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 $1\_{\{\delta\geq1\}}$?



Tail probability of random projection
$\newcommand{\R}{\mathbb{R}}
\renewcommand{\P}{\operatorname{\mathsf P}}
\newcommand{\Ga}{\Gamma}
\newcommand{\de}{\delta}$


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](https://en.wikipedia.org/wiki/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 $x\downarrow0$.)


