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Re: computing normalized beta weights

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Fri, 27 Aug 2004 16:05:13 +0100

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 ```Hi there - The idea is to pick out the standard deviation accounted for by the EV of interest. So the model is Y=b*X+e so, e.g. if you want to know the proportion of the standard deviation accounted for by EV1, a sensible thing to do would be set all the other elements in the "b" vector to be zero. i.e. do the element-wise product of c= [1 0 0 0 0 ... ] with b call that c.*b then the signal accounted for by this EV is [c.*b]*X and the standard deviation of this signal is sqrt([ [c.*b]*X ] * [ X' * [c.*b]' ]/(n-1) ) which is nearly what I said before :) Note that the interpretation of this "signal explained" is not as clear as this when for contrasts which are more complicated than e.g. [1 0 0 ... ]. Beware that contrasts are really about inferrence, and signal explained by a contrast is not always a meaningful concept.. T On Mon, 23 Aug 2004, Ramapriyan Pratiwadi wrote: > hi, > > sorry to regurgitate an old thread, but i don't quite understand the last > post. > > for the case of multiple PE's (and COPE's), the dimensions in the formula > sqrt(c'*b*X*X'*b'*c) do not match up; in my case, i have 320 time points and > 8 PE's (4 EV's +temporal derivatives). Therefore, the design matrix is > 320x8; if b is a vector, it would have to be 1x320 - how can that be? > > i tried transposing the design matrix, to make it 8x320 - then, the > multiplication becomes (1x1)(1x8)(8x320)(320x8)(8x1)(1x1), for each voxel. > However, when the result was divided by the standard deviation of the > filtered_func_data, the values were not normalized between -1 and 1. > > normally, we would divide the betas of the PE's by the beta of the constant > column; however, since there is not a constant column, the aforementioned > approached seems suitable. > > Thanks in advance, > > Ram > > ------------------------------------- > Brain Behavior Laboratory > > University of Pennsylvania > Philadelphia, Pennsylvania, USA > > > On Tue, 12 Aug 2003 14:39:47 +0100, Tim Behrens <[log in to unmask]> wrote: > > >Ok - there are various levels to this answer. > > > >It seems to be a sensible thing to do - effectively you want to know the > >amount amount of the data's standard deviation which is explained by a > >single COPE. > > > >1) If you've only got one PE, then this is relatively trivial. > >the std of the design can be computed easily from design.mat (ascii file > >containing design timeseries). Call this sx. The std of the data is > > > >avwmaths filtered_func_data -Tstd sy > > > >then the fractional deviation explained by your PE is just > > > >avwmaths PE -mul sx -div sy Beta_norm > > > >I think, in this case, you can do this with the unwhitened data as the > >whitening matrix is normalised. > > > >2) If you've got more than 1 PE, life is more complicated (and I don't > >think you can compute what you want with simple FSL commands ). > > > >You need to project the variance explained by all of your EVs onto a > >single COPE. > > > >if you assume the data is white and demeaned then and your Design is > >demeaned.. > > > >the std explained by your cope is: > > > >sqrt(c'*b*X*X'*b'*c)/dof > > > >c is cour contrast, b is your vector of PEs, X is your design, dof is your > >degrees of freedom > > > >if it's not white then > > > >sqrt(c'*b*k*X*X'*k'*b'*c)/dof > > > >k is the whitening matrix. > >(The whitening will change the projection) > > > >and the standard deviation of the whitened data is just std(k*Y) > > > >Dividing one by the other should give you what you want. > > > >Hate to say it, but you might need matlab!! > > > >cheers > >T > > > > > >------------------------------------------------------------------------------- > >Tim Behrens > >Centre for Functional MRI of the Brain > >The John Radcliffe Hospital > >Headley Way Oxford OX3 9DU > >Oxford University > >Work 01865 222782 > >Mobile 07980 884537 > >------------------------------------------------------------------------------- > > > >On Tue, 12 Aug 2003, Edward Vessel wrote: > > > >> Ok, well, here is something that might give you a feeling for the difference. > >> In a regression with only a single independent variable, the standardized > >> beta equals r (the Pearson's correlation). This would also be true (I think) > >> if all the variables were totally uncorrelated in a multiple regression. > >> > >> The analysis I have done is one in which I want to look at a correlation > as my > >> statistic, and the standardized Beta coefficient is one way to report a > >> factor loading which takes into account that factor's covariance w/ other > >> factors. It would be 1 if the activity of a voxel were perfectly predictable > >> from that factor, and 0 if that factor had no power to predict it. So, the > >> reason for having it is in the _interpretation_ of the statistic. > >> > >> Typically, it is computed as: > >> > >> Beta = b * (sx / sy) > >> > >> where > >> > >> Beta: standardized regression coefficient (-1 to 1) > >> > >> b: the unstandardized regression coefficient (can take any value), which is > >> the 'regulular' weight, sometimes confusingly referred to as beta but isn't > >> standardized) - probably a pe, which is equivalent to one of my copes in this > >> case > >> > >> sx: standard deviation of the EV > >> xy: standard deviation of the data > >> > >> Does that make any sense? > >> > >> Ed > >> > >> > -- ------------------------------------------------------------------------------- Tim Behrens Centre for Functional MRI of the Brain The John Radcliffe Hospital Headley Way Oxford OX3 9DU Oxford University Work 01865 222782 Mobile 07980 884537 ------------------------------------------------------------------------------- ```