I am reading a paper that is doing approximate inference over GP models. The idea is that when the likelihood is non-Gaussian than the inference of the posterior is not tractable and we need to use approximate inference schemes to compute the posterior distribution.

There is one bit in the paper that I do not follow. The author says:

In order to benefit from the advantages of MCMC it is necessary to develop efficient sampling strategies. This has proved to be particularly difficult in many GP applications that involve the estimation of a smooth latent function. Given that the latent function is represented by a discrete set of values, the posterior distribution over these function values can be highly correlated. The more discrete values used to represent the function, the worse the problem of high correlation becomes.

Perhaps this is very naive but why does the discreteness of the function increase the correlation?


2 Answers 2


The key word here is "smooth". Let's take a simple case where the values are on $[0,1]$ and we want to estimate a random function $f(x)$ with $x \in [0,1]$. If $f$ is differentiable, then the covariance function $k(x_t,x_s)$ will be differentiable in each argument. This is true whether the process is Gaussian or not, just as long as it has a finite covariance function.

When I sample more values on the interval, I get values that are closer together and this means that $f(x_{t+\epsilon})$ is almost a linear function of $f(x_t)$, and the covariance function at $x_{t+\epsilon}$ is almost a linear function of the covariance function at $x_t$. This means that the discrete covariance matrix of the sampled values will be almost singular.

All of this applies to Gaussian processes just as much as non-Gaussian processes.

Perhaps another way to look at this would be to observe that when you have a smooth function, sampling from the $x$ values has diminishing returns beyond a certain point. You get more values, but you don't get more information, since you are already well placed to estimate $f(x)$. This is the downside to all those neat convergence theorems that we learn in second year Calculus.

  • $\begingroup$ I get it now. Thank you for that clear explanation. $\endgroup$
    – Luca
    Commented Oct 3, 2015 at 12:43

In practice you might actually run into an extreme case of this problem: if you need to learn a function that you know is noiseless (i.e. no noise term added to the diagonal of the covariance matrix) and if your dataset happens to contain the same sample twice (with the same function value), then the covariance matrix will be degenerate, i.e. does not specify a proper probability distribution.

The argument you are asking about is just a softer version of this problem: there the inputs are not identical, but very close together (and more likely so if you sample a lot of them). The smoother a function is, the more correlated neighboring points will be, and hence the covariance matrix (and probability distribution) built from these points will be closer to degeneracy.


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