[R] comparing two regression models with different dependent variable

[Joris Meys]

As I explained, you cannot just test two models with different
dependent variables. You can model the difference you calculated, if
you think you can give a sensible interpretation to it. But be aware
of the fact that your model will tell you something about the relation
between your predictors and the calculated difference between the
predictor variables, not about the difference between both models. So
the hypothesis you test is whether or not the difference between the
learning vector of reward and the learning vector of punishment can be
explained by any of your predictor variables.

If I understand it right, the setup is as follows :
two groups of participants are given a certain task, and have to do
that repeatedly. One group gets a punishment for a mistake, the other
one a reward for a good answer. the response variable is the
difference between the correct answers before and after the learning
process.

If this is the case, the measurements for punishment and reward are
comparable. Then just combine both datasets, add a factor indicating
whether the score comes from reward or from punishment, take into
account the design (repeated measures, blocks, cross-over, ...), and
then you can test a multitude of hypotheses, eg :
1) if there is a main effect of your predictor variables
2) if there is a significant difference between the scores obtained by
reward and those obtained by punishment (which is the main effect of
the factor you constructed)
3) if the influence of your predictor variables is different for
scores obtained by reward or by punishment (which is tested by
interaction terms)

If you can't get it done with this information, please do consult a
statistician in your proximity. Statistics is not a detail, it's a
huge set of complex tools that require a thorough understanding of the
underlying processes. Don't expect to get some magical
one-size-fits-all answer.

Cheers
Joris

On Thu, Jun 10, 2010 at 11:39 AM, Or Duek <orduek at gmail.com> wrote:
> I'll try to add some more information regarding my experiment - maybe that
> would help clear things out.
> Instead of actually measuring the learning curve (i.e. number of correct
> responses per block) I created a variable that substract the number of
> correct answers from the last block with that of the first block.
> I did the same thing for reward and punishment.
> I also use the same predictors in both regression models.
> Until now I just created a new variable - learning vector of reward minus
> learning vector of punishment.
> By that I think I measure the difference.
> I just wanted to know if there's another option to compare a model with same
> predictors but different dependent variable.
>
>
> On Thu, Jun 10, 2010 at 12:33 PM, Joris Meys <jorismeys at gmail.com> wrote:
>>
>> This is only valid in case your X matrix is exactly the same, thus
>> when you have an experiment with multiple response variables (i.e.
>> paired response data). When the data for both models come from a
>> different experiment, it ends here.
>>
>> You also assume that y1 and y2 are measured in the same scale, and can
>> be substracted. If you take two models, one with response Y in meters
>> and one with response Y in centimeters, all others equal, your method
>> will find the models "significantly different"  whereas they are
>> exactly the same except for a scaling parameter. If we're talking two
>> different responses, the substraction of both responses doesn't even
>> make sense.
>>
>> The hypothesis you test is whether there is a significant relation
>> between your predictors and the difference of the "reward" response
>> and the "punishment" response. If that is the hypothesis of interest,
>> the difference can be interpreted in a sensible way, AND both the
>> reward learning curve and the punishment learning curve are measured
>> simultaneously for every participant in the study, you can
>> intrinsically compare both models by modelling the difference of the
>> response variable.
>>
>> As this is not the case (learning curves from punishment and reward
>> can never be made up simultaneously), your approach is invalid.
>>
>> Cheers
>> Joris
>>
>> On Thu, Jun 10, 2010 at 9:00 AM, Gabor Grothendieck
>> <ggrothendieck at gmail.com> wrote:
>> > We need to define what it means for these models to be the same or
>> > different.  With the usual lm assumptions suppose for i=1, 2 (the two
>> > models) that:
>> >
>> > y1 = a1 + X b1 + error1
>> > y2 = a2 + X b2 + error2
>> >
>> > which implies the following which also satisfies the usual lm
>> > assumptions:
>> >
>> > y1-y2 = (a1-a2) + X(b1-b2) + error
>> >
>> > Here X is a matrix, a1 and a2 are scalars and all other elements are
>> > vectors.  We say the models are the "same" if b1=b2 (but allow the
>> > intercepts to differ even if the models are the "same").
>> >
>> > If y1 and y2 are as in the built in anscombe data frame and x3 and x4
>> > are the x variables, i.e. columns of X, then:
>> >
>> >> fm1 <- lm(y1 - y2 ~ x3 + x4, anscombe)
>> >> # this model reduces to the following if b1 = b2
>> >> fm0 <- lm(y1 - y2 ~ 1, anscombe)
>> >> anova(fm0, fm1)
>> > Analysis of Variance Table
>> >
>> > Model 1: y1 - y2 ~ 1
>> > Model 2: y1 - y2 ~ x3 + x4
>> >  Res.Df    RSS Df Sum of Sq      F Pr(>F)
>> > 1     10 20.637
>> > 2      8 18.662  2    1.9751 0.4233 0.6687
>> >
>> > so we cannot reject the hypothesis that the models are the "same".
>> >
>> >
>> > On Wed, Jun 9, 2010 at 11:19 AM, Or Duek <orduek at gmail.com> wrote:
>> >> Hi,
>> >> I would like to compare to regression models - each model has a
>> >> different
>> >> dependent variable.
>> >> The first model uses a number that represents the learning curve for
>> >> reward.
>> >> The second model uses a number that represents the learning curve from
>> >> punishment stimuli.
>> >> The first model is significant and the second isn't.
>> >> I want to compare those two models and show that they are significantly
>> >> different.
>> >> How can I do that?
>> >> Thank you.
>> >>
>> >>        [[alternative HTML version deleted]]
>> >>
>> >> ______________________________________________
>> >> R-help at r-project.org mailing list
>> >> https://stat.ethz.ch/mailman/listinfo/r-help
>> >> PLEASE do read the posting guide
>> >> http://www.R-project.org/posting-guide.html
>> >> and provide commented, minimal, self-contained, reproducible code.
>> >>
>> >
>> > ______________________________________________
>> > R-help at r-project.org mailing list
>> > https://stat.ethz.ch/mailman/listinfo/r-help
>> > PLEASE do read the posting guide
>> > http://www.R-project.org/posting-guide.html
>> > and provide commented, minimal, self-contained, reproducible code.
>> >
>>
>>
>>
>> --
>> Joris Meys
>> Statistical consultant
>>
>> Ghent University
>> Faculty of Bioscience Engineering
>> Department of Applied mathematics, biometrics and process control
>>
>> tel : +32 9 264 59 87
>> Joris.Meys at Ugent.be
>> -------------------------------
>> Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php
>
>



-- 
Joris Meys
Statistical consultant

Ghent University
Faculty of Bioscience Engineering
Department of Applied mathematics, biometrics and process control

tel : +32 9 264 59 87
Joris.Meys at Ugent.be
-------------------------------
Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php