For every $x,y\in\mathbb R$ let
$$ V(x,y) \,\equiv\, a\,x^{2n} + b\,y^{2m} - \omega(x,y)\,$$
where $a,b>0$, $n,m\in\mathbb N$, $n\geq m\geq1$, and $\omega$ is such that $\omega(x,y)/(x^{2n}+y^{2m})\to 0 \ \ \textrm{as}\ |x|+|y|\to\infty\,.$
Moreover assume $V(x,y)$ is a convex polynomial.


Is it possible to estimate the following expectation:
$$ E(y) \,\equiv\, \frac{\int\_{-\infty}^{\infty}\,x^{2p}\,e^{-V(x,y)}\,dx}{\int\_{-\infty}^{\infty}\,e^{-V(x,y)}\,dx} \,=\, \frac{\int\_{-\infty}^{\infty}\,x^{2p}\;e^{-a\,x^{2n}\,+\,\omega(x,y)}\;dx}{\int\_{-\infty}^{\infty}\,e^{-a\,x^{2n}\,+\,\omega(x,y)}\;dx}$$
for $p\in\mathbb N$ (say $p=1$ for simplicity) and $|y|$ large?


In particular I am conjecturing that there exists a constant $C=C(p,V)>0$ such that:
$$ E(y) \leq C + C\,y^{2p} \;.$$


This can be checked by explicit computations when $n=m=1$ (Gaussian measure). I've tried some numerics with $n=2,3$, e.g., $V(x,y) = -x^4+ (x+y)^2$, and they seem to confirm the result.
Is it possible to prove it analytically or find a counterexample?



Estimate of $\frac{\int x^{2p}\,e^{-x^{2n}\,+\,\omega(x,y)}\;dx}{\int e^{-x^{2n}\,+\,\omega(x,y)}\;dx}$
$\newcommand{\R}{\mathbb R}\newcommand{\ep}{\varepsilon}
\newcommand{\om}{\omega}$We have to show that
\begin{equation\*}
 \frac{I\_1+I\_2}J\ll 1+y^{2p}, \tag{1}
\end{equation\*}
where
\begin{equation\*}
 I\_1:=\int\_{|x|\le|y|} dx\,x^{2p}e^{-ax^{2n}+\om(x,y)},
\end{equation\*}
\begin{equation\*}
 I\_2:=\int\_{|x|>|y|} dx\,x^{2p}e^{-ax^{2n}+\om(x,y)},
\end{equation\*}
\begin{equation\*}
 J:=\int\_\R dx\,e^{-ax^{2n}+\om(x,y)},
\end{equation\*}
and $A\ll B$ and $B\gg A$ mean that $A\le CB$ for some constant $C=C(p,V)\in(0,\infty)$.


For $|y|\ll1$, (1) follows by continuity and compactness. So, without loss of generality,
\begin{equation}
 |y|\to\infty. 
\end{equation}


We have
\begin{equation\*}
 I\_1\le y^{2p} \int\_{|x|\le|y|} dx\,e^{-ax^{2n}+\om(x,y)}\le y^{2p}J. \tag{2}
\end{equation\*}
Next, for some real $A=A(V)>1$ and all real $x$ and $y$ such that $|y|>A$ we have $|\om(x,y)|<a(x^{2n}+y^{2m})/4$ and hence
\begin{equation\*}
 J>\int\_A^{2A} dx\,e^{-ax^{2n}-a(x^{2n}+y^{2m})/4}\gg e^{-ay^{2m}/4}\ge e^{-ay^{2n}/4} \tag{3}
\end{equation\*}
and, in view of the conditions $|y|>A>1$ and $n\ge m\ge1$ (and the l'Hospital rule),
\begin{align\*}
 I\_2&\le\int\_{|x|>|y|} dx\,x^{2p}e^{-ax^{2n}+a(x^{2n}+y^{2m})/4} \\ 
 &\le\int\_{|x|>|y|} dx\,x^{2p}e^{-ax^{2n}/2} \\ 
 &\ll y^{2p+1-2n}e^{-ay^{2n}/2}. \tag{4}
\end{align\*}


Now (1) follows from (2), (3), (4).


