WiP: The Earth Is Round – Significance Testing¶

Note

This tutorial part is also available for download as an IPython notebook: [ipynb]

After performing a classification analysis one is usually interested in an evaluation of the results with respect to its statistical uncertainty. In the following we will take at a few possible approaches to get this from PyMVPA.

Let’s look at a typical setup for a cross-validated classification. We start by generating a dataset with 200 samples and 3 features of which only two carry some relevant signal. Afterwards we set up a standard leave-one-chunk-out cross-validation procedure for an SVM classifier. At this point we have seen this numerous times, and the code should be easy to read:

>>> # lazy import
>>> from mvpa2.suite import *
>>> # some example data with signal
>>> ds = normal_feature_dataset(perlabel=100, nlabels=2, nfeatures=3,
...                             nonbogus_features=[0,1], snr=0.3, nchunks=2)
>>> # classifier
>>> clf = LinearCSVMC()
>>> # data folding
>>> partitioner = NFoldPartitioner()
>>> # complete cross-validation setup
>>> cv = CrossValidation(clf,
...                      partitioner,
...                      errorfx=mean_match_accuracy,
...                      postproc=mean_sample())
>>> acc = cv(ds)

Exercise

Take a look at the performance statistics of the classifier. Explore how it changes with different values of the signal-to-noise (snr) parameter of the dataset generator function.

The simplest way to get a quick assessment of the statistical uncertainty of the classification accuracy is to look at the standard deviation of the accuracies across cross-validation folds. This can be achieved by removing the postproc argument of CrossValidation.

Another, slightly more informative, approach is to compute confidence intervals for the classification accuracy. We can do this by treating each prediction of the classifier as a Bernoulli trial with some success probability. If we further assume statistical independence of these prediction outcomes we can compute binomial proportion confidence intervals using a variety of methods. To implement this calculation we only have to modify the error function and the post processing of our previous analysis setup.

>>> # complete cross-validation setup
>>> cv = CrossValidation(
...         clf,
...         partitioner,
...         errorfx=prediction_target_matches,
...         postproc=BinomialProportionCI(width=.95, meth='jeffreys'))
>>> ci_result = cv(ds)
>>> ci = ci_result.samples[:, 0]
>>> ci < np.asscalar(acc) < ci
True

Instead of computing accuracies we use an error function that returns a boolean vector of prediction success for each sample. In the post processing this information is then used to compute the confidence intervals. We can see that the previously computed accuracy lies within the confidence interval. If the assumption of statistical independence of the classifier prediction success holds we can be 95% certain that the true accuracy is within this interval.

Exercise

Think about situations in which we cannot reasonably assume statistical independence of classifier prediction outcomes. Hint: What if the data in the testing dataset shows strong auto-correlation?

Null hypothesis testing¶

Another way of making statements like “Performance is significantly above chance-level” is Null hypothesis (aka H0) testing that PyMVPA supports for any Measure.

However, as with other applications of statistics in classifier-based analyses, there is the problem that we typically do not know the distribution of a variable like error or performance under the Null hypothesis (i.e. the probability of a result given that there is no signal), hence we cannot easily assign the adored p-values. Even worse, the chance-level or guess probability of a classifier depends on the content of a validation dataset, e.g. balanced or unbalanced number of samples per label, total number of labels, as well as the peculiarities of “independence” of training and testing data – especially in the neuroimaging domain.

Monte Carlo – here I come!¶

One approach to deal with this situation is to estimate the Null distribution using permutation testing. The Null distribution is estimated by computing the measure of interest multiple times using the original data samples but with permuted targets, presumably scrambling or destroying the signal of interest. Since quite often the exploration of all permutations is unfeasible, Monte-Carlo testing (see Nichols et al. (2002)) allows us to obtain a stable estimate with only a limited number of random permutations.

Given the results computed using permuted targets, we can now determine the probability of the empirical result (i.e. the one computed from the original training dataset) under the no signal condition. This is simply the fraction of results from the permutation runs that is larger or smaller than the empirical (depending on whether one is looking at performances or errors).

Here is our previous cross-validation set up:

>>> cv = CrossValidation(clf,
...                      partitioner,
...                      postproc=mean_sample(),
...                      enable_ca=['stats'])
>>> err = cv(ds)

Now we want to run this analysis again, repeatedly and with a fresh permutation of the targets for each run. We need two pieces for the Monte Carlo shuffling. The first is an instance of an AttributePermutator that will permute the target attribute of the dataset for each iteration. We will instruct it to perform 200 permutations. In a real analysis, the number of permutations will often be more than that.

>>> permutator = AttributePermutator('targets', count=200)

Exercise

The permutator is a generator. Try generating all 200 permuted datasets.

The second necessary component for a Monte-Carlo-style estimation of the Null distribution is the actual “estimator”. MCNullDist will use the already created permutator to shuffle the targets and later on report the p-value from the left tail of the Null distribution, because we are going to compute errors and we are interested in them being lower than chance. Finally, we also ask for all results from Monte-Carlo shuffling to be stored for subsequent visualization of the distribution.

>>> distr_est = MCNullDist(permutator, tail='left', enable_ca=['dist_samples'])

The rest is easy. Measures take an optional constructor argument null_dist that can be used to provide an instance of some NullDist estimator – and we have just created one! Because a cross-validation is nothing but a measure, we can assign it our Null distribution estimator, and it will also perform permutation testing, in addition to the regular classification analysis on the “real” dataset. Consequently, the code hasn’t changed much:

>>> cv_mc = CrossValidation(clf,
...                         partitioner,
...                         postproc=mean_sample(),
...                         null_dist=distr_est,
...                         enable_ca=['stats'])
>>> err = cv_mc(ds)
>>> cv.ca.stats.stats['ACC'] == cv_mc.ca.stats.stats['ACC']
True

Other than it taking a bit longer to compute, the performance did not change. But the additional waiting wasn’t in vain, as we get the results of the statistical evaluation. The cv_mc conditional attribute null_prob has a dataset that contains the p-values representing the likelihood of an empirical value (i.e. the result from analysing the original dataset) being equal or lower to one under the Null hypothesis, i.e. no actual relevant signal in the data. Or in more concrete terms, the p-value is the fraction of permutation results less than or equal to the empirical result.

>>> p = cv_mc.ca.null_prob
>>> # should be exactly one p-value
>>> p.shape
(1, 1)
>>> np.asscalar(p) < 0.1
True

Exercise

How many cross-validation analyses were computed when running cv_mc? Make sure you are not surprised that it is more than 200. What is the minimum p-value that we can get from 200 permutations?

Let’s practise our visualization skills a bit and create a quick plot to show the Null distribution and how “significant” our empirical result is. And let’s make a function for plotting to show off our Python-foo!

>>> def make_null_dist_plot(dist_samples, empirical):
...     pl.hist(dist_samples, bins=20, normed=True, alpha=0.8)
...     pl.axvline(empirical, color='red')
...     # chance-level for a binary classification with balanced samples
...     pl.axvline(0.5, color='black', ls='--')
...     # scale x-axis to full range of possible error values
...     pl.xlim(0,1)
...     pl.xlabel('Average cross-validated classification error')
>>>
>>> # make new figure ('_ =' is only used to swallow unnecessary output)
>>> _ = pl.figure()
>>> make_null_dist_plot(np.ravel(cv_mc.null_dist.ca.dist_samples),
...                     np.asscalar(err))
>>> # run pl.show() if the figure doesn't appear automatically

You can see that we have created a histogram of the “distribution samples” stored in the Null distribution (because we asked for it previously). We can also see that the Null or chance distribution is centered around the expected chance-level and the empirical error value is in the far left tail, thus relatively unlikely to be a Null result, hence the low-ish p-value.

This wasn’t too bad, right? We could stop here. But there is this smell....

Exercise

The p-value that we have just computed and the Null distribution we looked at are, unfortunately, invalid – at least if we want to know how likely it is to obtain our empirical result under a no-signal condition. Can you figure out why?

Avoiding the trap OR Advanced magic 101¶

Here is what went wrong: The dataset’s class labels (aka targets) were shuffled repeatedly, and for each iteration a full cross-validation of classification error was computed. However, the shuffling was done on the full dataset, hence target values were permuted in both training and testing dataset portions in each CV-fold. This basically means that for each Monte Carlo iteration the classifier was tested on new data/signal. However, we are actually interested in what the classifier has to say about the actual data, but when it was trained on randomly permuted data.

Doing a whole-dataset permutation is a common mistake with very beneficial side-effects – as you will see in a bit. Sadly, doing the permuting correctly (i.e. in the training portion of the dataset only) is a bit more complicated due to the data-folding scheme that we have to deal with. Here is how it goes:

>>> repeater = Repeater(count=200)

A repeater is a simple node that returns any given dataset a configurable number of times. We use this helper to configure the number of Monte Carlo iterations.

Exercise

A Repeater is also a generator. Try calling it with our dataset. What does it do? How can you get it to produce the 200 datasets?

The new permutator is again configured to shuffle the targets attribute. But this time only once and only for samples that were labeled as being part of the training set in a particular CV-fold. The partitions sample attribute is created by the NFoldPartitioner that we have already configured earlier (or any other partitioner in PyMVPA for that matter).

>>> permutator = AttributePermutator('targets',
...                                  limit={'partitions': 1},
...                                  count=1)

The most significant difference is that we are now going to use a dedicate measure to estimate the Null distribution. That measure is very similar to the cross-validation we have used before, but differs in an important twist: we use a chained generator to perform the data-folding. This chain comprises of our typical partitioner (marks one chunk as testing data and the rest as training, for all chunks) and the new one-time permutator. This chain-generator causes the cross-validation procedure to permute the training data only for each data-fold and leave the testing data untouched. Note, that we make the chain use the space of the partitioner, to let the CrossValidation know which samples attribute defines training and testing partitions.

>>> null_cv = CrossValidation(
...            clf,
...            ChainNode(
...                 [partitioner, permutator],
...                 space=partitioner.get_space()),
...            postproc=mean_sample())

Exercise

Create a separate chain-generator and explore what it does. Remember: it is just a generator.

Now we create our new and improved distribution estimator. This looks similar to what we did before, but we now use our dedicated Null cross-validation measure, and run it as often as repeater is configured to estimate the Null performance.

>>> distr_est = MCNullDist(repeater, tail='left',
...                        measure=null_cv,
...                        enable_ca=['dist_samples'])

On the “outside” the cross-validation measure for computing the empricial performance estimate is 100% identical to what we have used before. All the magic happens inside the distribution estimator.

>>> cv_mc_corr = CrossValidation(clf,
...                              partitioner,
...                              postproc=mean_sample(),
...                              null_dist=distr_est,
...                              enable_ca=['stats'])
>>> err = cv_mc_corr(ds)
>>> cv_mc_corr.ca.stats.stats['ACC'] == cv_mc.ca.stats.stats['ACC']
True
>>> cv_mc.ca.null_prob.samples <  cv_mc_corr.ca.null_prob.samples
array([[ True]], dtype=bool)

After running it we see that there is no change in the empirical performance (great!), but our significance did suffer (poor thing!). We can take a look at the whole picture by plotting our previous Null distribution estimate and the new, improved one as an overlay.

>>> make_null_dist_plot(cv_mc.null_dist.ca.dist_samples, np.asscalar(err))
>>> make_null_dist_plot(cv_mc_corr.null_dist.ca.dist_samples, np.asscalar(err))
>>> # run pl.show() if the figure doesn't appear automatically

It should be obvious that there is a substantial difference in the two estimates, but only the latter/wider distribution is valid!

Exercise

Keep it in mind. Keep it in mind. Keep it in mind.

The following content is incomplete and experimental¶

If you have a clue¶

There a many ways to further tweak the statistical evaluation. For example, if the family of the distribution is known (e.g. Gaussian/Normal) and provided via the dist_class parameter of MCNullDist, then permutation tests samples will be used to fit this particular distribution and estimate distribution parameters. This could yield enormous speed-ups. Under the (strong) assumption of Gaussian distribution, 20-30 permutations should be sufficient to get sensible estimates of the distribution parameters. Fitting a normal distribution would look like this. Actually, only a single modification is necessary (the dist_class argument), but we will also reduce the number permutations.

>>> distr_est = MCNullDist(Repeater(count=200),
...                        dist_class=scipy.stats.norm,
...                        tail='left',
...                        measure=null_cv,
...                        enable_ca=['dist_samples'])
>>> cv_mc_norm = CrossValidation(clf,
...                              partitioner,
...                              postproc=mean_sample(),
...                              null_dist=distr_est,
...                              enable_ca=['stats'])
>>> err = cv_mc_norm(ds)
>>> distr = cv_mc_norm.null_dist.dists()
>>> make_null_dist_plot(cv_mc_norm.null_dist.ca.dist_samples,
...                     np.asscalar(err))
>>> x = np.linspace(0,1,100)
>>> _ = pl.plot(x, distr.pdf(x), color='black', lw=2)

Family-friendly¶

When going through this chapter you might have thought: “Jeez, why do they need to return a single p-value in a freaking dataset?” But there is a good reason for this. Lets set up another cross-validation procedure. This one is basically identical to the last one, except for not averaging classifier performances across data-folds (i.e. postproc=mean_sample()).

>>> cvf = CrossValidation(
...         clf,
...         partitioner,
...         null_dist=MCNullDist(
...                     repeater,
...                     tail='left',
...                     measure=CrossValidation(
...                                 clf,
...                                 ChainNode([partitioner, permutator],
...                                           space=partitioner.get_space()))
...                     )
...         )

If we run this on our dataset, we no longer get a single performance value, but one per data-fold (chunk) instead:

>>> err = cvf(ds)
>>> len(err) == len(np.unique(ds.sa.chunks))
True

But here comes the interesting bit:

>>> len(cvf.ca.null_prob) == len(err)
True

So we get one p-value for each element in the datasets returned by the cross-validation run. More generally speaking, the distribution estimation happens independently for each value returned by a measure – may this be multiple samples, or multiple features, or both. Consequently, it is possible to test a large variety of measure with this facility.

Evaluating multi-class classifications¶

So far we have mostly looked at the situation where a classifier is trying to discriminate data from two possible classes. In many cases we can assume that a classifier that cannot discriminate these two classes would perform at a chance-level of 0.5 (ACC). If it does that we would conclude that there is no signal of interest in the data, or our classifier of choice cannot pick it up. However, there is a whole universe of classification problems where it is not that simple.

Let’s revisit the classification problem from the chapter on classifiers.

>>> from mvpa2.tutorial_suite import *
>>> ds = get_haxby2001_data_alternative(roi='vt', grp_avg=False)
>>> print ds.sa['targets'].unique
['bottle' 'cat' 'chair' 'face' 'house' 'scissors' 'scrambledpix' 'shoe']
>>> clf = kNN(k=1, dfx=one_minus_correlation, voting='majority')
>>> cv = CrossValidation(clf, NFoldPartitioner(), errorfx=mean_mismatch_error,
...                      enable_ca=['stats'])
>>> cv_results = cv(ds)
>>> print '%.2f' % np.mean(cv_results)
0.53

So here we have an 8-way classification problem, and during the cross-validation procedure the chosen classifier makes correct predictions for approximately half of the data points. The big question is now: What does that tell us?

There are many scenarios that could lead to this prediction performance. It could be that the fitted classifier model is very good, but only captures the data variance for half of the data categories/classes. It could also be that the classifier model quality is relatively poor and makes an equal amount of errors for all classes. In both cases the average accuracy will be around 50%, and most likely highly significant, given a chance performance of 1/8. We could now spend some time testing this significance with expensive permutation tests, or making assumptions on the underlying distribution. However, that would only give us a number telling us that the average accuracy is really different from chance, but it doesn’t help with the problem that the accuracy really doesn’t tell us much about what we are interested in.

Interesting hypotheses in the context of this dataset could be whether the data carry a signal that can be used to distinguish brain response patterns from animate vs. inanimate stimulus categories, or whether data from object-like stimuli are all alike and can only be distinguished from random noise, etc. One can imagine running such an analysis on data from different parts of the brain and the results changing – without necessarily having a big impact on the overall classification accuracy.

A lot more interesting information is available from the confusion matrix, a contingency table showing prediction targets vs. actual predictions.

>>> print cv.ca.stats.matrix
[[36  7 18  4  1 18 15 18]
[ 3 56  6 18  0  3  7  5]
[ 2  2 21  0  4  0  3  1]
[ 3 16  0 76  4  5  3  1]
[ 1  1  6  1 97  1  4  0]
[20  5 15  4  0 29 15 11]
[ 0  1  0  0  0  2 19  0]
[43 20 42  5  2 50 42 72]]

We can see a strong diagonal, but also block-like structure, and have to realize that simply staring at the matrix doesn’t help us to easily assess the likelihood of any of our hypotheses being true or false. It is trivial to do a Chi-square test of the confusion table...

>>> print 'Chi^2: %.3f (p=%.3f)' % cv.ca.stats.stats["CHI^2"]
Chi^2: 1942.519 (p=0.000)

... but, again, it doesn’t tell us anything other than that the classifier is not just doing random guesses. It would be much more useful if we could estimate how likely it is, given the observed confusion matrix, that the employed classifier is able to discriminate all stimulus classes from each other, and not just a subset. Even more useful would be if we could relate this probability to specific alternative hypotheses, such as an animate/inanimate-only distinction.

Olivetti et al. (2012) have devised a method that allows for doing exactly that. The confusion matrix is analyzed in a Bayesian framework regarding the statistical dependency of observed and predicted class labels. Confusions within a set of classes that cannot be discriminated should be independently distributed, while there should be a statistical dependency of confusion patterns within any set of classes that can all be discriminated from each other.

This algorithm is available in the BayesConfusionHypothesis node.

>>> cv = CrossValidation(clf, NFoldPartitioner(),
...                      errorfx=None,
...                      postproc=ChainNode((Confusion(labels=ds.UT),
...                                          BayesConfusionHypothesis())))
>>> cv_results = cv(ds)
>>> print cv_results.fa.stat
['log(p(C|H))' 'log(p(H|C))']

Most likely hypothesis to explain this confusion matrix:

>>> print cv_results.sa.hypothesis[np.argsort(cv_results.samples[:,1])[-1]]
[['bottle'], ['cat'], ['chair'], ['face'], ['house'], ['scissors'], ['scrambledpix'], ['shoe']]

Previously in part 8¶

Previously, while looking at classification we have observed that classification error depends on the chosen classification method, data preprocessing, and how the error was obtained – training error vs generalization estimates using different data splitting strategies. Moreover in attempts to localize activity using searchlight we saw that generalization error can reach relatively small values even when processing random data which (should) have no true signal. So, the value of the error alone does not provide sufficient evidence to state that our classifier or any other method actually learnt the mapping from the data into variables of interest. So, how do we decide what estimate of error can provide us sufficient evidence that constructed mapping reflects the underlying phenomenon or that our data carried the signal of interest?

Researchers interested in developing statistical learning methods usually aim at achieving as high generalization performance as possible. Newly published methods often stipulate their advantage over existing ones by comparing their generalization performance on publicly available datasets with known characteristics (number of classes, independence of samples, actual presence of information of interest, etc.). Therefore, generalization performances presented in statistical learning publications are usually high enough to obliterate even a slight chance that they could have been obtained simply by chance. For example, those classifiers trained on MNIST dataset of handwritten digits were worth reporting whenever they demonstrated average errors of only 1-2% while doing classification among samples of 10 different digits (the largest error reported was 12% using the simplest classification approach).

The situation is substantially different in the domain of neural data analysis. There classification is most often used not to construct a reliable mapping from data into behavioral variable(s) with as small error as possible, but rather to show that learnt mapping is good enough to claim that such mapping exists and data carries the effects caused by the corresponding experiment. Such an existence claim is conventionally verified with a classical methodology of null-hypothesis (H0) significance testing (NHST), whenever the achievement of generalization performance with statistically significant excursion away from the chance-level is taken as the proof that data carries effects of interest.

The main conceptual problem with NHST is a widespread belief that having observed the data, the level of significance at which H0 could be rejected is equivalent to the probability of the H0 being true. I.e. if it is unlikely that data comes from H0, it is as unlikely for H0 being true. Such assumptions were shown to be generally wrong using deductive and Bayesian reasoning since P(D|H0) not equal P(H0|D) (unless P(D)==P(H0)). Moreover, statistical significance alone, taken without accompanying support on viability and reproducibility of a given finding, was argued more likely to be incorrect.

What differs multivariate analysis from univariate is that it

• avoids multiple comparisons problem in NHST
• has higher flexibility, thus lower stability

Multivariate methods became very popular in the last decade of neuroimaging research partially due to their inherent ability to avoid multiple comparisons issue, which is a flagman of difficulties while going for a fishing expedition with univariate methods. Performing cross-validation on entire ROI or even full-brain allowed people to state presence of so desired effects without defending chosen critical value against multiple-comparisons. Unfortunately, as there is no such thing as free lunch, ability to work with all observable data at once came at a price for multivariate methods.

The second peculiarity of the application of statistical learning in psychological research is the actual neural data which researchers are doomed to analyze. As we have already seen from previous tutorial parts, typical fMRI data has

• relatively low number of samples (up to few thousands in total)
• relatively large dimensionality (tens of thousands)
• small signal-to-noise ratio
• non-independent measurements
• unknown ground-truth (either there is an effect at all, or if there is – what is inherent bias/error)
• unknown nature of the signal, since BOLD effect is not entirely understood.

In the following part of the tutorial we will investigate the effects of some of those factors on classification performance with simple (or not so) examples. But first lets overview the tools and methodologies for NHST commonly employed.

Statistical Tools in Python¶

scipy Python module is an umbrella project to cover the majority of core functionality for scientific computing in Python. In turn, stats submodule covers a wide range of continuous and discrete distributions and statistical functions.

Exercise

Glance over the scipy.stats documentation for what statistical functions and distributions families it provides. If you feel challenged, try to figure out what is the meaning/application of rdist().

The most popular distribution employed for NHST in the context of statistical learning, is binom for testing either generalization performance of the classifier on independent data could provide evidence that the data contains the effects of interest.

Note

scipy.stats provides function binom_test(), but that one was devised only for doing two-sides tests, thus is not directly applicable for testing generalization performance where we aim at the tail with lower than chance performance values.

Exercise

Think about scenarios when could you achieve strong and very significant mis-classification performance, i.e. when, for instance, binary classifier tends to generalize into the other category. What could it mean?

binom whenever instantiated with the parameters of the distribution (which are number of trials, probability of success on each trial), it provides you ability to easily compute a variety of statistics of that distribution. For instance, if we want to know, what would be the probability of having achieved 57 of more correct responses out of 100 trials, we need to use a survival function (1-cdf) to obtain the weight of the right tail including 57 (i.e. query for survival function of 56):

>>> from scipy.stats import binom
>>> binom100 = binom(100, 1./2)
>>> print '%.3g' % binom100.sf(56)
0.0967

Apparently obtaining 57 correct out 100 cannot be considered significantly good performance by anyone. Lets investigate how many correct responses we need to reach the level of ‘significance’ and use inverse survival function:

>>> binom100.isf(0.05) + 1
59.0
>>> binom100.isf(0.01) + 1
63.0

So, depending on your believe and prior support for your hypothesis and data you should get at least 59-63 correct responses from a 100 trials to claim the existence of the effects. Someone could rephrase above observation that to achieve significant performance you needed an effect size of 9-13 correspondingly for those two levels of significance.

Exercise

Plot a curve of effect sizes (number of correct predictions above chance-level) vs a number of trials at significance level of 0.05 for a range of trial numbers from 4 to 1000. Plot %-accuracy vs number of trials for the same range in a separate plot. TODO

Dataset Exploration for Confounds¶

Unfortunately it is impossible to detect and warn about all possible sources of confounds which would invalidate NHST based on a simple parametric binomial test. As a first step, it is always useful to inspect your data for possible sources of samples non-independence, especially if your results are not strikingly convincing or too provocative. Possible obvious problems could be:

• dis-balanced testing sets (usually non-equal number of samples for each label in any given chunk of data)
• order effects: either preference of having samples of particular target in a specific location or the actual order of targets

To allow for easy inspection of dataset to prevent such obvious confounds, summary() function (also a method of any Dataset) was constructed. Lets have yet another look at our 8-categories dataset:

>>> from mvpa2.tutorial_suite import *
>>> ds = get_haxby2001_data(roi='vt')
>>> print ds.summary()
Dataset: 16x577@float64, <sa: chunks,run,runtype,subj,targets,task,time_coords,time_indices>, <fa: voxel_indices>, <a: imgaffine,imghdr,imgtype,mapper,voxel_dim,voxel_eldim>
stats: mean=11.5788 std=13.7772 var=189.811 min=-49.5554 max=97.292

Counts of targets in each chunk:
chunks\targets bottle cat chair face house scissors scrambledpix shoe
---  ---  ---   ---  ---     ---        ---      ---
0+2+4+6+8+10      1    1    1     1    1       1          1        1
1+3+5+7+9+11      1    1    1     1    1       1          1        1

Summary for targets across chunks
targets  mean std min max #chunks
bottle      1   0   1   1     2
cat       1   0   1   1     2
chair      1   0   1   1     2
face       1   0   1   1     2
house      1   0   1   1     2
scissors     1   0   1   1     2
scrambledpix   1   0   1   1     2
shoe       1   0   1   1     2

Summary for chunks across targets
chunks   mean std min max #targets
0+2+4+6+8+10   1   0   1   1      8
1+3+5+7+9+11   1   0   1   1      8
Sequence statistics for 16 entries from set ['bottle', 'cat', 'chair', 'face', 'house', 'scissors', 'scrambledpix', 'shoe']
Counter-balance table for orders up to 2:
Targets/Order O1                |  O2                |
bottle:     0 2 0 0 0 0 0 0  |   0 0 2 0 0 0 0 0  |
cat:      0 0 2 0 0 0 0 0  |   0 0 0 2 0 0 0 0  |
chair:     0 0 0 2 0 0 0 0  |   0 0 0 0 2 0 0 0  |
face:      0 0 0 0 2 0 0 0  |   0 0 0 0 0 2 0 0  |
house:     0 0 0 0 0 2 0 0  |   0 0 0 0 0 0 2 0  |
scissors:    0 0 0 0 0 0 2 0  |   0 0 0 0 0 0 0 2  |
scrambledpix:  0 0 0 0 0 0 0 2  |   1 0 0 0 0 0 0 0  |
shoe:      1 0 0 0 0 0 0 0  |   0 1 0 0 0 0 0 0  |
Correlations: min=-0.52 max=1 mean=-0.067 sum(abs)=5.7

You can see that labels were balanced across chunks – i.e. that each chunk has an equal number of samples of each target label, and that samples of different labels are evenly distributed across chunks. TODO...

Counter-balance table shows either there were any order effects among conditions. In this case we had only two instances of each label in the dataset due to the averaging of samples across blocks, so it would be more informative to look at the original sequence. To do so avoiding loading a complete dataset we would simply provide the stimuli sequence to SequenceStats for the analysis:

>>> attributes_filename = os.path.join(pymvpa_dataroot, 'attributes_literal.txt')
>>> attr = SampleAttributes(attributes_filename)
>>> targets = np.array(attr.targets)
>>> ss = SequenceStats(attr.targets)
>>> print ss
Sequence statistics for 1452 entries from set ['bottle', 'cat', 'chair', 'face', 'house', 'rest', 'scissors', 'scrambledpix', 'shoe']
Counter-balance table for orders up to 2:
Targets/Order O1                           |  O2                           |
bottle:    96  0  0  0  0  12  0  0  0  |  84  0  0  0  0  24  0  0  0  |
cat:      0 96  0  0  0  12  0  0  0  |   0 84  0  0  0  24  0  0  0  |
chair:     0  0 96  0  0  12  0  0  0  |   0  0 84  0  0  24  0  0  0  |
face:      0  0  0 96  0  12  0  0  0  |   0  0  0 84  0  24  0  0  0  |
house:     0  0  0  0 96  12  0  0  0  |   0  0  0  0 84  24  0  0  0  |
rest:     12 12 12 12 12 491 12 12 12  |  24 24 24 24 24 394 24 24 24  |
scissors:    0  0  0  0  0  12 96  0  0  |   0  0  0  0  0  24 84  0  0  |
scrambledpix:  0  0  0  0  0  12  0 96  0  |   0  0  0  0  0  24  0 84  0  |
shoe:      0  0  0  0  0  12  0  0 96  |   0  0  0  0  0  24  0  0 84  |
Correlations: min=-0.19 max=0.88 mean=-0.00069 sum(abs)=77

Order statistics look funky at first, but they would not surprise you if you recall the original design of the experiment – blocks of 8 TRs per each category, interleaved with 6 TRs of rest condition. Since samples from two adjacent blocks are far apart enough not to contribute to 2-back table (O2 table on the right), it is worth inspecting if there was any dis-balance in the order of the picture conditions blocks. It would be easy to check if we simply drop the ‘rest’ condition from consideration:

>>> print SequenceStats(targets[targets != 'rest'])
Sequence statistics for 864 entries from set ['bottle', 'cat', 'chair', 'face', 'house', 'scissors', 'scrambledpix', 'shoe']
Counter-balance table for orders up to 2:
Targets/Order O1                       |  O2                       |
bottle:    96  2  1  2  2  3  0  2  |  84  4  2  4  4  6  0  4  |
cat:      2 96  1  1  1  1  4  2  |   4 84  2  2  2  2  8  4  |
chair:     2  3 96  1  1  2  1  2  |   4  6 84  2  2  4  2  4  |
face:      0  3  3 96  1  1  2  2  |   0  6  6 84  2  2  4  4  |
house:     0  1  2  2 96  2  4  1  |   0  2  4  4 84  4  8  2  |
scissors:    3  0  2  3  1 96  0  2  |   6  0  4  6  2 84  0  4  |
scrambledpix:  2  1  1  2  3  2 96  1  |   4  2  2  4  6  4 84  2  |
shoe:      3  2  2  1  3  0  1 96  |   6  4  4  2  6  0  2 84  |
Correlations: min=-0.3 max=0.87 mean=-0.0012 sum(abs)=59

TODO

Exercise

Generate few ‘designs’ consisting of varying condition sequences and assess their counter-balance. Generate some random designs using random number generators or permutation functions provided in numpy.random and assess their counter-balance.

Some sources of confounds might be hard to detect or to eliminate:

• dependent variable is assessed after data has been collected (RT, ACC, etc) so it might be hard to guarantee equal sampling across different splits of the data.
• motion effects, if motion is correlated with the design, might introduce major confounds into the signal. With multivariate analysis the problem becomes even more sever due to the high sensitivity of multivariate methods and the fact that motion effects might be impossible to eliminate entirely since they are strongly non-linear. So, even if you regress out whatever number of descriptors describing motion (mean displacement, angles, shifts, etc.) you would not be able to eliminate motion effects entirely. And that residual variance from motion spread through the entire volume might contribute to your generalization performance.

Exercise

Inspect the arguments of generic interface of all splitters Splitter for a possible workaround in the case of dis-balanced targets.

Therefore, before the analysis on the actual fMRI data, it might be worth inspecting what kind of generalization performance you might obtain if you operate simply on the confounds (e.g. motion parameters and effects).

Hypothesis Testing¶

Note

When thinking about what critical value to choose for NHST keep such guidelines from NHST inventor, Dr.Fisher in mind. For significance range ‘0.2 - 0.5’ he says: “judged significant, though barely so; ... these data do not, however, demonstrate the point beyond possibility of doubt”.

Ways to assess by-chance null-hypothesis distribution of measures range from fixed, to estimated parametric, to non-parametric permutation testing. Unfortunately not a single way provides an ultimate testing facility to be applied blindly to any chosen problem without investigating the appropriateness of the data at hand (see previous section). Every kind of Measure provides an easy way to trigger assessment of statistical significance by specifying null_dist parameter with a distribution estimator. After a given measure is computed, the corresponding p-value(s) for the returned value(s) could be accessed at ca.null_prob.

“Applications of permutation testing methods to single subject fMRI require modelling the temporal auto-correlation in the time series.”

Exercise

Try to assess significance of the finding on two problematic categories from 8-categories dataset without averaging the samples within the blocks of the same target. Even non-parametric test should be overly optimistic (forgotten exchangeability requirement for parametric testing, such as multiple samples within a block for a block design)... TODO

Independent Samples¶

Since “voodoo correlations” paper, most of the literature in brain imaging is seems to became more careful in avoiding “double-dipping” and keeping their testing data independent from training data, which is one of the major concerns for doing valid hypothesis testing later on. Not much attention is given though to independence of samples aspect – i.e. not only samples in testing set should be independent from training ones, but, to make binomial distribution testing valid, testing samples should be independent from each other as well. The reason is simple – number of the testing samples defines the width of the null-chance distribution, but consider the limiting case where all testing samples are heavily non-independent, consider them to be a 1000 instances of the same sample. Canonical binomial distribution would be very narrow, although effectively it is just 1 independent sample being tested, thus ... TODO

Statistical Treatment of Sensitivities¶

Note

Statistical learning is about constructing reliable models to describe the data, and not really to reason either data is noise.

Note

How do we decide to threshold sensitivities, remind them searchlight results with strong bimodal distributions, distribution outside of the brain as a true by-chance. May be reiterate that sensitivities of bogus model are bogus

Moreover, constructed mapping with barely above-chance performance is often further analyzed for its sensitivity to the input variables.