2
$\begingroup$

$\newcommand\R{\mathbb R}$ 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 $k<n$, 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.

enter image description here


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.

$\endgroup$
4
  • $\begingroup$ 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). $\endgroup$ Dec 3, 2021 at 8:59
  • $\begingroup$ @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. $\endgroup$ Dec 3, 2021 at 14:47
  • $\begingroup$ 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. $\endgroup$ Dec 4, 2021 at 18:20
  • $\begingroup$ @AnthonyQuas : How do you prove this? Maybe, you can expand your comment into an answer? $\endgroup$ Dec 4, 2021 at 22:24

2 Answers 2

1
$\begingroup$

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: $\mathbb P(X_1^2+\ldots+X_k^2\ge \frac{t^2}{1-t^2}(X_{k+1}^2+\ldots+X_n^2))$.

$\endgroup$
1
  • $\begingroup$ Thank you for your answer. $\endgroup$ Dec 5, 2021 at 1:50
2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Thank you for your answer. $\endgroup$ Dec 5, 2021 at 1:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.