# Meaning of syntax $N(\mathbf{y} \mid \mathbf{0}, \mathbf{K})$ (multivariate normal distribution)

So I'm reading notes on Gaussian Processses, and came across syntax $p(\mathbf{y} \mid \text{stuff}) = N(\mathbf{y} \mid \mathbf{0}, \mathbf{K})$ for multivariate normal distribution, and I'm not exactly sure how to decipher it the $\mathbf{y} |$ bit. I suppose it doesn't mean "mean $\mathbf{y}$ is conditional on zero vector", because I don't think that makes sense. Assuming this is standard notation for something and I'm just unfamiliar with it, what it means? E.g. how it differs from $N(\mathbf{0}, \mathbf{K})$?

This means that $$\mathbf{y}$$ (a $$d$$-dimensional random variable) conditional on stuff has the distribution $$\mathrm{N}(\mathbf{0},\mathbf{K})$$, i.e., normal with mean $$\mathbf{0}\in\mathbf{R}^d$$ and covariance $$\mathbf{K}$$ (a $$d\times d$$ matrix). The notation points a difference between a (conditional) probability distribution of a random variable ($$\mathrm{N}(\mathbf{0},\mathbf{K})$$) and the corresponding (conditional) probability density function $$\mathrm{N}(\mathbf{y} \mid \mathbf{0},\mathbf{K})$$.
So, we write $$$$\mathbf{y} \mid \textrm{stuff} \sim \mathrm{N}(\mathbf{0},\mathbf{K})$$$$ to say that $$\mathbf{y}$$ conditional on stuff is normally distributed with mean $$\mathbf{0}$$ and covariance $$\mathbf{K}$$. This in turn implies that the probability density function of the conditional distribution of $$\mathbf{y}$$ given stuff is
$$$$p(\mathbf{y} \mid \textrm{stuff}) = \mathrm{N}(\mathbf{y} \mid \mathbf{0},\mathbf{K})= (2\pi)^{-d/2}\,|\mathbf{K}|^{-1/2}\,e^{-\frac{1}{2}\,\mathbf{y}^T\mathbf{K}^{-1}\mathbf{y}}.$$$$
• @JHBonarius Thanks, added remarks about the dimensions (also changed $k$ to $d$) – Juho Kokkala Aug 10 at 6:31