The field $\psi$ is centered and jointly <Gaussian random field>[Gaussian]. Since the <Gaussian free field> has covariance $\mathbb E[\varphi_x\varphi_y]=g(x,y)$ and $h_x=g(x,0)/g(0,0)$,
$$
\begin{aligned}
\mathbb E[\psi_x\psi_y]
&=g(x,y)-h_xg(0,y)-h_yg(x,0)+h_xh_yg(0,0)\\
&=g(x,y)-\frac{g(x,0)g(0,y)}{g(0,0)}.
\end{aligned}
$$
Also $\psi_0=0$ and
$$
\mathbb E[\psi_x\varphi_0]=g(x,0)-h_xg(0,0)=0.
$$
Joint Gaussianity turns this zero <covariance> into independence. Therefore $\psi$ is a <Pinned Gaussian free field> at $0$, independent of $\varphi_0$, with covariance equal to the Green function of the walk killed on hitting $0$.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 226 / 2 / d / Solution

{c}

Pinning a Gaussian free field at a vertex $o$ gives the field with value zero at $o$ and covariance
$$
g^{\{o\}^c}(x,y)=g(x,y)-\frac{g(x,o)g(o,y)}{g(o,o)}.
$$
It is independent of the removed scalar Gaussian mode at $o$.


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