I would like to compare a parametric function to a Gaussian Process. This may sound weird, but read on:

Data description. I am looking at projections of a 3D object. However I expect a certain amount of smoothness in the true object, so it seems natural to use a GP to represent it, thinking of each image pixel as a (partial) observation of every voxel in the final object. So I've fitted my GP.

Now I want to do inference on that GP. Why? Because even though I have inferred the shape, I need to know what it actually is made of (FYI, I'm inferring a molecule's components from the observed outline). This means I need to fit a parametric function (actually a Gaussian Mixture Model) to a GP. So I would like to write some kind of likelihood of the GP given a GMM.

I'm open to suggestions, including using something besides a GP to represent my data. But note that it is impractical to fit the final model directly to raw data, because that model has many additional priors so it takes a long time to sample. The intermediate data processing is a kind of data reduction.

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    $\begingroup$ What do you mean by "a more extreme function"? Greater $L^2$ norm or something? $\endgroup$
    – Ben Kuhn
    Dec 9, 2014 at 1:18
  • $\begingroup$ Possibly :). I guess I was thinking there should be some generalization of the multivariate t-test that gives you the probability of observing a function given a GP. $\endgroup$
    – cgreen
    Dec 9, 2014 at 1:22
  • $\begingroup$ The t-test does not give the probability of observing a particular t-statistic: that would be zero. Likewise, for any Gaussian process the probability of observing any particular instance is always zero. What, then, are you trying to ask? How to write down the likelihood of some observations? How to conduct some kind of statistical test? If so, what are you trying to test? $\endgroup$
    – whuber
    Dec 19, 2014 at 19:03
  • $\begingroup$ Attempted to clarify... $\endgroup$
    – cgreen
    Dec 20, 2014 at 4:15
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    $\begingroup$ Yeah it's similar to a Radon transform, actually a "ray transform" which is 3D-to-2D. So you really are integrating along an axis to get the projection. It's formed by an electron microscope, so what you are seeing is the "shadow" of the coulomb field around the atoms formed by scattered electrons. They have a chance of scattering throughout the molecule, hence the ray transform. $\endgroup$
    – cgreen
    Dec 20, 2014 at 22:44

1 Answer 1


I looked into the in the case of a 1D GP and the parametric function $y = f(x)$ a few months ago. This was my findings:

In order to get the probability of a function, $y = f(x)$, we have to perform a line integral through the probability distribution provided by the GP. This line integral would therefore be the probability of the curve given the GP. Unfortunately, we can't do this easily as there is no way of writing the GP in closed form to perform this operation.

For this reason a sample based approach is the best option (or at least this was what I believed). I thought of to methods to consider. The prediction of the GP at a point is $O(n^2)$ for every additional prediction after training, so if $n$ is small it is cheep to sample the function and the GP at many $x$'s and Monte Carlo methods make sense.

When $n$ is very large it is expensive to perform many samples. In this case it makes sense to perform bayesian quadrature to estimate this line integral. In this situation I looked at using normal BQ sampling policies but did think at the time that there maybe an opportunity for novel acquisition functions for this exact situation.

  • $\begingroup$ Also @cgreen if you are interested in taking this type of thing further I might have some useful information that I was looking at back then. If you want to talk about this at a later stage my university email is on the home page of my website (link in my profile) so feel free to get in touch. $\endgroup$
    – j__
    Dec 20, 2014 at 22:23
  • $\begingroup$ Yes, this makes a lot of sense. Do you have any references on this approach? It's quite clear, though I'm a little fearful now of the computation time (since it's likely I will have to make the calculation frequently In order to optimize the model). Thanks! $\endgroup$
    – cgreen
    Dec 22, 2014 at 0:30

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