Let $Q$ be a probability measure on $\mathbb{R}$. Let $$Q\_h(dy) = e^{y \cdot h} Q(dy) / M(h) \quad \text{where} \quad M(h) = \int e^{y \cdot h} Q(dy)$$ defined for $h \in (-c,\infty)$ with some $c > 0$. Let $Y\_h \sim Q\_h$ and $V(h) := EY\_h^2 - (EY\_h)^2$. Then, does it follow that $$\exists c' > 0,\ h\_0 > -c\ \text{ s.t. }\ \forall h \geq h\_0,\ V(h) \geq e^{-c' h} \text{?}$$



Variance lower bound for natural exponential family
The desired result obviously fails to hold if the probability measure $Q$ is degenerate, that is, supported on a singleton set. Indeed, then $V(h)=0$ for all $h$.


It is much harder to construct a counterexample with a non-degenerate $Q$. The idea of such a counterexample, given below, is to make the distribution $Q$ highly lacunary, in order to make the tilted distribution $Q\_h$ very close to a degenerate one, for large $h$.


Indeed, let
\begin{equation\*}
 Q:=\sum\_{i\ge1}p\_i\delta\_{2^i},
\end{equation\*}
where $\delta\_a$ is the Dirac distribution supported on the set $\{a\}$, $p\_i:=Ce^{-4^i}$, and $C$ is the normalizing constant. Then
\begin{equation\*}
 V(h)=\frac IJ, \tag{10}\label{10}
\end{equation\*}
where
\begin{equation\*}
 I:=I(h):=\frac12\sum\_{i,j\ge1}(2^j-2^i)^2z\_i(h)z\_j(h) \\ 
 \le S:=S(h):=\sum\_{1\le i<j}4^j z\_i(h)z\_j(h), \tag{20}\label{20}
\end{equation\*}
\begin{equation\*}
 J:=J(h):=M(h)^2=\Big(\sum\_{i\ge1}z\_i(h)\Big)^2, \tag{30}\label{30}
\end{equation\*}\begin{equation\*}
 z\_i(h):=e^{h2^i}p\_i.
\end{equation\*}


For large natural $k$, let now
\begin{equation\*}
 h=h\_k:=2^{k+1}, \tag{35}\label{35}
\end{equation\*}
so that
\begin{equation\*}
 z\_i(h\_k)=y\_i:=y\_{k,i}:=C\exp(2^{k+1+i}-4^i). 
\end{equation\*}
Then  

\begin{equation\*}
 \frac{y\_{i+1}}{y\_i}=\exp(2^{k+i}(2-3\times2^{i-k})) \\ \le\exp(-2^{k+i})\le\exp(-2^{i+1}) 
 \text{ for natural $i\ge k$} \tag{40}\label{40}
\end{equation\*}
and
\begin{equation\*}
 \frac{y\_{i+1}}{y\_i}\ge\exp(2^{k-1+i})\ge1 \\
 \text{ for natural $i<k$.} \tag{50}\label{50}
\end{equation\*}
In particular, \eqref{30} and \eqref{40} imply
\begin{equation\*}
 \frac{y\_{k+1}}{y\_k}\le\exp(-4^k)\quad\text{and}\quad \frac{y\_k}{y\_{k-1}}\ge\exp(4^{k-1}). \tag{60}\label{60} 
\end{equation\*}


Now recall \eqref{20} and write
\begin{equation\*}
 S=S\_1+S\_2+S\_3, \tag{70}\label{70}
\end{equation\*}
where
\begin{equation\*}
 S\_1:=\sum\_{i,j\colon k\le i<j}4^jy\_iy\_j, \quad 
 S\_2:=\sum\_{i,j\colon 1\le i<j\le k}4^jy\_iy\_j, \quad 
 S\_3:=\sum\_{i,j\colon i<k<j}4^jy\_iy\_j. 
\end{equation\*}
In view of \eqref{40}, \eqref{50}, and \eqref{60},
\begin{equation\*}
 S\_1\le c\_1y\_ky\_{k+1}\le c\_1 e^{-4^k}y\_k^2, 
\end{equation\*}
where $c\_1:=1+\sum\_{j\ge1}j4^j e^{-2^j}\in(0,\infty)$;
\begin{equation\*}
 S\_2\le k^2 4^k y\_{k-1}y\_k\le k^2 4^k e^{-4^{k-1}}y\_k^2\le e^{-4^{k-2}}y\_k^2, 
\end{equation\*}
since $k$ is large;
\begin{equation\*}
 S\_3\le c\_2 k y\_{k-1}y\_k\le e^{-4^{k-2}}y\_k^2, 
\end{equation\*}
where $c\_2:=\sum\_{j\ge1}4^j e^{-2^j}\in(0,\infty)$.


So, by \eqref{20} and \eqref{70},
\begin{equation\*}
 I\le e^{-4^{k-3}}y\_k^2.
\end{equation\*}
On the other hand, clearly, $J\ge y\_k^2$. Thus, recalling \eqref{35},
\begin{equation\*}
 V(h)=V(h\_k)\le e^{-4^{k-3}}<e^{-c'2^{k+1}}=e^{-c'h\_k}
\end{equation\*}
for any real $c'>0$ and all large enough $k$. $\quad\Box$


