[Eeglablist] Is ITC biased by trial numbers?
Bradley Voytek
bradley.voytek at gmail.com
Mon Nov 22 11:20:38 PST 2010
Zara:
The best approach would be to do resampling statistics.
So the question you want to know is: does your stimulus affect ITC in
some task-dependent manner? So for each subject you have an ITC value
for each time point, for each condition.
Let's say each condition has 100 trials per subject, except for the
condition shown by the red line, which has 50. Let's say you want to
compare condition red to blue. What you do now is you take your 100
blue trials, calculate ITC, and do the same separately for your 50 red
trials. Your concern is that the number of trials biases the ITC.
To address this, calculate a difference ITC (red - blue) for each
subject, giving you a real ITC difference. What's important to you
here are the trial labels: red v. blue. If you want to see if your
difference between conditions is real effect of trial label (red v.
blue), or an artifact of number of samples, you can resample your
data.
Take your 100 blue trials, 50 red trials, and create a vector of 150
trials for subject 1. Now, randomly select 100 trials from that vector
of 150 total trials, and calculate ITC from that. This gives you a
"surrogate blue" ITC. Do the same for the remaining 50 trials in the
vector to get a "surrogate red". Subtract the two to get a surrogate
difference. Repeat at least 500 times (though I recommend 10000).
Now you have a distribution (at each time point) of 10000 surrogate
ITC differences *from your actual data*. That is, you now know, given
your real data, what is the probability of getting an ITC difference
between red and blue just based on the actual data? In other words,
does the number of samples bias the ITC estimate?
Statistically, you can calculate a z-score from this distribution at
each time point, for each subject, by taking the REAL ITC difference
at that time point, subtracting the mean ITC difference from the 10000
surrogate values at that same time point, and dividing by the standard
deviation of the distribution of the 10000 values from that same time
point. From this z-score you know the probability (or significance) of
your difference, and whether it is biased by the number of trials.
Of course, because ITC is a bounded value in the range of [0,1],
technically you might want to normalize all the ITC values, but
because you're calculating a difference, in the end you should be ok.
::brad
2010/11/19 Zara Bergström <zmb25 at cam.ac.uk>:
> Dear EEG experts,
>
> is the ITC measure as implemented by EEGLAB biased towards lower trial
> numbers (i.e. higher itc when fewer trials are used in the computation, as
> some measures of phase coherence supposedly are), and if so, how do you deal
> with that issue when comparing conditions with different trial numbers? Do
> you think it is appropriate to compute baseline corrected ITC, which might
> help?
>
> I analysed epoched datasets (-1-2s using default pre-stimulus baseline for
> ERSP) with wavelets using newtimef to get complex ITC values (using the
> default 'phasecoher' option), converted the ITC to real numbers between 0-1
> (absitc=sqrt(real(itc).^2+imag(itc).^2)), and averaged these into
> participant x condition x time x frequency ITC matrices for use in group
> level statistics.
>
> The attached line plot shows grand average (24 subjects) ITC averaged across
> the alpha band (8-12 hz) for four conditions. The condition with the fewest
> average trial numbers (red line) has significantly higher ITC than the other
> conditions throughout the trial, even before stimulus onset, which cannot be
> explained by psychological factors since these conditions were presented
> randomly intermixed. If I however were to subtract the average baseline
> period ITC from the post-stimulus data, it seems that the difference would
> disappear. Would that be an appropriate step to take here?
>
> Do you think this pattern is caused by a trial number bias, or have I done
> something wrong in the analysis pipeline?
>
> Any thoughts would be very much appreciated.
>
> Thanks very much for you time,
>
> Zara Bergstrom
>
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