# Tail probability of random projection

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 $$l such that when $$l\to n$$, the upper bound tends to $$1_{\{\delta\geq1\}}$$?

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 $$$$p_\de:=\P\Big(\sum_1^l Y_i^2\le\de^2\Big);$$$$ we are assuming that $$\de\in(0,1)$$. Next, we may write $$$$Y_i=X_i\Big/\sqrt{\sum_1^n X_i^2},$$$$ where the $$X_i$$'s are iid $$N(0,1)$$, whence
$$$$p_\de=\P\Big(\frac{\sum_1^l X_i^2}{\sum_{l+1}^n X_i^2}\le\frac{\de^2}{1-\de^2}\Big) =F_{l,n-l}\Big(\frac{\de^2}{1-\de^2}\frac{n-l}l\Big) \sim c_{n,l}\de^l$$$$ as $$\de\downarrow0$$, where $$F_{l,n-l}$$ is the cdf of the $$F$$ distribution with $$l,n-l$$ degrees of freedom and $$$$c_{n,l}:=\frac2{l\,B\left(l/2,(n-l)/2\right)} =\frac{2\Ga(n/2)}{l\,\Ga(l/2)\Ga((n-l)/2)}.$$$$ (Here we used the easily verifiable fact that, if the pdf $$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$$.)