# Anti-concentration of Gaussian when conditioning on event

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., $$\delta = 0.01$$).

$$\newcommand{\si}{\sigma}\newcommand{\Si}{\Sigma}\newcommand{\R}{\mathbb R}\newcommand{\ep}{\varepsilon}$$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 $$\R^n$$, $$\begin{equation*} E_r:=E_{v,\Si,r}:=\{x=(x_1,\dots,x_n)\in\R^n\colon\sum_1^n(x_i-v_i)^2/\si_i^2 and the $$\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| 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 $$P(s>u_i-v_i>0)\le P(-s. So, \begin{align*} P(u\in E_\ep,f\cdot u\ge f\cdot v)&\ge P(\max_1^n|(u_i-v_i)/\si_i|<\ep/\sqrt n,f\cdot u\ge f\cdot v) \\ &\ge\prod_1^n P(0 Further, for all $$i\in[n]$$ and all real $$s>0$$ $$\begin{equation*} P(|u_i-v_i| and $$\begin{equation*} P(0 since $$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).