# On an angle distribution of a random linear subspace of a given dimension

$$\newcommand\R{\mathbb R}$$ Let $$u$$ be a fixed unit vector in $$\R^n$$, and let $$\Pi_u$$ be the hyperplane in $$\R^n$$ with normal vector $$u$$. Let $$B$$ be the (say open) unit ball in $$\R^n$$ centered at the origin. For a natural $$k, let $$V$$ be a random linear subspace, of dimension $$k$$, uniformly distributed on the Grassmannian manifold $$\mathbf{Gr}(k,n)$$ of all linear subspaces of $$\R^n$$ of dimension $$k$$.

Is there a tractable expression of the probability $$p_{n,k;t}:=\mathsf P(V\cap(tu+\Pi_u)\cap B\ne\emptyset)$$ for $$t\in(0,1)$$, or at least good lower and upper bounds on this probability?

Clearly, $$p_{n,k;t}$$ depends only on $$n,k,t$$.

The case $$k=1$$ is easy: then the probability $$p_{n,k;t}$$ is the ratio of the area of a spherical cap to the area of the sphere. Here is a picture for $$n=3$$, $$k=1$$, $$u=(0,0,1)$$, and $$t=\cos\pi/4$$, showing part of the shifted (hyper)plane $$tu+\Pi_u$$, the spherical cup that this shifted (hyper)plane cuts off the unit sphere, and part of a realization of the random one-dimensional subspace $$V$$ (blue), such that the event $$V\cap(tu+\Pi_u)\cap B\ne\emptyset$$ occurs. Of course, the Grassmannian manifold $$\mathbf{Gr}(k,n)$$ is in a one-to-one correspondence with the set, say $$\mathcal{P}(k,n)$$, of the matrices $$P$$ of the orthogonal projectors of $$\R^n$$ of rank $$k$$, which are characterized by the conditions $$P^2=P=P^\top$$ and $$\text{tr}\,P=k$$, where $$\text{tr}$$ denotes the trace. However, a problem with this approach is to parameterize $$\mathcal{P}(k,n)$$. More generally, a problem is to find a good atlas for the Grassmannian manifold.

• I'm not sure if you already knew this, but one (non-unique) way of representing an element of the Grassmannian is as a $k\times n$ matrix with orthogonal rows (this is giving you an orthonormal basis of the subspace). You can pick a sample a uniform random element of the Grassmannian by sampling a uniform random $k\times n$ matrix with orthonormal rows in this way. If this randomly sampled matrix is called $A$, your question is equivalent to asking whether $\|Au\|>t$. (In the case that $k=1$, it's easy to see this is the same as normalized area of the spherical cap). Dec 3, 2021 at 8:59
• @AnthonyQuas : Thank you for your comment. Yes, I did have this in mind. Part of the problem with this approach is the non-uniqueness of this representation. Dec 3, 2021 at 14:47
• Here is another, possibly more tractable, expression: your probability is exactly equal to $\mathbb P(x_1^2+\ldots+x_k^2\ge t^2)$ where $x$ is uniformly distributed on the unit sphere, or also to $\mathbb P(X_1^2+\ldots+X_k^2\ge t^2(X_1^2+\ldots+X_n^2))$ where the $X_i$ are independent standard normal random variables. Dec 4, 2021 at 18:20
• @AnthonyQuas : How do you prove this? Maybe, you can expand your comment into an answer? Dec 4, 2021 at 22:24

This is an expansion of some comments that I made earlier. At the time that I post this, an answer has already been posted by @jlewk, taking some of the material in the comments further. I include this answer because I feel it is quite down-to-earth.

First, observe that if $$A$$ is an $$k\times n$$ matrix of orthogonal row vectors, then the unit ball of the row space intersects the plane $$tu+\Pi_u$$ if and only if there is a $$v\in R^k$$ of norm 1 such that $$v^TAu\ge t$$ if and only if $$\|Au\|\ge t$$.

Next, if $$A$$ is taken to be $$PR$$ where $$P$$ is the $$k\times n$$ matrix with 1's on the diagonal and 0's elsewhere and $$R$$ is a uniform random variable taking values in $$O(n)$$, then the distribution of the row space is $$O(n)$$ invariant, so that it is the uniform distribution on the $$k$$-dimensional Grassmannian.

Hence the probability that the intersection is non-empty is the probability that $$\|PRu\|\ge t$$. However since $$Ru$$ is uniformly distributed on the unit sphere, this is the probability that $$\|Px\|\ge t$$ where $$x$$ is uniformly distributed on the unit sphere.

Since a vector with independent normal entries is isotropic, this is equal to the probability that $$\|PX\|\ge t\|X\|$$ where $$X$$ is a standard normal random variable, that is $$\mathbb P(X_1^2+\ldots+X_k^2\ge t^2(X_1^2+\ldots+X_n^2))$$. Finally, this can be rearranged as an inequality involving two $$\chi^2$$ distributions: $$\mathbb P(X_1^2+\ldots+X_k^2\ge \frac{t^2}{1-t^2}(X_{k+1}^2+\ldots+X_n^2))$$.

This expands comments by AnthonyQuas. Start with the observation from Anthony Quas that the event of interest is $$\|Au\|>t$$ for an orthogonal projection matrix $$A$$ whose image is distributed according to the Grassmanian. For instance one can realize $$A$$ as $$X(X^TX)^{-1}X^T$$ where $$X\in R^{n\times k}$$ has iid $$N(0,1)$$ entries. Now $$RX=^dX$$ (equality in distribution) for any rotation matrix $$R\in O(n)$$ independent of $$X$$, and $$RAR^T=^d A$$. Taking $$R$$ distributed according to the Haar measure, $$P(\|Au\|>t) = P(\|RAR^Tu\|>t) = P(\|Av\|>t)$$ where $$v\sim$$ is uniformly distributed on the unit sphere and independent of $$A$$. Finally, realize $$v$$ as $$v=g/\|g\|$$ for standard normal vector $$g$$. Then the event of interest is $$\|Ag\|^2 > t^2 \|g\|^2 = t^2 (\|Ag\|^2 + \|(I-A)g\|^2)$$ and the question reduces to $$P((1-t^2)\| Ag\|^2 - t^2\|(I-A)g\|^2>0)$$. The Hanson-Wright inequality applied to the quadratic form $$(1-t^2)\| Ag\|^2 - t^2\|(I-A)g\|^2$$ written as $$g^TMg$$ for $$M=(1-t^2)A-t^2(I-A)$$ gives $$P\Big( |(1-t^2)\| Ag\|^2 - t^2\|(I-A)g\|^2 - trace[M]| >2\sqrt{x}\|M\|_F +2 x \|M\|_{op} \Big) > 2e^{-x^2/2}$$ for any $$x>0$$. Then $$trace[M]=k(1-t)^2 - (n-k)t^2$$ and the right-hand side inside the probability sign is the sum of two terms $$2 \sqrt{x[k(1-t^2)^2 + (n-k)t^4} + 2 x \max(1-t^2, t^2)$$ and the first term will typically dominate.