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<div>Hi Matti,</div>
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<div>sure-sorry and thanks for your quick reply. So here is what I am doing using the average covariance matrix (so data covariance).</div>
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<div>cfg=[];<br/>
cfg.covariance='yes';<br/>
cfg.channel=data.label;<br/>
avg_data=ft_timelockanalysis(cfg, data);</div>
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<div>cfg=[];<br/>
cfg.method='lcmv';<br/>
cfg.grid=lf;<br/>
cfg.vol=hdm;<br/>
cfg.lcmv.keepfilter='yes';<br/>
cfg.lcmv.fixedori= 'yes'; <br/>
cfg.lcmv.lambda= '10%'; %0% rank deficient data-use stronger regularization??<br/>
lcmv_avg=ft_sourceanalysis(cfg, avg_data);</div>
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<div>Best,</div>
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<div>Luca</div>
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<div style="margin:0 0 10px 0;"><b>Gesendet:</b> Montag, 20. August 2018 um 15:10 Uhr<br/>
<b>Von:</b> "Matti Stenroos" <matti.stenroos@aalto.fi><br/>
<b>An:</b> "Luca Kaiser" <luca.kaiser@web.de><br/>
<b>Betreff:</b> Re: [FieldTrip] Set regularization parameter after SSS</div>
<div name="quoted-content">Dear Luca,<br/>
<br/>
I think you'd need to tell a bit more, as the meaning of "lambda"<br/>
depends on the algorithm you are using. Also "the covariance matrix" is<br/>
not uniquely defined --- are you talking about noise covariance or<br/>
measurement covariance?<br/>
<br/>
In general, the eigenvalue idea you had read does not make sense --- the<br/>
eigenvalue text deals with condition numbers, while lambda does, in<br/>
general, not relate directly to that. The advice might have been that<br/>
"set lambda so that the ratio of largest and smallest eigenvalues is<br/>
1000"...<br/>
<br/>
Cheers,<br/>
Matti<br/>
also working with low-rankingMax-filtered MEG and coincidentally<br/>
having a lot of covariance metrics on screen at the moment<br/>
<br/>
<br/>
<br/>
On 2018-08-20 15:39, Luca Kaiser wrote:<br/>
> Dear FieldTrip community,<br/>
> I am using ft_sourceanalysis on SSS preprocessed (neuromag) data. I<br/>
> wonder if there are any suggestions on how to set lambda in this case?<br/>
> My covariance matrix is rank deficient (rank 60, 306 sensors). I read in<br/>
> the mailing list that I would need to look at the eigenvalues of XX' and<br/>
> set lambda to be 1/1000 of the largest eigenvalue. However, I do not<br/>
> really understand why to divide by 1/1000?<br/>
> Every help would be very much appreciated,<br/>
> Thanks!<br/>
> Luca<br/>
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