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Subject:

Re: linear contrast with only three levels in a parametric design

From:

Christian Beckmann <[log in to unmask]>

Reply-To:

FSL - FMRIB's Software Library <[log in to unmask]>

Date:

Mon, 30 Oct 2006 09:43:10 +0000

Content-Type:

text/plain

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text/plain (100 lines)

Hi,

just to add to jespers email: if you are worried about EV2 and it's  
role in the linear trend simply add the relevant contrasts for the  
quadratic trends [-1 2 -1] and [1 -2 1]. Also, it's easy to  
explicitly check for %BOLD changes within supra-thresholded clusters  
using Featquery - that way you can quantify the changes between  
conditions.
cheers
christian

On 30 Oct 2006, at 10:04, Jesper Andersson wrote:

> Hi Guido,
>
>> I am analysing some data where participants were exposed to a  
>> stimulus
>> on three different intensity levels.
>> So far,  these levels are coded in the design matrix as three  
>> seperate
>> EVs. it seems to me that it is problematic to test for a linear  
>> effect
>> across ALL three levels, by using a contrast [-1 0 1] (assuming,
>> intensity increases from EV1 to EV3). However, this is the approach
>> suggested in the last 2 paragraphs on this fls web page:
>> http://www.fmrib.ox.ac.uk/fsl/feat5/glm.html.
>> My problem with the [-1 0 1] contrast is that the result of this test
>> is totally independent of the PE for EV2. The contrast would also get
>> significant, if the activation under medium intensity stimulation is
>> lower (higher) than under minimum (maximum) intensity stimulation.  
>> [-1
>> 0 1] seems only to test, if activation increases from the minimum to
>> the maximum level.
>
> I agree that it can seem a little confusing, but it is actually  
> correct.
> I think that the important thing to realise here is that the value of
> the second beta says nothing about _linear_ effect.
>
> First of all, what the contrast weight vector does is simply to  
> create a
> linear combination of the betas. And the betas you can think of as the
> "activity" during the three different conditions.
>
> So, say for example that the betas are [1 2 3], the contrast [-1 0 1]
> would then give you 3-1=2, i.e. indicate the presence of a linear
> effect. Now say instead that the betas are [1 4 3], the contrast would
> again give you 2 and indicate a linear effect. Now this seems a little
> confusing, but if you sit down and plot the points [1 1], [2 4] and [3
> 3] (which is what we are really looking at) and then draw the best
> regression line through those points you will see that there is a  
> linear
> trend. In fact you can change the middle beta (i.e. the [2 4]  
> point) to
> any old value you like (e.g. [2 -4]) and you would still draw exactly
> the same regression line.
>
> Hence, the second point carries no information about the presence or
> absence of a linear trend.
>
> It does however carry information about a possible quadratic trend.
>
>>
>> It seems to me that in order to test for a linear effect, one would
>> need to use a different design matrix, where one  EVs codes for the
>> presence of a stimulus and a second EV codes for the intensity (e.g.,
>> with 1,2,3 for the linear assumption.). A linear effect could than be
>> tested with a contrast on the second EV. Would this be the proper way
>> to do it?
>
> You could in principle recode your design to have one ev to look for
> overall effecs and one that looks for linear trends. However, to make
> the "linear regressor" look exclusively for linear trends you need to
> orthogonalise it w.r.t. the the "overall effect regressor". So, lets
> saye we have three graded "tasks" and a baseline and lets say our two
> regressors look like [0 1 1 1 0 1 1 1] and [0 1 2 3 0 1 2 3] (i.e. a
> rather shortish experiment). Note also that the mean has been
> subtracted, which is equivalent to adding a regressor [1 1 1 1 1 1  
> 1 1].
> These three regressors are now quite correlated (e.g. the "linear"
> regressor can be used to model parts of the "overall" effect).  
> However,
> the orthogonalised version of the linear regressor is [0 -1 0 1 0 -1 0
> 1]. I.e., the scans under "condition 2" are again (correctly) ignored.
>
> Just for accuracy I should mention that even if you did not explicitly
> perform the orthogonalisation any t/F-test you performed would still
> pertain to the orthogonalised version (though this is slightly less
> intuitive).
>
> Good luck Jesper

--
  Christian F. Beckmann
  Oxford University Centre for Functional
  Magnetic Resonance Imaging of the Brain,
  John Radcliffe Hospital, Headington, Oxford OX3 9DU, UK
  Email: [log in to unmask] - http://www.fmrib.ox.ac.uk/~beckmann/
  Phone: +44(0)1865 222551 Fax: +44(0)1865 222717

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