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.



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> **Question 1.** *Is there a nontrivial lower-bound for $\alpha:=\mathbb E[e^{-b(v^\top X - c)^2}]$ in terms of $b$ and $c$ ?*
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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
$$
\alpha = \mathbb E\_{Z \sim N(0,1)}[e^{-b(Z-c)^2}] = \sqrt{\frac{1}{1 + 2b}}e^{-bc^2/(1 + 2b)}.
$$



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> **Question 2.** *Same question as **Question 1**, additional condition that $X$ is isotropic.*
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Lower-bound for $\mathbb E[e^{-b(v^\top X - c)^2}]$, when $X$ is log-concave in high-dimensions
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 $0$.




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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.)


