MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).
show/hide this revision's text 3 added 112 characters in body

What is wrong with the following simple definition? Let $g_1$ and $g_2$ be two independent standard Gaussian variables. For $t=(a,b)\in S^1$ (so $a^2+b^2=1$), let $B_t=ag_1+bg_2$. You get a Gaussian process whose distribution is invariant under rotation. Each $B_t$ is a standard Gaussian variable and the variance of $B_t-B_s$ is $\|t-s\|^2$.

Following Ori's remark below, concerning the 2-dimensionality, maybe a better simple suggestion is the following: Take two independent Brownian motions on $(-\infty, \infty)$, $C_a,D_a$ ($C_0=D_0=0$) and for $t=(a,b)\in S^1$ define $B_t=C_a+D_b$.

On third thought, this is probably just a Brownian motion on $R^2$ (zero at the origin) restricted to $S^1$.

show/hide this revision's text 2 added 256 characters in body; deleted 3 characters in body

What is wrong with the following simple definition? Let $g_1$ and $g_2$ be two independent standard Gaussian variables. For $t=(a,b)\in S^1$ (so $a^2+b^2=1$), let $B_t=ag_1+bg_2$. You get a Gaussian process whose distribution is invariant under rotation. Each $B_t$ is a standard Gaussian variable and the variance of $B_t-B_s$ is $\|t-s\|^2$.

Following Ori's remark below, concerning the 2-dimensionality, maybe a better simple suggestion is the following: Take two independent Brownian motions on $(-\infty, \infty)$, $C_a,D_a$ ($C_0=D_0=0$) and for $t=(a,b)\in S^1$ define $B_t=C_a+D_b$.

show/hide this revision's text 1

What is wrong with the following simple definition? Let $g_1$ and $g_2$ be two independent standard Gaussian variables. For $t=(a,b)\in S^1$ (so $a^2+b^2=1$), let $B_t=ag_1+bg_2$. You get a Gaussian process whose distribution is invariant under rotation. Each $B_t$ is a standard Gaussian variable and the variance of $B_t-B_s$ is $\|t-s\|^2$.