Confusion related to predictive distribution of gaussian processes I have this confusion related to the predictive distribution of gaussian process. I was reading this paper 

I didn't get how the integration gave that result. What is P(u*|x*,u). Also how come the covariance of the posterior distribution is $\sigma^2(\sigma^2I+K)^{-1}K$
 A: $P(u*|x*,u) ~ N(u(x*)$, $\sigma^2$), directly from the definition of $u*$.
Notice that integration of two Gaussian pdf is normalized. It can be shown from the fact that
$$
\int_{-\infty}^{\infty}P(u^*|x^*, u)du^* =\int_{-\infty}^{\infty}\int_{u}P(u^*|x^*, u)P(u|s)dudu^*
=\int_{u}P(u|s)\int_{-\infty}^{\infty}P(u^*|x^*, u)du^*du
=\int_{u}P(u|s)\int_{-\infty}^{\infty}N(u^*-u(x*); 0, \sigma^2)du^*du
=\int_{u}P(u|s)du\int_{-\infty}^{\infty}N(u^*; 0, \sigma^2)du^*
=1
$$
With normalization out of the way, 
$\int_{u}P(u^*|x^*, u)P(u|s)du$ is integrated by the following tips:  


*

*Substitute the 2 normal pdf into the equation and eliminate the terms independent of $u$, as we have already shown normalization.

*Using the completing the square trick for integrating multivariate exponential, i.e., construct a multivariate normal pdf with the remaining exponential terms. Refer to this youTube video.

*Eventually you are left with an exponential in terms of $u^*$, it can be observed that this is again a factor away from a normal pdf. Again, the proof of normalization gives us confidence that the final form is indeed a normal pdf. The pdf is the same as the one given in the original post.
A: The detailed derivations of the equations for the conditional distribution of a Gaussian process can be found in chapter 2 and appendix A of the book [Rasmussen2005].
Take a look at (Eq. 2.23, 2.24) and above, which are based on the Gaussian identities (A.6) and the matrix property (A.11).

[Rasmussen2005] C. E. Rasmussen and C. Williams. Gaussian Processes for Machine Learning. MIT Press, 2005.
