Consider the model $$ y(\mathbf{X}) = f(\mathbf{X}) + \epsilon, $$ where $\mathbf{X}$ is a given $n\times D$ matrix, and where $\epsilon \sim \mathcal{N}(0, \sigma^{2}I_{N})$ is iid Gaussian and is the noise.
First, by modeling the function $f$ as a GP, such that $f(\cdot)\sim GP(0, K(\cdot, \cdot))$ where K is a chosen covariance function. It can be shown (see for instance Rasmussen & Williams, 2006) that the prediction for a new point $\mathbf{X}$ is normal distributed so that $$ f(\mathbf{x})|y(\mathbf{X}),\mathbf{X},\mathbf{x} \sim \mathcal{N}(K(\mathbf{x}, \mathbf{X})[K(\mathbf{X}, \mathbf{X}) + \sigma^{2}I_N]^{-1}\mathbf{Y}, K(\mathbf{x}, \mathbf{x})-K(\mathbf{x}, \mathbf{X})[K(\mathbf{X}, \mathbf{X}) + \sigma^{2}I_N]^{-1}K(\mathbf{x}, \mathbf{X})).$$
Now, instead of modeling the function with a GP, one wants to minimize the following Kernel Ridge Regression (KRR) equation: $$\min_{f} \sum\limits_{i=1}^{N}(f(\mathbf{X}_i) - y(\mathbf{X}_i))^{2} + \lambda||f||^{2}_{H_K},$$ where $H_K$ is the RKHS associated with the kernel $K$. The solution of this minimization is: $$ f(\mathbf{x}) = K(\mathbf{x}, \mathbf{X})[K(\mathbf{X}, \mathbf{X}) + \lambda I_N]^{-1}\mathbf{Y}, $$ which is equal to the predictive mean of the Gaussian process (under the assumption that $\lambda = \sigma_N^{2}$).
Now my questions:
It is shown in Rasmussen & Williams, 2006 that if one takes the exponential of minus the KKR equation given above, that is: $$ exp(-\frac{1}{2\lambda}\sum\limits_{i=1}^{N}(f(\mathbf{X}_i) - y(\mathbf{X}_i))^{2}) \times exp(-\frac{1}{2}||f||^{2}_{H_K})$$ it is equal to the product of a Gaussian likelihood (the least squares term) times a Gaussian process prior on $f$. Why does the term $\exp(-\frac{1}{2}||f||^{2}_{H_K})$ correspond to a Gaussian process prior on $f$?
How do you show that $$\max_f exp(-\frac{1}{2\lambda}\sum\limits_{i=1}^{N}(f(\mathbf{X}_i) - y(\mathbf{X}_i))^{2}) \times exp(-\frac{1}{2}||f||^{2}_{H_K})$$ is equal to the predictive mean of the Gaussian process prior?
