Hi Chris, I think possibly there's confusion because you can't view things
in this way - you can't necessarily just mutiply the EVs by the contrasts,
to get "new EVs" (in the way that you're doing) and interpret things in a
simple way - that's not how the maths works. Instead, think of the
underlying multiple regression model:
Let's replace each group of 5 data/model rows with a single row; this
doesn't change the logic. Also, let's replace all the separate subject
means with a single all-mean EV. So you have the model:
|A| | 1| | 1| |1|
|B| = a*|-1| + b*| 0| + c*|1|
|C| | 0| |-1| |1|
in other words:
A = a+b+c
B = -a+c
C = -b+c
hence A-B=2a+b etc.....
Hope this is clearer? Sorry if not! Cheers, Steve.
On Sat, 10 Sep 2005, Chris Kelly wrote:
> Thanks for the quick response, Steve. Your explanation makes sense to me at face value; maybe
> the issue I'm getting hung up on is how you can model B without considering A or C, and vice
> versa.
>
> If you combine EV's 1 and 2, as you have done, to model A, then you get [2 2 2 2 2 -1 -1 -1 -1 -1
> -1 -1 -1 -1 -1]. This model says to me: "Where is there A significantly greater than B and C?" and
> so it makes sense that A, thus, is a+b. But then if you take B to be -a, you have a model that looks
> like [-1 -1 -1 -1 -1 1 1 1 1 1 0 0 0 0 0]; indeed, there are ones for the second set of five trials,
> but I don't see how this considers the third set of trials, since those are set to 0. Wouldn't the
> correct model be, as above, [-1 -1 -1 -1 -1 2 2 2 2 2 -1 -1 -1 -1 -1]?
>
> Sorry if this is all obvious, but I want to do this analysis and would really like to understand.
>
>
>
> >OK - what we're saying is that, due to the subject-wise demeaning, the
> >responses to the three conditions A, B and C must add up to zero - i.e.
> >have zero mean across the different conditions. In the example, the first
> >5 scans on average contain the level of condition A in the data (relative
> >to this zero mean across conditions), the next 5 scans comprise condition
> >B and the final five are condition C.
> >
> >Hence the first 2 EVs, which both have value 1 for the first 5 scans,
> >between them model condition A: so after estimating the PEs, if we define
> >PE1=a and PE2=b, the modelled data for those first 5 scans is 1*a + 1*b,
> >i.e., A=a+b.
> >
> >Likewise, for the next five scans, EV2 is 0, and EV1 is -1, hence the
> >modelling of the data is: B=-a. Likewise, C=-b.
> >
> >Hopefully this clarifies things?
> >
> >> Also, I'm not sure why the following setup would not work:
> >>
> >> EV1=[1 1 1 1 1 0 0 0 0 0 0 0 0 0 0]
> >> EV2=[0 0 0 0 0 1 1 1 1 1 0 0 0 0 0]
> >
> >This would not work because there is now nothing in the model to model
> >condition C, hence there can't be a good model fit!
> >
> >Cheers, Steve.
> >
> >--
> > Stephen M. Smith DPhil
> > Associate Director, FMRIB and Analysis Research Coordinator
> >
> > Oxford University Centre for Functional MRI of the Brain
> > John Radcliffe Hospital, Headington, Oxford OX3 9DU, UK
> > +44 (0) 1865 222726 (fax 222717)
> >
> > [log in to unmask] http://www.fmrib.ox.ac.uk/~steve
> >===========================================================
> ==============
>
--
Stephen M. Smith DPhil
Associate Director, FMRIB and Analysis Research Coordinator
Oxford University Centre for Functional MRI of the Brain
John Radcliffe Hospital, Headington, Oxford OX3 9DU, UK
+44 (0) 1865 222726 (fax 222717)
[log in to unmask] http://www.fmrib.ox.ac.uk/~steve
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