What happens to the Gaussian volume of a Borel set when it is translated? Let $\gamma_n$ be the standard Gaussian measure on $\mathbb R^n$, $A \subseteq \mathbb R^n$ be Borel and $c \in \mathbb R^n$. Define the translate $A_c := c + A := \{c+a \mid a \in A\} = \{x \in \mathbb R^n \mid x-c \in A\}$. Assume $A$ is not too "small", e.g  $\gamma_n(A) \ge 1/2$.
Question


*

*What are good lower (and perhaps upper) bounds for $\gamma_n(A_c)$ ?


Observations
For a half-space $H = \{x \in \mathbb R^n \mid u^Tx \le v\}$, one has $H_c = \{x \in \mathbb R^n \mid u^Tx \le v + u^Tc\}$, another half-space with Gaussian volume $\gamma_n(H_c) = \Phi\left(\frac{v + u^Tc}{\|u\|}\right) = \Phi\left(\Phi^{-1}(\gamma_n(H)+\frac{u^Tc}{\|u\|}\right)$, where $\Phi$ is the 1D Gaussian CDF.

Update
Let $H$ be a half-space with same Gaussian volume as $B$, i.e $\gamma_n(H)=\gamma_n(A)$.
Question: Under which minimal / reasonable conditions do we have $\gamma_n(A_c) \ge \gamma_n(H_c)$ ?
 A: Let 
\begin{equation}
 g(t):=\gamma_n(tu+c+A)=P(Z\in tu+c+A)=\int_{c+A}(2\pi)^{-n/2}e^{-|z+tu|^2/2}\,dz,
\end{equation}
where $t\in\mathbb R$, $u$ is a unit vector in $\mathbb R^n$, $Z$ is a standard Gaussian random vector in $\mathbb R^n$, and $|\cdot|$ is the Euclidean norm. Then 
\begin{equation}
 g'(0)=-\int_{c+A}(2\pi)^{-n/2}z\cdot u\,e^{-|z|^2/2}\,dz
 =-p\,u\cdot E(Z|Z\in c+A), 
\end{equation}
where $p:=P(Z\in c+A)=\gamma_n(c+A)$ and $z\cdot u$ is the dot product of $z$ and $u$. 
That is, the rate of change of $\gamma_n(c+A)$ with respect to $c$ in the direction of a unit vector $u$ equals the following: 

$\gamma_n(c+A)$ times the $u$-coordinate $u\cdot E(Z|Z\in c+A)$ of the center $E(Z|Z\in c+A)$ of the standard Gaussian mass over the set $c+A$.  

A: It turns out that Neyman-Pearson theory helps get a nontrivial inequality.
Notations. For a p.s.d matrix $M$ of size $p$, consider the inner product on $\mathbb R^p$
defined by $\langle x,z \rangle_M := x^TMz$. This induces a norm defined by
$\|x\|_M:=\sqrt{\langle x,x \rangle_M}$.

Theorem (Neyman-Pearson for translated multivaritate Gaussians).
    Let $\beta \in \mathbb R$, $\delta\in \mathbb R^p$ and $A$ be a
    Borell subset of $\mathbb R^p$. Let $X \sim \mathcal N(0,\Sigma)$ and
    $Y:=X+\delta$. Consider the half-space
    \begin{eqnarray}
    H=\{z \in \mathbb R^p \mid \langle\delta,z\rangle_{\Sigma^{-1}} \le \beta\}.
 \end{eqnarray} 
  
  
*
  
*If $\mathbb P(X \in A) \ge \mathbb P(X \in H)$, then $\mathbb P(Y \in A) \ge
    \mathbb P(Y \in H)$.
  
*If $\mathbb P(X \in A) \le \mathbb P(X \in H)$, then $\mathbb P(Y \in A) \le
    \mathbb P(Y \in H)$.    

Proof. The log of the ratio of the densities of $Y$ and $X$ is given by
  \begin{eqnarray*}
    \begin{split}
    \log(f_Y(z))-\log(f_X(z)) &=
    -\frac{1}{2}(z-\delta)^T\Sigma^{-1}(z-\delta)-\frac{1}{2}z^T\Sigma^{-1}z\\
    &= \langle \delta,z \rangle_{\Sigma^{-1}}-\frac{1}{2}\|\delta\|_{\Sigma^{-1}}^2.
    \end{split}
  \end{eqnarray*}
  Thus $f_Y(z) \le t f_X(z)$ iff $\langle \delta,z
  \rangle_{\Sigma^{-1}}-\frac{1}{2}\|\delta\|_{\Sigma^{-1}}^2 \le \log(t)$.
  Define $t := e^{\beta-\frac{1}{2}\|\delta\|_{\Sigma^{-1}}^2}$. Then $S_t=H$,
  and we can apply the Neyman-Pearson Lemma (see Appendix below) to get the claimed results. $\Box$

Appendix
The following Lemma is a modern formulation of the celebrated Neyman-Pearson Lemma.

Lemma (Neyman-Pearson 1933).
    Let $A$ be an event in a probability space $\mathcal Z$, and let $X$, $Y$ be random
    variables on $\mathcal Z$, with densities $f_X$ and $f_Y$ respectively.
    Finally, let $t > 0$ and define
    \begin{eqnarray}
    S_t:= \{z \in \mathcal Z \mid f_Y(z) \le tf_X(z)\}.
  \end{eqnarray}
    We have the following:
  
  
*
  
*If $\mathbb P(X \in A) \ge \mathbb P(X \in S_t)$, then $\mathbb P(Y \in
    A) \ge \mathbb P(Y \in S_t)$.
  
*If $\mathbb P(X \in A) \le \mathbb P(X \in S_t)$, then $\mathbb P(Y \in
    A) \le \mathbb P(Y \in S_t)$.    

Proof.
 Suppose $\mathbb P(X \in A) \ge \mathbb P(X \in S_t)$, and let $A^c$ be the set
  complement of $A$ in $\mathcal Z$. One computes
  \begin{eqnarray*}
    \begin{split}
      &\mathbb P(Y \in A)-\mathbb P(Y \in S_t)=\int 1_A(z)f_Y(z)dz-\int_{S_t}f_Y(z)dz\\
      &= \int_{S_t} 1_A(z)f_Y(z)dz+\int_{S_t^c} 1_A(z)f_Y(z)dz-\left(\int_{S_t}
        1_A(z)f_Y(z)dz+\int_{S_t} 1_{A^c}(z)f_Y(z)dz  \right)\\
      &=\int_{S_t^c} 1_A(z)f_Y(z)dz-\int_{S_t} 1_{A^c}(z)f_Y(z)dz\\
      &\ge t\left(\int_{S_t^c} 1_A(z)f_X(z)dz-\int_{S_t}
        1_{A^c}(z)f_X(z)dz\right),\text{ by definition of }S_t\\
      &= t(\mathbb P(X \in A)-\mathbb P(X \in S_t)) \ge 0,\text{ by assumption}.
    \end{split}
  \end{eqnarray*}
  Thus $\mathbb P(Y \in A) \ge \mathbb P(Y \in S_t)$. Similarly, one proves
  the second part of the claim with "$\ge$" replaced with "$\le$". $\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\Box$
