# Closed form Karhunen-Loeve/PCA expansion for gaussian/squared-exponential covariance

The Gaussian, or squared exponential covariance is $k_{SE}(s,t) = \exp \left\{ -\frac{1}{2l} (s - t)^2 \right\}$. It is a common covariance function used in Gaussian processes. The Karhunen-Loeve expansion is an orthonormal decomposition of sample paths of a Gaussian process. If $g(t)$ is a sample path from a Gaussian process with mean 0 and covariance $k(s,t)$, then $g(t) = \sum_{i=1}^\infty \xi_i f_i (t)$ where the eigenfunctions $f_i(t)$ are deterministic functions determined by $k$ and eigenvalues $\xi_i$ are the standard normals.

My question is, does there exist a closed form expression for the $f_i$ corresponding to $k_{SE}$?

According to [1], closed form expressions for $f_i$ are known for exponential covariances ($k(s,t) = \exp \left\{ -|s - t| \right\}$), band-limited stationary processes (finite sums of trigonometric functions), and Brownian motion.

[1] Huang, S. P. and Quek, S. T. and Phoon, K. K., Convergence study of the truncated Karhunen–Loeve expansion for simulation of stochastic processes, International Journal for Numerical Methods in Engineering (2001), http://dx.doi.org/10.1002/nme.255

You could try following first the standard way of deriving the solution for $k(s,t) = e^{-|s-t|}$ given for instance in [1] and then try to reapply this to your case. If you don't have the access to [1] I can give an outline here. Generally you have to carefully differentiate
$\int k_{SE}(s,t)f_i(s)dt = \lambda_i f_i(s)$
with respect to $t$ and see if it simplifies to an ODE for $f_i$. Your ODE is going to be more complicated but should be solvable (haven't done the calculation myself!).