 Solution

Put $a=1/4$ and $E-a=\{\gamma -a:\gamma \in E\}$. Both events consist of loops surrounding $0$; $E-a$ lies in the disc of radius $3/4$, so its loops remain in $D$. Translation invariance and the pointed representation implySince $C>0$, cancel it. Outer boundaries commute with translation, and $\Gamma (B)\in E-a$ is exactly $\Gamma (B+a)\in E$. HenceThe half-disc condition ensures that translating a relevant filled Brownian trace still lies inside the original unit-disc cutoff. This proves the asserted event identity; it does not claim translation invariance of the entire measure rooted at $0$.