Let $v$ be a given vector with $\|v\|_{\Sigma^{-1}} \leq 1$, where $\Sigma$ is a positive semi-definite matrix and $\|v\|_{\Sigma^{-1}} = \sqrt{v^\top\Sigma v}$. Meanwhile, let $u$ be a random vector drawn from $N(0,\Sigma^{-1})$. We know that for any given vector $\phi$, it holds that $$ P(\phi^\top u > \phi^\top v) > c $$ where $c$ is a positive absolute constant that depends on $v$.

Question: Does a similar anti-concentration property still hold when we additionally condition on the event $ \mathcal{E} = \{u \in \mathcal{C}\} $, where $\mathcal{C}$ is a given set such that $v \in \mathcal{C}$ and moreover $v$ is in the interior of $\mathcal{C}$? In other words, does it hold that $$ P(\phi^\top u > \phi^\top v \,|\, \mathcal{E})> c', $$ where $c'$ is a positive absolute constant that depends on $v$ and $\mathcal{C}$? One particular example of $\mathcal{C}$ that I am interested in is $$ \mathcal{C} = \{u: \| u + w \| \leq 1 \}, $$ where $w$ is a given vector such that $\| v + w \| \leq 1-\delta$ with $\delta > 0$ (e.g., $\delta = 0.01$).


1 Answer 1


$\newcommand{\si}{\sigma}\newcommand{\Si}{\Sigma}\newcommand{\R}{\mathbb R}\newcommand{\ep}{\varepsilon}$The answer is: in general, no -- even for convex $\mathcal C$.

Indeed, let $C:=\mathcal C=(-\infty,1)\times\R$, $v=(0,0)$, $f:=\phi=(1,0)$, and $\Si=\begin{pmatrix}\si^2&0\\0&1\end{pmatrix}$, with $\si\to\infty$. Then $$P(u\in C)\ge P\big(u\in(-\infty,0)\times\R\big)=1/2$$ and \begin{equation*} P(u\in C,f\cdot u>f\cdot v)=P(1>f\cdot u>0)\to0, \end{equation*} so that $P(f\cdot u>f\cdot v\,|\,u\in C)\to0$.

This example is written for dimension $2$, but can be easily modified for any dimension $\ge1$.

On a positive note, let us also show that the answer becomes yes if $C$ is assumed to be bounded (as in your "particular example") and, of course, $f\ne0$. Indeed, then there are real $\ep>0$ and $R>\ep$ (depending only on $C$ and $v$) such that $B_v(\ep)\subseteq C\subseteq B_v(R)$, where $B_v(r)$ is the open ball of radius $r$ centered at $v$. Replacing $u,v,C$ by $\Si^{1/2}u,\Si^{1/2}v,\Si^{1/2}C$, respectively, we see that it is enough to to prove the following: \begin{equation*} \frac{P(u\in E_\ep,f\cdot u\ge f\cdot v)}{P(u\in E_R)}\ge c_{\ep,R,n} \tag{1} \end{equation*} for some $c_{\ep,R,n}>0$ depending only on $\ep$, $R$, and $n:=\dim C$, where $u=(u_1,\dots,u_n)\sim N(0,I_n)$, $v=(v_1,\dots,v_n)$, $\|v\|\le1$, $\|\cdot\|$ is the Euclidean norm on $\R^n$, \begin{equation*} E_r:=E_{v,\Si,r}:=\{x=(x_1,\dots,x_n)\in\R^n\colon\sum_1^n(x_i-v_i)^2/\si_i^2<r^2\}, \end{equation*} and the $\si_i$'s are some positive real numbers determined by $\Si$.

We have \begin{equation*} P(u\in E_R)\le P(\max_1^n|(u_i-v_i)/\si_i|<R) =\prod_1^n P(|u_i-v_i|<R\si_i). \tag{2} \end{equation*} Next, the inequality $f\cdot u\ge f\cdot v$ for $f=(f_1,\dots,f_n)$ will hold if for each $i\in[n]:=\{1,\dots,n\}$ the sign of $u_i-v_i$ is the same as the sign of $f_i$. Also, for all $i\in[n]$ and all real $s>0$ without loss of generality $v_i\ge0$ and hence $P(s>u_i-v_i>0)\le P(-s<u_i-v_i<0)$. So, \begin{align*} P(u\in E_\ep,f\cdot u\ge f\cdot v)&\ge P(\max_1^n|(u_i-v_i)/\si_i|<\ep/\sqrt n,f\cdot u\ge f\cdot v) \\ &\ge\prod_1^n P(0<u_i-v_i<\ep\si_i/\sqrt n). \tag{3} \end{align*} Further, for all $i\in[n]$ and all real $s>0$ \begin{equation*} P(|u_i-v_i|<s)\ll\min(1,s) \end{equation*} and \begin{equation*} P(0<u_i-v_i<s)\gg\min(1,s), \end{equation*} since $0\le v_i\le\|v\|\le1$, where $a\ll b$ and $b\gg a$ mean $a\le Cb$ for some universal real constant $C>0$. Now (1) follows from (2) and (3).


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