Is there a common method used to determine how many training samples are required to train a classifier (an LDA in this case) to obtain a minimum threshold generalization accuracy?

I am asking because I would like to minimize the calibration time usually required in a brain-computer interface.

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    $\begingroup$ user2030669, @cbeleites answer below is superb but as a rough rule of thumb: you need at least 6 times the number of cases (samples) as features. $\endgroup$
    – BGreene
    Mar 7, 2013 at 14:48
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    $\begingroup$ ... in each class. I've also seen recommendations of 5p and 3p / class. $\endgroup$ Mar 7, 2013 at 20:02

2 Answers 2


The search term you are looking for is "learning curve", which gives the (average) model performance as function of the training sample size.

Learning curves depend on a lot of things, e.g.

  • classification method
  • complexity of the classifier
  • how well the classes are separated.

(I think for two-class LDA you may be able to derive some theoretical power calculations, but the crucial fact is always whether your data actually meets the "equal COV multivariate normal" assumption. I'd go for some simulation on for both LDA assumptions and resampling of your already existing data).

There are two aspects of the performance of a classifier trained on a finite sample size $n$ (as usual),

  • bias, i.e. on average a classifier trained on $n$ training samples is worse than the classifier trained on $n = \infty$ training cases (this is usually meant by learning curve), and
  • variance: a given training set of $n$ cases may lead to quite different model performance.
    Even with few cases, you may be lucky and get good results. Or you have bad luck and get a really bad classifier.
    As usual, this variance decreases with incresing training sample size $n$.

Another aspect that you may need to take into account is that it is usually not enough to train a good classifier, but you also need to prove that the classifier is good (or good enough). So you need to plan also the sample size needed for validation with a given precision. If you need to give these results as fraction of successes among so many test cases (e.g. producer's or consumer's accuracy / precision / sensitivity / positive predictive value), and the underlying classification task is rather easy, this can need more independent cases than training of a good model.

As a rule of thumb, for training, the sample size is usually discussed in relation to model complexity (number of cases : number of variates), whereas absolute bounds on the test sample size can be given for a required precision of the performance measurement.

Here's a paper, where we explained these things in more detail, and also discuss how to constuct learning curves:
Beleites, C. and Neugebauer, U. and Bocklitz, T. and Krafft, C. and Popp, J.: Sample size planning for classification models. Anal Chim Acta, 2013, 760, 25-33.
DOI: 10.1016/j.aca.2012.11.007
accepted manuscript on arXiv: 1211.1323

This is the "teaser", showing an easy classification problem (we actually have one easy distinction like this in our classification problem, but other classes are far more difficult to distinguish): teaser sample size planning paper

We did not try to extrapolate to larger training sample sizes to determine how much more training cases are needed, because the test sample sizes are our bottleneck, and larger training sample sizes would let us construct more complex models, so extrapolation is questionable. For the kind of data sets I have, I'd approach this iteratively, measuring a bunch of new cases, showing how much things improved, measure more cases, and so on.

This may be different for you, but the paper contains literature references to papers using extrapolation to higher sample sizes in order to estimate the required number of samples.

  • $\begingroup$ would using a regularization scheme for my LDA allow me to work with a smaller training set? $\endgroup$
    – Lunat1c
    Mar 7, 2013 at 9:55
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    $\begingroup$ @user2036690, A more parsimonious model (less features) would need less training samples. A regularization scheme would not affect the number of samples needed, if would just reduce the impact of less important features. Some kind of feature rationalization may allow a smaller training set $\endgroup$
    – BGreene
    Mar 7, 2013 at 15:50
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    $\begingroup$ Yet data-driven feature selection needs enormous amounts of samples as each model comparison is in fact a statistical test. Feature selection by expert knowledge, however, can help immediately. @BGreene: can you expand why regularization cannot help reducing the sample size requirements (e.g. considering a ridge on an ill-conditioned covariance matrix)? IMHO it cannot do wonders, but it can help. $\endgroup$ Mar 7, 2013 at 17:27
  • $\begingroup$ Well without getting into an epic discussion, I was referring to Friedman's regularization formulation rather than ridge or other penalized regression. But either way coefficients are not reduced to zero as in Lasso so the dimensionality is unaffected which would as a result have no effect on the sample size needed to avoid an ill posed matrix as you refer to above. Apologies if this seems rambling $\endgroup$
    – BGreene
    Mar 7, 2013 at 19:49
  • $\begingroup$ @BGreene: no rambling, I did ask back. The interesting question is: how much are the overall d.f./complexity reduced by setting coefficients to zero in a data-driven way. Anyways, we're drifting into a different story... $\endgroup$ Mar 7, 2013 at 20:06

Asking about training sample size implies you are going to hold back data for model validation. This is an unstable process requiring a huge sample size. Strong internal validation with the bootstrap is often preferred. If you choose that path you need to only compute the one sample size. As @cbeleites so nicely stated this is often an "events per candidate variable" assessment, but you need a minimum of 96 observations to accurately predict the probability of a binary outcome even if there are no features to be examined [this is to achieve of 0.95 confidence margin of error of 0.1 in estimating the actual marginal probability that Y=1].

It is important to consider proper scoring rules for accuracy assessment (e.g., Brier score and log likelihood/deviance). Also make sure you really want to classify observations as opposed to estimating membership probability. The latter is almost always more useful as it allows a gray zone.


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