1
$\begingroup$

I am reading the chapter on Gaussian process regression from this book.

As I understand it, Gaussian processes are extending multivariate Gaussian to infinity.

While multivariate Gaussians are distributions over vectors, Gaussian processes are distributions over infinite dimensional vectors, so basically functions.

Lets say we are given an input $X$ with corresponding output values $y$. We are also given a test set $X_{*}$ and we want to predict the output values $y_{*}$.

We assume that both $y$ and $y_{*}$ are Gaussian distributed.

The joint prior distribution is given by:

$\begin{bmatrix} y\\ y_{*} \end{bmatrix} \sim \mathcal{N} \begin{pmatrix} 0,& \begin{bmatrix} K(X,X) & K(X,X_{*})\\ K(X_{*},X) & K(X_{*},X_{*}) \end{bmatrix} \end{pmatrix}.$

where $K$ is the covariance matrix.

In order to get the posterior distribution we condition the prior distribution on the observations $y$.

$y_{*}|X_{*},X,y \sim \mathcal{N}(\mu, \Sigma)$

where $\mu = K(X_{*},X)K(X,X)^{-1}y$ and $\Sigma = K(X_{*},X_{*})-K(X_{*},X)K(X,X)^{-1}K(X,X_{*})$.

In order to predict the $y_{*}$ values, we can just sample from the posterior distribution.

I believe that I understand it this far, please correct me if I am wrong.

But how does a drawn sample look like?

$y_{*}|X_{*},X,y$ is a distribution over the vector where each component $y_{*i}$ is the corresponding prediction for the input $\mathbf{x}_{*i} \in X_{*}$, correct?

$\mu$ is a vector that contains a $\mu_{i}$ for every $\mathbf{x}_{*i} \in X_{*}$, lets say $\mu$ is $k$ x $1$.

So $y_{*}|X_{*},X,y$ is a distribution over a $k$ x $1$ vector ?

That doesn't sound right. I think that every sample should be a function.

I am bit confused.

$\endgroup$
  • 1
    $\begingroup$ Every prediction $y_{*,i}$ is the function evaluated at $x_{*,i}$, hence a number. This is why you end up with a $k$ vector when you evaluate the function at the $k$ test points in $X_*$. $\endgroup$ – combo Dec 23 '17 at 22:13

Your Answer

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

Browse other questions tagged or ask your own question.