[FieldTrip] Statistical test of robustness of a graph measure based on reduced amount of nodes

Darren Price Darren.Price at mrc-cbu.cam.ac.uk
Wed Oct 12 19:00:42 CEST 2016


Hi

Sorry, I don’t have time to write a long answer right now, but in short you don’t need to select a subset of channels. In fact this would not be a valid null since those subset of channels are really just picking up roughly the same signal as the unselected channels. One type of valid null (given certain assumptions) can be computed by phase randomisation of the original data (to create what is known as a surrogate dataset). The general idea is to Fourier transform each channel, randomise the phase (but not the amplitude) of each frequency component, using the same phase randomisation for each channel, and then inverse Fourier transform each channel. Don’t try to do this yourself unless you know what you are doing, as there are a couple of big pitfalls to avoid. What you will be left with is a random time domain signal, that has the same frequency components, and also (crucially) the same covariance structure between channels (so that the rank of the data is the same as your original data in each iteration), although you might not want this, and it’s not strictly necessary. If you want a proper reference then here it is. http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.73.951 .

If you want an example of how to do this in Matlab. I can provide it. It’s actually just a few lines of code. Let me know and I will send the code over.

P.S if you want your results to be reproducible you should always set your random seed at the start of your script.

Good luck.

Thanks
Darren


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MRC Cognition & Brain Sciences Unit
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From: fieldtrip-bounces at science.ru.nl [mailto:fieldtrip-bounces at science.ru.nl] On Behalf Of Ta Dinh, Son
Sent: 12 October 2016 16:06
To: fieldtrip at science.ru.nl
Subject: Re: [FieldTrip] Statistical test of robustness of a graph measure based on reduced amount of nodes

Hey Matthew,

Thanks for the answer, but the question is exactly how to actually build a representative null distribution. As the calculation using all (64) electrodes is deterministic, it can’t really be used to create a distribution, it would just be a vector of 1000 x 1 exact same value.
The graph measure is called hub disruption index and was introduced here: Achard, S., et al. (2012). "Hubs of brain functional networks are radically reorganized in comatose patients." PNAS.
To put it in a nutshell, it compares a subject against a group of controls, thereby giving a single value for every subject (in comparison to the control group).

I hope this has cleared up the context a bit.

Best
Son

Son Ta Dinh, M.Sc.
PhD student in Human Pain Research
Klinikum rechts der Isar
Technische Universität München
Munich, Germany
Phone: +49 89 4140 7664<tel:%2B49%2089%204140%207664>
http://www.painlabmunich.de/

Von: Nickel, Moritz
Gesendet: Mittwoch, 12. Oktober 2016 16:37
An: Ta Dinh, Son <son.ta.dinh at tum.de<mailto:son.ta.dinh at tum.de>>
Betreff: Fwd: [FieldTrip] Statistical test of robustness of a graph measure based on reduced amount of nodes


---------- Forwarded message ----------
From: Matt Gerhold <matt.gerhold at gmail.com<mailto:matt.gerhold at gmail.com>>
Date: 2016-10-05 13:31 GMT+02:00
Subject: Re: [FieldTrip] Statistical test of robustness of a graph measure based on reduced amount of nodes
To: FieldTrip discussion list <fieldtrip at science.ru.nl<mailto:fieldtrip at science.ru.nl>>

Hi Son,

What you are explaining sounds like resampling to build a distribution under the null hypothesis. You would need to make sure that your random draws are representative in some way of an instance where the test statistic (graph theoretic measure) is truly zero, i.e. representative of the null hypothesis. There is no info on your measure, so one can't comment any further on how one would achieve this.

Once you have the bootstrapped distribution you compute the proportion of values above the test statistic and those below the test statistic--the test statistic is the measure you got from the actual sample, not the bootstrapped distribution.

Then it depends whether you use a two-tail or one-tail test and the direction of the hypothesized effect: for a one-tail test you could potentially take the proportion of the distribution above equal to the test statistic, that would be your p-value. For two tailed-tests take the min value of the two-proportions as your p-value and remember to divide alpha by 2 to test for significance.

That, in a nutshell, is a simple approach; however, there are other ways to go about this.

Matthew

On Wed, Oct 5, 2016 at 6:54 AM, Ta Dinh, Son <son.ta.dinh at tum.de<mailto:son.ta.dinh at tum.de>> wrote:
Dear Fieldtrippers,

the general problem we are facing is one of statistics. In particular, we are trying to test the robustness of a graph measure when reducing the amount of nodes it is computed with. In our case, we use the EEG electrodes as nodes.

We are trying to find out whether a graph measure differs significantly from zero over a group of subjects. The exact calculation of the measure is rather complicated to explain, suffice it to say that every subject has exactly one scalar value in the end. Computation of this measure using 64 electrodes is straightforward and we can easily calculate a p-value and/or a confidence interval.
When we calculate based on only 32 electrodes however, we draw 32 electrodes randomly. Therefore, we need to repeat this computation many times (let’s say 1000 times). So we then get [1000 x number of subjects] values, or 1000 p-values/confidence intervals.
How do we statistically test whether the measure is robustly different from 0? Is it too naive to simply assume that if the confidence interval does not contain 0 in at least 950 of the 1000 computations then it is robustly different from 0?

Any help would be greatly appreciated!

Best regards,
Son

Son Ta Dinh, M.Sc.
PhD student in Human Pain Research
Klinikum rechts der Isar
Technische Universität München
Munich, Germany
Phone: +49 89 4140 7664<tel:%2B49%2089%204140%207664>
http://www.painlabmunich.de/


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