A rank-one finite-dimensional Gaussian process is a tight Banach prior

With $v=(1,2)$, the mean is zero and $E[X_i{X}_j]=v_i{v}_j$. The kernel $k(i,j)=v_i{v}_j$ is positive semidefinite because for all $a_1,a_2$, ${\sum }_{i,j}{a}_i{a}_{j}k(i,j)=(a_1+2a_2{)}^2\ge 0$. Thus [F1] gives the same possibly singular Gaussian process law as the explicit vector.

The map $z\mapsto (z,2z)$ is continuous, so its pushforward of the standard normal law is a Borel probability on $B$. For every positive integer $R$, the events $\{\mathrm{\mid }Z\mathrm{\mid }\le R\}$ increase to the whole space; continuity from below [F4] therefore lets us choose $R$ with $P(\mathrm{\mid }Z\mathrm{\mid }\le R)>1-\delta$. The image segment $\{(z,2z):\mathrm{\mid }z\mathrm{\mid }\le R\}$ is compact and has probability above $1-\delta$, so the law is tight.

For ${\mathrm{\ell }}_a(x)=a_1{x}_1+a_2{x}_2$, ${\mathrm{\ell }}_a(X)=(a_1+2a_2)Z$, and [F2] gives $C{\mathrm{\ell }}_a=E[{\mathrm{\ell }}_a(X)X]=(a_1+2a_2)(1,2)\mathrm{.}$ The range is exactly $\mathrm{span}\{(1,2)\}$. If $h=c(1,2)$, then every representative with $a_1+2a_2=c$ has covariance norm squared $E[{\mathrm{\ell }}_a(X{)}^2]=c^2$, so $\mathrm{\parallel }h{\mathrm{\parallel }}_H=\mathrm{\mid }c\mathrm{\mid }$.

The law is concentrated on this line. Every open ball around $c(1,2)$ meets the line in an interval of positive normal probability, while a point outside the line has a sufficiently small ball disjoint from it. Hence the support is the line, equal to the RKHS. Since $\mathrm{\parallel }X{\mathrm{\parallel }}^2=5Z^2$, [F2] and [F5] give $E{e}^{a\mathrm{\parallel }X{\mathrm{\parallel }}^2}=E{e}^{5a{Z}^2}=\frac{1}{\sqrt{2\pi }}{\int }_{\mathbb{R}}{e}^{-(1\mathrm{/}2-5a)z^2}dz=(1-10a{)}^{-1\mathrm{/}2}<\mathrm{\infty }$ for $0<a<1\mathrm{/}10$.