Hi Steve, thanks for the answer and I'll have a look at the suggested
chapter.
I'm familiar enough with matlab and would be understand the varcope
from the matlab code you gave
varcope(i)=c'*pinvX*pinvX'*c*sigsq;
Could you just short let me know what is c (beta Values?), what is X
(Model or Data?) and sigsq I assume is the variance (is it the
variance of the data?) ?
Thanks, Michael
On 15-Jul-09, at 12:15 AM, Steve Smith wrote:
> Hi - for the GLM equations for contrasts I recommend you read (for
> example) Keith Worsley's chapter in our FMRI book.
> In matlab-speak, the equation is
> varcope(i)=c'*pinvX*pinvX'*c*sigsq;
> Cheers.
>
> On 14 Jul 2009, at 03:22, Michael Scheel wrote:
>
>> Hi Jesper, thanks for your answer. A short follow up question. I'm
>> still a bit unsure about the Variance of a contrast. I know that
>> Variance is the Sum of the Squared Deviations from the Mean. However
>> I just don't get if for contrasts and I guess my question really is:
>> How is the VARCOPE calculated?
>>
>> Here is what I think it is with a simple example:
>> Let's say I have a voxel with the following 25 values over time.
>> 1 1 8 9 2 1 1 2 9 8 1 2 1 2 8 8 1 1 2 2 9 9 1 2 1
>>
>> my EV1 is
>> 0 0 1 1 0 0 0 0 1 1 0 0 0 0 1 1 0 0 0 0 1 1 0 0 0
>>
>> my EV2 is
>> 0 0 0 0 0 1 1 0 0 0 0 1 1 0 0 0 0 1 1 0 0 0 0 0 0
>>
>> taken together
>> 1 1 8 9 2 1 1 2 9 8 1 2 1 2 8 8 1 1 2 2 9 9 1 2 1
>> 0 0 1 1 0 0 0 0 1 1 0 0 0 0 1 1 0 0 0 0 1 1 0 0 0
>> 0 0 0 0 0 1 1 0 0 0 0 1 1 0 0 0 0 1 1 0 0 0 0 0 0
>>
>> one can clearly see that EV1 covariates with the data while EV2
>> doesn't.
>>
>> I set ev1 and ev2 up as a model matrix and got the inverse (Moore-
>> Penrose-Pseudoinverse) of it and then multiplied the inverse with
>> the data.
>> My parameter estimate for EV1 is 8.50 and for EV2 is 1.33.
>>
>> I specified a contrast - Contrast1 EV1>EV2: 1 -1
>> COPE in this case is the same as PE1 minus PE2. Right?
>> So COPE1 would be (8.5*1) + (-1*1.33) = 7.2
>>
>> Now I multiply my PEs with my contrast and get [8.50, -1.33]
>> then I multiply [8.50, -1.33] with my model matrix and see what the
>> model predicts. In my case that would be:
>> 0.00 0.00 8.50 8.50 0.00 -1.33 -1.33 0.00 8.50 8.50 0.00 -1.33
>> -1.33 0.00 8.50 8.50 0.00 -1.33 -1.33 0.00 8.50 8.50 0.00 0.00 0.00
>>
>> When I calculate the difference between predicted_data and actual
>> data I get
>> 1.00 1.00 -0.50 0.50 2.00 2.33 2.33 2.00 0.50 -0.50 1.00 3.33 2.33
>> 2.00 -0.50 -0.50 1.00 2.33 3.33 2.00 0.50 0.50 1.00 2.00 1.00
>>
>> then I square every element and sum it up - in my case 72. I think
>> that is VARCOPE1 - is that right?
>>
>> Now I could calculate e.g. a t-test:
>> t = Parameter_Estimate / Standard_Error
>> t = PE / ((VARCOPE)^0.5 / (degrees_of_freedom)^0.5)
>> t = 7.2 / (72^0.5 / 23^0.5) = 4.07
>> That seems a rather low t-value for this made up data so I think I'm
>> doing something wrong here.
>>
>> Sorry for these basic questions but I at least want to get the
>> basics right. Thanks, Michael
>>
>>
>> On 19-Jun-09, at 12:14 PM, Jesper Andersson wrote:
>>
>>> Very simple answer,
>>>
>>>> Hi, I have a very simple question.
>>>> I know that PE means parameter estimate.
>>>> What does COPE and VARCOPE stand for?
>>>
>>> COPE=COntrast of Parameter Estimates.
>>> VARCOPE=VARiance of COntrast of Parameter Estimates.
>>>
>>> Jesper
>>
>
>
> ---------------------------------------------------------------------------
> Stephen M. Smith, Professor of Biomedical Engineering
> Associate Director, Oxford University FMRIB Centre
>
> FMRIB, JR 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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