I did quite a few numerical computations and think the following is true, but I cannot prove it:


Let $\varphi(x):=\sum\_{i=1}^n \varphi\_i(x\_i)$ where $x=(x\_1,...,x\_n) \in \mathbb{R}^n$ and $\varphi\_i \in C^{\infty}$ are even scalar convex functions such that $\varphi''$ is strictly increasing on $[0,\infty).$


We then define a probability measure (which under apppropriate normalization) is defined as $$p\_y(x) \propto e^{\langle y, x\rangle}e^{-\varphi(x) } \ dx. $$


Can we show that for all unit vectors $z \in \mathbb{R}^n$ we have for all $y \in \mathbb{R}^n$


$$Var\_{p\_0}(\langle z,X\_0 \rangle\_{\mathbb{R}^n}) \ge Var\_{p\_y}(\langle z,X\_y \rangle\_{\mathbb{R}^n})?$$


In other words, the variance of $\langle z,X\_y\rangle$ where $X\_y$ is distributed according to $p\_y$ is maximized at $y=0$ for any unit vector $z.$


Is this a known theorem or somehow easy to show?-Any pointers are highly appreciated and please let me know if there are any questions.



Variance of random variable decreasing in parameter
Your probability measure is a product measure, so by
$$\text{Var}\_y(\langle z,X\rangle) = \sum\_{i=1}^nz\_i^2\text{Var}\_{y\_i}(X\_i)$$
everything reduces to the 1d case. Let $q\_y(dx)=e^{xy-\varphi(x)-C(y)}dx$ be one of the marginals, where $y\in\mathbb{R}$ and $C(y)$ is chosen such that $q\_y$ is normalized, and denote by $U\_y$ the 1d r.v. with distributon $q\_y$. It can be shown that $C(y)-C(0)$ is the cumulant-generating function of $U\_0$, and $\text{Var}\_y(U\_y)=C''(y)$. Thus the variance has a local maximum at $y=0$ iff the third cumulant $\kappa\_0^{(3)}$ of $U\_0$ vanishes and the fourth cumulant $\kappa\_0^{(4)}$ is negative.
Indeed, if we denote by $m\_y^{(k)}$ the $k$'th moment of $U\_y$, we have
$$\kappa\_0^{(3)}=m\_0^{(3)}-3m\_0^{(1)}m\_0^{(2)}+2\left(m\_0^{(1)}\right)^3=0,$$
because $p\_0$ is symmetric and therefore the first and third moments are zero. Similarly, dropping the odd-numbered moments, we have
$$\kappa\_0^{(4)}=m\_0^{(4)}-3\left(m\_0^{(2)}\right)^2.$$
Therefore assuming that $\varphi(x)>\lambda x^2$ a.e. for some $\lambda>0$ $(\star)$, then
\begin{align\*}
\kappa\_0^{(4)}&=\int\_{\mathbb{R}^2}\left(x^4-3x^2y^2\right)e^{-\varphi(x)-\varphi(y)-2C(0)}dx dy \\
&< e^{-2C(0)}\int\_{\mathbb{R}^2}\left(x^4-3x^2y^2\right)e^{-\lambda x^2-\lambda y^2}dx dy\\
&=0,
\end{align\*}
because the last expression is proportional to the fourth cumulant of a Gaussian.
In order to show that this is the *global* maximum, let's show that $C''(y)$ is concave, i.e. $C^{(4)}(y)<0$ for all $y\in\mathbb{R}$. In fact, since
$$C(y)=\log\int\_{-\infty}^{\infty}e^{xy-\varphi(x)}dx,$$
$C^{(4)}(y)$ is the 4'th cumulant of $U\_y$, i.e.
\begin{align\*}
C^{(4)}(y)&=m\_y^{(4)}-4m\_y^{(3)}m\_y^{(1)}-3\left(m\_y^{(2)}\right)^2+12m\_y^{(2)}\left(m\_y^{(1)}\right)^2-6\left(m\_y^{(1)}\right)^4\\
&\propto\int\_{\mathbb{R}^4}\left(x\_1^4-4x\_1^3x\_2-3x\_1^2x\_2^2+12x\_1^2x\_2x\_3-6x\_1x\_2x\_3x\_4\right)e^{\sum\_{i=1}^4(x\_iy-\varphi(x\_i))}dx\\
&<\int\_{\mathbb{R}^4}\left(x\_1^4-4x\_1^3x\_2-3x\_1^2x\_2^2+12x\_1^2x\_2x\_3-6x\_1x\_2x\_3x\_4\right)e^{\sum\_{i=1}^4(x\_iy-\lambda x\_i^2)}dx\\
&=0,
\end{align\*}
again because the last expression is proportional to the fourth cumulant of a Gaussian (with non-zero mean).


$(\star)$ This is slightly different than your 'increasing convexity' assumption, but seems sufficiently close.


