Lower-bound for $\mathbb E[e^{-b(v^\top X - c)^2}]$, when $X$ is log-concave in high-dimensions

Question

Let $d$ be a large positive integer. Fix a unit-vector $v \in \mathbb R^d$, and scalars $b,c \in \mathbb R$ with $b > 0$. Let $X$ be a log-concave random vector in $\mathbb R^d$ normalized so that $\mathbb E[(1/d)\|X\|^2] = 1$, WLOG.

Question 1. Is there a nontrivial lower-bound for $\alpha:=\mathbb E[e^{-b(v^\top X - c)^2}]$ in terms of $b$ and $c$ ?

For example, if $X$ is distributed according to $N(0,I_d)$, then $Z:=v^\top X$ has distribution $N(0,1)$, and so direct integration gives $$ \alpha = \mathbb E_{Z \sim N(0,1)}[e^{-b(Z-c)^2}] = \sqrt{\frac{1}{1 + 2b}}e^{-bc^2/(1 + 2b)}. $$

Question 2. Same question as Question 1, additional condition that $X$ is isotropic.

Answer 1

The answer to Question 1 is no. Indeed, let $X=\sqrt d\,V v$, where $V\sim N(0,1)$. Then \begin{equation} Ee^{-b(v\cdot X-c)^2} = \frac1{ \sqrt{1 + 2bd}} e^{-bc^2/(1 + 2bd)}\to0 \end{equation} (as $d\to\infty$). So, the only lower bound on $Ee^{-b(v\cdot X-c)^2}$ in general is the trivial bound $0$.

---

The answer to Question 2 is yes, even without the log-concavity condition. Indeed, let \begin{equation} l_{b,c}:=\inf_{|y|<\sqrt2}\frac{e^{-b(y-c)^2}}{2-y^2} =\min_{|y|<\sqrt2}\frac{e^{-b(y-c)^2}}{2-y^2}>0. \end{equation} Then $e^{-b(y-c)^2}\le l_{b,c}(2-y^2)$ for all real $y$ and hence \begin{equation} Ee^{-b(v\cdot X-c)^2} \ge l_{b,c}(2-E(v\cdot X)^2)=l_{b,c} \end{equation} -- because, if $X$ is isotropic and $E(1/d)\|X\|^2=1$, then $E(v\cdot X)^2=\|v\|^2=1$. So, $l_{b,c}>0$ is a nontrivial lower bound on $Ee^{-b(v\cdot X-c)^2}$ in the "isotropic" case.

(It is not hard to see that $l_{b,c}\ge e^{-2b-2bc^2}$. The latter lower bound is not far off. Indeed, using the Prékopa–Leindler inequality, one can see that $Ee^{-b(v\cdot X-c)^2}\le e^{-b-bc^2}$ if $EX=0$, $E(1/d)\|X\|^2=1$, $X$ is isotropic, and the pdf of $X$ is log-concave.)