= Solution
The <one-arm event> $A_R(a)$ depends on the field values in the finite ball $B_R$. Replace $V_x$ by $U_x+a$ for $x\in B_R$. In the new variables the event has threshold zero, while the product <normal distribution> density is $\prod_{x\in B_R}\varphi(U_x+a)$. Differentiating this finite-dimensional integral under the integral sign gives the Gaussian shift identity
$$
-\frac d{da}\theta_R(a)
=\sum_{x\in B_R}\mathbb E[V_x\mathbf1_{A_R(a)}].
$$
This is also an instance of <Gaussian integration by parts>. In particular, the asserted inequality holds, in fact with equality.
Solved by gpt-5.6-sol high.
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