>>> compared to the order directly below it, to develop a series of
>>> likelihoood ratios that describe the relative improvement of the fit.
>>>
>>> Does this sound appropriate?
>>>
>>> Cheers,
>>>
>>> Mike
>>>
>>> --
>>> Mike Lawrence
>>> Department of Psychology
>>> Dalhousie University
>>>
>>> Looking to arrange a meeting? Check my public calendar:
>>> http://tr.im/mikes_public_calendar
>>>
>>> ~ Certainty is folly... I think. ~
>>>
>>>
>>>
>>> On Sun, Oct 24, 2010 at 6:41 AM, Jarrod Hadfield <j.hadfield at ed.ac.uk> wrote:
>>>> Hi Mike,
>>>>
>>>> You would be better off trying out something like polynomials or splines,
>>>> For example:
>>>>
>>>>  fit1 = lmer(
>>>>     formula = response ~ (1|individual)+poly(grade_as_numeric,n),
>>>>     , data = my_data
>>>>     , family = gaussian
>>>>  )
>>>>
>>>> where n is the order of the polynomial. n=1 would fit the same model as your
>>>> original fit1, although the covariate (and the regression parameter) would
>>>> be scaled by some number. When n=6 the model would be a reparameterised
>>>> version of your model fit2. When 1<n<6 you would be working with a
>>>> non-linear relationship in between these two extremes, although the model is
>>>> still linear in the parameters.
>>>>
>>>> Cheers,
>>>>
>>>> Jarrod
>>>>
>>>>
>>>>
>>>>
>>>>
>>>>
>>>>
>>>>
>>>> Quoting Mike Lawrence <Mike.Lawrence at dal.ca>:
>>>>
>>>>> Hi folks,
>>>>>
>>>>> I have developmental data collected across several grades (1-6). I
>>>>> would like to be able to assess whether there are any linear or
>>>>> non-linear trends across grade. Does it make sense to run a first lmer
>>>>> treating grade as continuous, obtain the residuals, then run a second
>>>>> lmer treating grade as a factor? That is:
>>>>>
>>>>> fit1 = lmer(
>>>>>    formula = response ~ (1|individual)+grade_as_numeric
>>>>>    , data = my_data
>>>>>    , family = gaussian
>>>>> )
>>>>> my_data\$resid = residuals(fit1)
>>>>> fit2 = lmer(
>>>>>    formula = resid ~ (1|individual)+grade_as_factor
>>>>>    , data = my_data
>>>>>    , family = gaussian
>>>>> )
>>>>>
>>>>>
>>>>> As I understand it, fit1 will tell me if there are any linear trends
>>>>> in the data, while fit2 will tell me if there are any non-linear
>>>>> trends in the data in addition to the linear trends obtained in fit1.
>>>>>
>>>>> If this is sensible, how might I apply it to a second binomial
>>>>> response variable given that the residuals from a binomial model are
>>>>> not 0/1?
>>>>>
>>>>> Cheers,
>>>>>
>>>>> Mike
>>>>>
>>>>> --
>>>>> Mike Lawrence
>>>>> Department of Psychology
>>>>> Dalhousie University
>>>>>
>>>>> Looking to arrange a meeting? Check my public calendar:
>>>>> http://tr.im/mikes_public_calendar
>>>>>
>>>>> ~ Certainty is folly... I think. ~
>>>>>
>>>>> _______________________________________________
>>>>> R-sig-mixed-models at r-project.org mailing list
>>>>> https://stat.ethz.ch/mailman/listinfo/r-sig-mixed-models
>>>>>
>>>>>
>>>>
>>>>
>>>>
>>>> --
>>>> The University of Edinburgh is a charitable body, registered in
>>>> Scotland, with registration number SC005336.
>>>>
>>>> _______________________________________________
>>>> R-sig-mixed-models at r-project.org mailing list
>>>> https://stat.ethz.ch/mailman/listinfo/r-sig-mixed-models
>>>>
>>>
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>>
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>

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