Deje que los datos de la matriz $\pmb X$ ser una colección de $n$ $p$-vectores $x_i\sim\mathcal{E}_p(\mu,\pmb \varSigma)$ donde $\mathcal{E}_p(\mu,\pmb \varSigma)$ es un cuadrado integrable, continua elíptica densidad en $\mathbb{R}^p$ con vectores ubicación de $\mu$ y de dispersión de la matriz $\pmb\varSigma$$n>p$. A continuación, vamos a $\{S_m\}_{m=1}^M$ $p$ índices extraídos de $\{1:n\}$$M=\binom{n}{p}$. A continuación, vamos a $A_m=\{x_i|i\in S_m\}$ y $y_m\in\mathbb{R}^p:\pmb A_m'y_m=1_p$ ($1_p$ denota una $p$-vector).
$$\underset{1\le m\le M}{\max}\;\underset{1\le i\le n}{\max}\;(x_i'y_m-1)^2||y_m||^{-2}\text{ is bounded}$$
I'm looking for a reference to a proof (presumably of a more general variant) of (or a counter-example to) this statement.
EDIT1:
a '$p$ index' is a subset of $p$ elements of $\{1:n\}$ drawn randomly and without replacement. There are $\binom{n}{p}$ distinct such $p$-subsets of $\{1:n\}$. $m\in\{1,\ldots,M\}$. $x_i$ is indeed a row of $\pmb X$. $\pmb A_m$ is a $p$ by $p$ matrix formed of $p$ rows of $\pmb X$ with indexes in $S_m$.
EDIT2:
following Mike McCoy's question. For a given $y_m$ the inner product $x_i'y_m=1$ only for the $x_i$ with index $i\en S_m$. Here is a short R code to illustrate.
n<-20
p<-3
X<-matrix(rnorm(n*p),nc=p)
S_m<-sample(1:n,p)
A_m<-X[S_m,]
y_m<-solve(A_m,rep(1,p)) #equation of the hyperplane through the members of S_m
dis<-tcrossprod(X%*%y_m-1)/as.numeric(crossprod(y_m))
stated otherwise, the claim is that the $\max$ (over all possible $M$ subsets $S_m$) of the sup norm of dis is bounded whenever $\pmb X$ is drawn from a square integrable and continuous density.
EDIT3:
to address Mike McCoy's questions. By bounded, I mean that
$$\exists U(\mu, \pmb \varSigma):\underset{1\le m\le M}{\max}\;\underset{1\le i\le n}{\max}\;(x_i'y_m-1)^2||y_m||^{-2}\le U(\mu, \pmb \varSigma)\;a.s.$$
for some positive scalar $U(\mu \pmb \varSigma)$ depending only on $\pmb \varSigma$ and $\mu$. De nuevo, sospecho que este teorema (o incluso una versión más general de la misma) ha sido probada. Yo simplemente no puede encontrar la referencia.
