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., $\delta = 0.01$).



Anti-concentration of Gaussian when conditioning on event
$\newcommand{\si}{\sigma}\newcommand{\Si}{\Sigma}\newcommand{\R}{\mathbb R}\newcommand{\ep}{\varepsilon}$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).


