Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 226 / 3 / c

Table of contents
- Solution c

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}} φ (U_{x} + a)

. Differentiating this finite-dimensional integral under the integral sign gives the Gaussian shift identity

- \frac{d}{d a} θ_{R} (a) = \sum_{x \in B_{R}} E [V_{x} 1_{A_{R} (a)}] .

This is also an instance of Gaussian integration by parts. In particular, the asserted inequality holds, in fact with equality.

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