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I'm reading the GPML book and in Chapter 2 (page 15), it tells how to do regression using Gaussian Process(GP), but I'm having a hard time figuring how it works.

In Bayesian inference for parametric models, we first choose a prior on the model parameters $\theta$, that is $p(\theta)$; second, given the training data $D$, we compute the likelihood $p(D|\theta)$; and finally we have the posterior of $\theta$ as $p(\theta|D)$, which will be used in the predictive distribution $$p(y^*|x^*,D)=\int p(y^*|x^*,\theta)p(\theta|D)d\theta$$, and the above is what we do in Bayesian inference for parametric models, right?

Well, as said in the book, GP is non-parametric, and so far as I understand it, after specifying the mean function $m(x)$ and the covariance function $k(x,x')$, we have a GP over function $f$, $$f \sim GP(m,k)$$, and this is the prior of $f$. Now I have a noise-free training data set $$D=\{(x_1,f_1),...,(x_n,f_n)\}$$, I thought I should compute the likelihood $p(D|f)$ and then the posterior $p(f|D)$, and finally use the posterior to make predictions.

HOWEVER, that's not what the book does! I mean, after specifying the prior $p(f)$, it doesn't compute the likelihood and posterior, but just go straight forward to the predictive prediction.

Question:

1) Why not compute the likelihood and posterior? Just because GP is non-parametric, so we don't do that?

2) As what is done in the book (page 15~16), it derives the predictive distribution via the joint distribution of training data set $\textbf f$ and test data set $\textbf f^*$, which is termed as joint prior. Alright, this confuses me badly, why joint them together?

3) I saw some articles call $f$ the latent variable, why?

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  • $\begingroup$ Personally, I don't think GP regression belongs to Bayesian inference, since it doesn't follow the steps in Bayesian approach. The so-called predictive distribution in GP is derived by jointing the training and test data in the prior, and then conditioning on the training data, it doesn't make use of likelihood or posterior. $\endgroup$
    – avocado
    Commented Jan 26, 2014 at 7:29

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and the above is what we do in Bayesian inference for parametric models, right?

The book is using Bayesian model averaging, which is the same for parametric models or any other Bayesian method, given that you have posterior over your parameters.

Now I have a noise-free training data set

It doesn't need to be 'noise-free'. See later pages.

HOWEVER, that's not what the book does! I mean, after specifying the prior p(f), it doesn't compute the likelihood and posterior, but just go straight forward to the predictive prediction.

See this: https://people.cs.umass.edu/~wallach/talks/gp_intro.pdf

I believe, in page 17 we have the prior, and later the likelihood. I believe if you write the derivations, and find the posterior, and then average over the posterior for prediction (like in the weight-space view) it will result in the same equations as in page 19 for mean and covariance.

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  • $\begingroup$ Thanks for your reply, but I've already seen that many books don't mention Bayesian at all, they just compute the conditional distribution $p(f^*|f)$, and say this is the posterior, what the hell? $\endgroup$
    – avocado
    Commented Feb 2, 2014 at 9:23
  • $\begingroup$ Finding the conditional is basically using the Bayes formula. Writing stuff in the conventional Bayesian formulation is a little cumbersome for GPs; they just refer to finding the conditional and .... $\endgroup$
    – Daniel
    Commented Feb 2, 2014 at 19:09
  • $\begingroup$ AFAIK, the conditional is computed this way, $p(x|y)=p(x,y)/p(y)$, but Bayes formula is $p(x|y)=p(y|x)p(x)/p(y)$. I don't see why finding conditional is using Bayes formula, could you please be more specific? $\endgroup$
    – avocado
    Commented Feb 3, 2014 at 9:50
  • $\begingroup$ And as you said in the comment, "writing stuff in the conventional Bayesian formulation is cumbersome for GPs", by conventional Bayesian formulation, did you mean, first compute the posterior $p(f|D)$, then compute the predictive distribution $p(f^*|D)=\int p(f^*|f)p(f|D)df$. $\endgroup$
    – avocado
    Commented Feb 3, 2014 at 9:56

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