On the interpretation of kernels as the covariance functions of stochastic processes, whcih is one way to define stochastic processes.

Suppose we have a real-valued stochastic process

\[\{\mathsf{x}(t)\}_{t\in \mathcal{T}} \] For now we may as well take the index to be \(\mathcal{T}\subseteq\mathbb{R}^D\), or at least a nice metric space.

The covariance kernel of \(\mathsf{x}\) is a function

\[\begin{aligned}\kappa:&\mathcal{T}\times \mathcal{T}&\to& \mathbb{R}\\ &s,t&\mapsto&\operatorname{Cov}(\mathsf{x}(s),\mathsf{x}(t)). \end{aligned}\]

This is covariance in the usual sense, to wit,

\[\begin{aligned} \operatorname{Cov}(\mathsf{x}(s),\mathsf{x}(t)) &:=\mathbb{E}[\mathsf{x}(s)-\mathbb{E}[\mathsf{x}(t)]] \mathbb{E}[\mathsf{x}(t)-\mathbb{E}[\mathsf{x}(t)]]\\ &=\mathbb{E}[\mathsf{x}(s)\mathsf{x}(t)]- \mathbb{E}[\mathsf{x}(s)]\mathbb{E}[\mathsf{x}(t)]\\ \end{aligned}\]

These are useful objects. In spatial statistics, Gaussian processes, kernel machines and covariance estimation we are concerned with such covariances between values of stochastic processes at different values of their indices. The Karhunen—Loève transform decomposes stochastic processes into a basis of eigenfunctions of the covariance kernel operator. You might consider them in terms of processes defined through convolution instead.

Any process with finite second moments has a covariance function. They are especially renowned for Gaussian process methods, since Gaussian processes are uniquely specified by their mean function and covariance kernels, and also have the usual convenient algebraic properties by virtue of being Gaussian.

TODO: relate to representer theorems.

## Covariance kernels of some example processes

### A simple Markov chain

Consider a homogeneous continuous time Markov process taking values in \(\{0,1\}\). Suppose it has transition rate matrix

\[\left[\begin{array}{cc} 0 & \lambda\\ \lambda & 0 \end{array}\right] \] and moreover, that we start the chain from the stationary distribution, \([\frac 1 2\; \frac 1 2]^\top,\) which implies that \(\operatorname{Cov}(0, t)=\operatorname{Cov}(s, s+t)\) for all \(s\), and further, that \(\mathbb{E}[\mathsf{x}(t)]=\frac 1 2 \,\forall t\). So we know that \(\operatorname{Cov}(s,s+t)=\mathbb{E}[\mathsf{x}(0)\mathsf{x}(t)]- \frac 1 4.\) What is \(\mathbb{E}[\mathsf{x}(0)\mathsf{x}(t)]\)?

\[\begin{aligned} \mathbb{E}[\mathsf{x}(0)\mathsf{x}(t)] &=\mathbb{P}[\{\mathsf{x}(0)=1\}\cap\{\mathsf{x}(t)=1\}]\\ &=\mathbb{P}[\text{number of jumps on \([0,t]\) is even}]\\ &=\mathbb{P}[\mathsf{z}\text{ is even}]\text{ where } \mathsf{z}\sim\operatorname{Poisson}(\lambda t)\\ &=\sum_{k=0}^{\infty}\frac{(\lambda t)^{2k} \exp(-\lambda t)}{(2k)!}\\ &= \exp(-\lambda t) \sum_{k=0}^{\infty}\frac{(\lambda t)^{2k}}{(2k)!}\\ &= \exp(-\lambda t)(\exp(-\lambda t) + \exp(\lambda t))/2 &\text{Taylor expansion}\\ &= \frac{\exp(-2\lambda t)}{2} + \frac{1}{2} \end{aligned}\]

From this we deduce that \(\operatorname{Cov}(s,s+t)=\frac{\exp(-2\lambda t)}{2} + \frac{1}{4}.\)

Question: what functions are admissible as covariance kernels for Markov chains?

### The Hawkes process

Covariance kernels are also important in various point processes. Notably, the Hawkes process was introduced in terms of its covariance. 🚧

### Gaussian processes

Handling certain functions in terms of their covariances is particularly convenient. Specifically, Gaussian processes are simple in this context. Such processes are

## General real covariance kernels

A function \(K:\mathcal{T}\times\mathcal{T}\to\mathbb{R}\) can be a covariance kernel if

- It is symmetric in its arguments \(K(s,t)=K(t,s)\) (more generally, conjugate symmetric — \(K(s,t)=K^*(t,s)\), but I think maybe my life will be simpler if I ignore the complex case for the moment.)
- It is positive semidefinite.

That positive semidefiniteness means that for arbitrary real numbers \(c_1,\dots,c_k\) and arbitrary indices \(t_1,\dots,t_k\)

\[ \sum_{i=1}^{k} \sum_{j=1}^{k} c_{i} c_{j} K(t_{i}, t_{j}) \geq 0 \]

The interpretation here is since we need the covariances induced by the finite dimensional distributions fo this process to be consistent, it is necessary that

\[ \operatorname{Var}\left\{c_{1} X_{\mathbf{t}_{1}}+\cdots+c_{k} X_{\mathbf{t}_{k}}\right\}= \sum_{i=1}^{k} \sum_{j=1}^{k} c_{i} c_{j} K\left(\mathbf{t}_{i}, \mathbf{t}_{j}\right) \geq 0 \]

This arises from the constraint on the covariance of \(\operatorname{Var}(\mathbf X\in \mathbb{R}_+^d\) which requires that for \(\mathbf {b}\in \mathbb{R}^d\)

\[ \operatorname {var} (\mathbf {b} ^{\top}\mathbf {X} ) =\mathbf {b} ^{\top}\operatorname {var} (\mathbf {X} )\mathbf {b} \]

Question: What can we say about this covariance if every element of \(\mathbf X\) is non-negative?

Amazingly (to me), this necessary condition will also be sufficient to make something a covariance kernel. In practice designing covariance functions using positive definiteness is tricky; the space of positive definite kernels is implicit. What we normally do is find a fun class that guarantees positive definiteness and riffle through that. Most of the rest of this notebook is devoted to such classes.

## Bonus: complex covariance kernels

I talked in terms of real kernels above because I generally observer real measurements of processes. But often complex covariances arise in a natural way too.

A function \(K:\mathcal{T}\times\mathcal{T}\to\mathbb{C}\) can be a covariance kernel if

- It is symmetric
*conjugate*symmetric in its arguments — \(K(s,t)=K^*(t,s)\), - It is positive semidefinite.

That positive semidefiniteness means that for arbitrary complex numbers \(c_1,\dots,c_k\) and arbitrary indices \(t_1,\dots,t_k\)

\[ \sum_{i=1}^{k} \sum_{j=1}^{k} c_{i} \overline{c_{j}} K(t_{i}, t_{j}) \geq 0. \]

Every analytic real kernel should also be a complex kernel, right? 🏗

## Kernel zoo

For some examples of covariance kernels, see the kernel zoo.

## Learning kernels

See learning kernels.

## Non-positive kernels

As in, kernels which are not positive-definite. For example, [Ong et al. (2004);SahaLearning2020] 🏗

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