# Gaussian Mean Width

Consider $K\subset S^{n-1}$. Its Gaussian Mean Width is defined to be $$\mathbb{E}\,\sup_{x\in K}\vert \big<g, x\big>\vert$$ where $g$ is a standard Gaussian Random Vector in $\mathbb{R}^{n}$. I think that this is somehow supposed to be an averaging of the distance between planes needed to bound $K$, but I'm not sure how this definition actually does this. Is this definition dependent on $g$? Can anyone explain the motivation for this definition for me?

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Does $S^{n-1}$ stand for $n-1$ dimensional hypersphere defined by $x_1^2+x_2^2+\cdots+x_n^2 = 1$? –  Sasha May 23 '12 at 21:43
Yes, $S^{n-1}$ is the $n-1$ dimensional unit ball in $\mathbb{R}^{n}$. –  Alex Lapanowski May 23 '12 at 21:47
@Sasha: But the rotation is not allowed to depend on $x$, hence your argument breaks down, see my answer. –  Did May 24 '12 at 6:36
@Didier I removed my comment, however I am still much confused. Should I replace $| \langle u, x \rangle |$ with even powers, they are easily seen to dependent only on $\langle x, x \rangle$ which is the same for all $x \in S^{n-1}$. Numerical simulation also shows that $\mathbf{E}(| \langle u, x \rangle |)$ has the same value for all $x \in S^{n-1}$. In fact argument I made can be translated to carrying out the expectation. Let $\vec{Z} = {z_1,z_2,\ldots,z_n}$ be standard normal vector. For every $\vec{x} \in S^{n-1}$ I can introduce hyperspherical coordinates.... –  Sasha May 24 '12 at 12:49
@Didier ... such that $\vec{x}$ is aligned with the north pole. Then $\mathbf{E}(|\langle u, x \rangle|) = \frac{1}{B((n-1)/2,1/2)} \int_0^\pi | \cos(\theta) | \sin^{n-2} \theta \mathrm{d} \theta = \frac{\Gamma(n/2)}{\Gamma(n/2+1/2)} \frac{1}{\sqrt{\pi}}$. –  Sasha May 24 '12 at 12:53

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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Thank you for your response. –  Alex Lapanowski May 24 '12 at 18:09
Can $c_n$ here be seen as a normalized rotation invariant measure on the sphere? Thanks! –  sleeve chen Nov 28 '14 at 19:55
@Did: Besides, $g=ru$, can I think that $r$ stands for radius, and $u$ stands for the $e^{jq}$,Thanks! –  sleeve chen Nov 28 '14 at 20:02
@sleevechen No, $c_n$ is a number, not a measure on the sphere. Yes, $g=ru$ holds with $r=\|g\|$ and $u=g/\|g\|$ hence, in dimension 2, $u$ would indeed be $e^{i\varphi}$. –  Did Nov 28 '14 at 21:34
@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). –  Did Nov 29 '14 at 7:28