Inner product of the spherical cap and Gaussian

Question

Let $d\in \mathbb{N}$ and $\eta \sim N(0,I_d)$ where $N(0,I_d)$ is the gaussian distribution with the covariance matrix of $I_d$. Also, define a spherical cap as follows. Fix $v \in \mathbb{S}^{(d-1)}$ and $\theta \in (0,1)$ where $\mathbb{S}^{(d-1)}$ is the unit sphere in $\mathbb{R}^d$. Let $$ C_{v,\theta} = \{x\in \mathbb{S}^{(d-1)}: \langle x,v \rangle \geq \theta \}. $$

I am interested in an upperbound on the following probability: $$ \mathbb{E}_{\eta \sim N(0,I_d)}\left[ \sup_{x\in C_{v,\theta}} \langle \eta,x \rangle \right]. $$

Easy Upperbound

We can write \begin{align} \mathbb{E}_{\eta \sim N(0,I_d)}\left[ \sup_{x\in C_{v,\theta}} \langle \eta,x \rangle \right] &\leq \mathbb{E}_{\eta \sim N(0,I_d)}\left[ \sup_{x\in \mathbb{S}^{(d-1)}} \langle \eta,x \rangle \right]\\ &= \mathbb{E}_{\eta \sim N(0,I_d)}\left[ \lVert \eta \rVert \right]\\ &\leq \sqrt{d} \end{align}

How can I improve this naive bound?

Answer 1

$\newcommand\th\theta\newcommand\R{\Bbb R}$Assume that $d\ge2$. Without loss of generality $v=(1,0,\dots,0)$. Identify then $\eta$ with $X=(X_1,X_2)$, where $X_1$ and $X_2$ are independent random elements of $\R$ and $\R^{d-1}$, respectively, such that $X_1\sim N(0,1)$ and $X_2\sim N(0,I_{d-1})$.

Then, by simple plane geometry considerations, $$s_d:=s_d(\th):=\sup_{x\in C_{v,\th}} \langle \eta,x \rangle =\|X\|\,f(R),$$ where $\|\cdot\|$ denotes the Euclidean norm on $\R^m$ for any $m$, $$f(R):=1(R>\th)+(\th R+\sqrt{1-\th^2}\,\sqrt{1-R^2})\,1(R\le\th),$$ $$R:=\frac{X_1}{\|X\|}.$$

By (say) the law of large numbers, $\|X\|\to\infty$ and hence $R\to0$ and $f(R)\to\sqrt{1-\th^2}$ in distribution (as $d\to\infty$). So, by dominated convergence, $Ef(R)\to\sqrt{1-\th^2}$. Therefore and because $R$ is independent of $\|X\|$ and because $E\|X\|\sim\sqrt d$, we conclude that $$Es_d=E\|X\|\,Ef(R)\sim\sqrt{(1-\th^2)d}.$$

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Moreover, since $R^2$ has the beta distribution with parameters $1/2,(d-1)/2$ and $\|X\|^2$ has the chi square distribution with $d$ degrees of freedom, we get the exact expression for $Es_d$: $$Es_d \\ =\frac1{2\sqrt{2\pi}}\bigg[-2 \theta \left(1-\theta ^2\right)^{\frac{d-1}{2}}- B_{\theta ^2}\left(\frac{1}{2},\frac{d-1}{2}\right)d +B_{\theta ^2}\left(\frac{1}{2},\frac{d-1}{2}\right) \\ +(d-1) \sqrt{1-\theta ^2} \left(B_{\theta ^2}\left(\frac{1}{2},\frac{d}{2}\right)+B\left(\frac{1}{2},\frac{d}{2}\right)\right) \\ +(d-1)B\left(\frac{1}{2},\frac{d-1}{2}\right)\bigg],$$ where $B(\cdot,\cdot)$ is the beta function and $B_\cdot(\cdot,\cdot)$ is the incomplete beta function. These functions are easy to analyze.

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For $\rho_\th(d):=Es_d/\sqrt{(1-\th^2)d}=Es_d(\th)/\sqrt{(1-\th^2)d}$, here are the graphs $\{(d,\rho_{0.1}(d))\colon2\le d\le100\}$ (orange) and $\{(d,\rho_{0.9}(d))\colon2\le d\le100\}$ (blue):

Answer 2

Write $\eta=\|\eta\|\hat{\eta}$; then $\hat{\eta}$ is uniformly distributed on $\mathbb{S}^{d-1}$ and independent of $\|\eta\|$. We have $$ \mathbb{E}\left[\sup_{x\in C_{v,\theta}}(\eta,x)\right]=\mathbb{E}\left[\|\eta\|\sup_{x\in C_{v,\theta}}(\hat{\eta},x)\right]=\sqrt{d}\cdot\mathbb{E}\left[\sup_{x\in C_{v,\theta}}(\hat{\eta},x)\right]\stackrel{d\to\infty}{\sim}\sqrt{1-\theta^2}\sqrt{d}. $$ Indeed, the supremum in the expectation is equal to $1$ when $\hat{\eta}\in C_{v,\theta}$ and to $\cos(\nu(\hat{\eta})-\nu)$ otherwise, where $\nu(\hat{\eta})=\arccos((\hat{\eta},v))$ is the angle between $\hat{\eta}$ and $v$, and $\nu=\arccos\theta$. For large $d$, $\nu(\hat{\eta})$ is concentrated near $\pi/2$, so the cosine in question is concentrated near $\cos(\pi/2-\nu)$. Plugging in effective concentration bounds, you can get a non-asymptotic inequality.