# 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 , 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.

 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  and then try to reapply this to your case. If you don't have the access to  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!).