1
$\begingroup$

I am a neuroscientist, and have neuronal response measurements for a large range of stimuli delivered to my subjects. I use this data to construct receptive fields (RFs), which show the mean response magnitude as a function of stimulus value. So, two RFs, for two different neurons, when they are shown stimuli from a 2-d stimulus space, might look like this:

two receptive fields

I have multiple presentations of each stimulus (here I am representing each stimulus as an $(x,y)$ pair: you can consider it as the location of a light presented to the eye, and the height is the mean neuronal response to the light presented at that location).

Note that the two receptive fields above are not probabilities, but the height represents the mean response to the stimulus.

I want to test whether the receptive field centers are different for the two neurons. That is, is the location of the peak in the above two bumps different? Let's say we assume the two bumps are Gaussian in shape: what is the best way to test the null hypothesis that the two RF centers are identical?

If all else fails, I could do some kind of bootstrap, where I sample with replacement from the set of all $(x_i, y_i, r_i)$ triplets from each group. But even then, I would be back to square one of figuring out what statistical test to apply to these two samples of the original data.

Potentially Relevant Links:

$\endgroup$
  • 2
    $\begingroup$ Neuroscientist here as well. Instead of bootstrapping, you could do a permutation-based test. Choose a test statistic, e.g. fit a Gaussian to each cell and take distance $d$ between the peaks. Then pool all your $(x,y,r)$ measurements for both cells, randomly split them in two (as if these two groups were two cells), compute the same statistic $d$, and repeat many times. You will get a null distribution of your distances $d$ and can directly get a p-value for your actual empirical $d$ being larger than expected under null. $\endgroup$ – amoeba Apr 13 '15 at 13:10
  • $\begingroup$ That is very clean and simple: nice idea! You should consider turning into an answer. :) Have you seen it done that way before? I.e., is there a reference out there on this, specifically wrt RFs? I would encourage you to turn this into an answer :) $\endgroup$ – neuronet Apr 13 '15 at 13:14
  • $\begingroup$ I don't have a reference, sorry. I am not working with RFs myself, so I don't really know what people do, this was just the first/easiest thing that came to my mind. I might post it as an answer a bit later. $\endgroup$ – amoeba Apr 13 '15 at 14:13
  • $\begingroup$ One nice thing about your answer is that it is relatively easy to extend to N>2 neurons (or N>2 recording sessions with the same neuron). And it can be applied to essentially any descriptive statistic you care to compare. $\endgroup$ – neuronet Apr 13 '15 at 14:30
2
$\begingroup$

I am posting my earlier comment as an answer.

The easiest thing to do would probably be a permutation-based test. Choose a test statistic, e.g. fit a Gaussian to each receptive field and take distance $d$ between the peaks as your test statistic. Then pool all your $(x,y,r)$ measurements for both cells, randomly split them in two (as if these two groups were two cells) in the same proportion as in the actual data, compute the same statistic $d$, and repeat many times with different random splits. You will get a null distribution $f_0(d)$ of your distances $d$, and can directly get a $p$-value for your actual empirical $d$ being larger than expected under null.

This is a very general framework that can work in almost any circumstances.

PS. Wikipedia insists that it is only called "permutation test" when you actually loop over all possible permutations; otherwise it should (apparently) be called "approximate permutation tests, Monte Carlo permutation tests or random permutation tests". I find it weird; my understanding is that people do call it "permutation tests", and I don't see anything wrong with it.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.