Why does the predictive distribution involve no $\textit{new}$ data?

Question

Given data $(\textbf{X,Y})$, a Gaussian Process $\textbf{F}$ that is normally distributed and weights $\textbf{W}_1$ and $\textbf{b}$ also from a distribution, how come is the following distribution

\begin{align} p(\textbf{Y}|\textbf{X}) = \int p(\textbf{Y}|\textbf{F})p(\textbf{F}|\textbf{W}_1,\textbf{b},\textbf{X})p(\textbf{W}_1)p(\textbf{b}) \label{ref1} \end{align}

a predictive distribution?

Because I thought that a predictive distribution is always, given the data $(\textbf{X,Y})$, the propability of the target label $y^*$ given the new data point $x^*$ data, such as

\begin{align*} p(y^*|x^*,\textbf{X,Y}) = \int p(y^*|x^*,\pmb{\omega}) p(\pmb{\omega}|\textbf{X,Y}) \, d \pmb{\omega} \end{align*}

for some parameters $\pmb{\omega}$.

Currently Im investigating this paper, and in it (section 3.1.) I tripped over the above, for me incomprehensible, predictive distribution.

Anyone has a trick up their sleeve and knows by chance whats going on here?

Cheers

Answer 1

It's all about bayesian terminology.

In bayesian statistics the name 'predictive distribution' is used for new data points and for old (training) ones as well. It is an established diagnositic procedure to compare observed y values to their predictive distribution, and in-sample data is used more often than not.