Why is squared exponential kernel often used in Gaussian Process regression when the most standard case is time-like X?

Question

I might be confused about something.

Consider doing inference on $Y'\mid X',Y,X$ using standard Gaussian Process Regression with 1d $Y$ and 1d $X$. Suppose $X$ is time-like (target is stationary or shift invariant and that the covariance grows like $\Delta X$. This would imply a good kernel is something like $k(x, x') \propto \frac{e^{- \frac{\alpha}{|x - x'|}}}{| x - x'|}$ yet we often see the squared exponential used as a default for stationary GPR kernels which is $e^{- \frac{(x - x')^2}{2 \sigma^2}}$.

Another way of asking this question is what kernel corresponds to the GPR being a brownian bridge density?

Answer 1

I don't quite follow what the reasoning is in the question, but here are some thoughts that don't fit in a comment.

Gaussian process regression uses a Gaussian Process (GP) as the prior on the unknown function. That is, \begin{align} f &\sim \text{GP}(\mu, K), \\ y_i &= f(x_i) + \varepsilon_i, \quad i=1,\dots,n \end{align} where $\mu$ and $K$ are the mean and covariance functions of the GP, $\{x_i\}$ is a deterministic sequence of points the function is evaluated at and $\{\varepsilon_i\}$ is an i.i.d. sample from $N(0, \sigma^2)$.

The kernel you are referring to is the covariance function $K$. For example,

• For a Brownian motion in $[0,\infty)$ , we have $K(s,t) = \min(s,t)$.
• For a Brownian bridge in $[0,1]$, we have $K(s,t) = \min(s,t) - st$.

For the second assertion, e.g., see this. (Neither process is stationary.)

Neither has much to do with the Gaussian kernel (the exponential kernel) mentioned in the question.