Suppose $v\in R^n$ is a constant unit vector. $P_l$ is a random projection matrix to an $l$ dimensional subspace of $R^n$ which is uniformly sampled from $G(l,R^n)$ which is the collection of all $l$-dimensional subspace in $R^n$. What is the upper bound of the following: $$\mathbb{P}(P_lv\leq\delta)$$ What is the order of the above probability as $\delta\to0$?

Also, when $l=n$, the above probability is the indicator function $1_{\{\delta\geq1\}}$. Can we derive an upper bound of the above probability for $l<n$ such that when $l\to n$, the upper bound tends to $1_{\{\delta\geq1\}}$?


1 Answer 1


$\newcommand{\R}{\mathbb{R}} \renewcommand{\P}{\operatorname{\mathsf P}} \newcommand{\Ga}{\Gamma} \newcommand{\de}{\delta}$

In view of the spherical symmetry of the distribution of the $l$-dimensional subspace, we can fix it to be, say, the span of the first $l$ vectors of the standard basis of $\R^n$ and, accordingly, let $v=:(Y_1,\dots,Y_n)$ be a random vector uniformly distributed on the unit sphere $S_{n-1}$. So, the probability in question equals \begin{equation} p_\de:=\P\Big(\sum_1^l Y_i^2\le\de^2\Big); \end{equation} we are assuming that $\de\in(0,1)$. Next, we may write \begin{equation} Y_i=X_i\Big/\sqrt{\sum_1^n X_i^2}, \end{equation} where the $X_i$'s are iid $N(0,1)$, whence
\begin{equation} p_\de=\P\Big(\frac{\sum_1^l X_i^2}{\sum_{l+1}^n X_i^2}\le\frac{\de^2}{1-\de^2}\Big) =F_{l,n-l}\Big(\frac{\de^2}{1-\de^2}\frac{n-l}l\Big) \sim c_{n,l}\de^l \end{equation} as $\de\downarrow0$, where $F_{l,n-l}$ is the cdf of the $F$ distribution with $l,n-l$ degrees of freedom and \begin{equation} c_{n,l}:=\frac2{l\,B\left(l/2,(n-l)/2\right)} =\frac{2\Ga(n/2)}{l\,\Ga(l/2)\Ga((n-l)/2)}. \end{equation} (Here we used the easily verifiable fact that, if the pdf $f$ of a nonnegative r.v. $X$ is such that for some real $c,p>0$ we have $f(x)\sim cx^{p-1}$ as $x\downarrow0$, then for the cdf $F$ of $X$ we have $F(x)\sim cx^p/p$ as $x\downarrow0$.)


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