Consider $K\subset S^{n-1}$. Its Gaussian Mean Width is defined to be $ \mathbb{E}\,\sup_{x\in K}\vert \big
Gaussian Mean Width
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0@Sasha Aaaah... so, THIS was the source of the confusion. Makes sense now. – 2012-05-24
1 Answers
Recall that the distribution of $g$ is rotationally invariant, more precisely, $g=ru$ where $r\gt0$ and $u$ in $S^{n-1}$ are independent random variables such that $u$ is uniformly distributed on $S^{n-1}$ (the distribution of $r$ is explicit and depends on $n$).
Hence $\langle g,x\rangle=r\langle u,x\rangle$ and, by independence, the gaussian mean width of $K$ is also $ \gamma_n(K)=c_n\cdot\mathrm E\left(\sup\limits_{x\in K}|\langle u,x\rangle|\right), $ where $c_n=\mathrm E(r)$ is a constant which depends only on the dimension $n$.
Now, for every realization of $u$, one looks for the point $x$ in $K$ such that $u$ and $x$ are as best aligned as possible and $\gamma_n(K)$ is the average over $u$ of the result, scaled by $c_n$.
If $K$ is gaussian wide in the sense that for nearly every $u$ in $S^{n-1}$ there exists some $x$ in $K$ such that $\langle u,x\rangle$ is nearly $+1$ or nearly $-1$, then the average is close to $1$ and $\gamma_n(K)$ is close to $c_n$. On the contrary, if $K$ is sparse in the sense that for some $u$ in $S^{n-1}$, no good $x$ in $K$ exists, then the average over $u$, and as a consequence $\gamma_n(K)$ itself, are smaller.
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1@sleevechen That$u$is uniformly distributed on the sphere means exactly that its distribution is *the unique* "rotation invariant measure on the sphere" with mass 1 (but, contrarily to what your last comment suggests, it is not a measure with a density on the real line). – 2014-11-29