# Emperical mean when using Gaussian Process regression

Say one has $$n = 100$$ values of an unknown function $$f$$ depending on one random variable $$x$$. One can compute its emperical mean as: $$\mathbb{E}_n[f] = \frac{1}{n} \sum\limits_{i = 1}^{n}f(x_i)$$. Now, one builds a surrogate model using a Gaussian Process Regression using these $$n$$ values denoted by $$g$$. Suppose we know the pdf of the input variable $$x$$ and we want to compute the emperical mean obtained by using the GPR model through the Monte-Carlo process $$\mathbb{E}_N[g] = \sum\limits_{i = 1}^{N}g(x_i)$$ with $$N >>1$$. Do we know if $$\mathbb{E}_N[g] = \mathbb{E}_n[f]$$ ?

• It's not clear to me what you're asking. Is $g$ is a GP that is fit to $\{(x_i, f(x_i)) : i=1,\dots, n\}$? What are you sampling for the Monte-Carlo process? Are you sampling from the GP posterior? Or new points $X_i$?
– jld
Commented Mar 25, 2021 at 18:16
• After re-reading what I wrote, it is not very clear. But yes, it is a GP that fits ${(x_i,f(x_i)):i=1,…,n}$. Then using this metamodel, we compute the mean from new points $X_i$ following the pdf. I guess we will not get the same mean but it might be possible to quantify the difference. Commented Mar 25, 2021 at 18:44
• ok, that helps. Are you then looking at the mean of the GP at a particular arbitrary input, say $x_0$, or more like the mean value over the whole support of the process?
– jld
Commented Mar 25, 2021 at 18:50
• Each $g(x_i)$ in $\mathbb{E}[g]$ is computed as the mean of the GP at $x_i$. My main question is that say you have 2 models, $f$ and $g$, where $f$ is very complicated and very expensive and where $g$ is a GP regression model build using $n$ observations of $f$. Moreover, say you know the pdf of the input variable $x$, you can compute the mean -through MC for instance- using the metamodels $f$ and $g$.The problem is evaluating $f$ is much more expensive than using $g$. Thus, I would like to know if it is possible to get a bound between the true mean $\mathbb{E}[f]$ and $\mathbb{E}_N[g]$. Commented Mar 25, 2021 at 19:28
• Or between $\mathbb{E}[f]$ and $\mathbb{E}[g]$ (the true mean of the $g$) if we use many many evaluations of $g$ because it is very cheap. Commented Mar 25, 2021 at 19:34