Denote $P^m$ as the probability distribution of a d-dimensional Brownian motion starting at $m \in D \subset \mathbb{R}^d$. Let $x_0 \in \partial D$. Let $\tau_D$ be the random hitting time and suppose we know $\tau_D < h$. Let $\delta >0$
Why is it true that: $$P^x\bigg(\sup_{t \in [0,h]}|B_t-x_0|\geq \delta \bigg) \leq P^0 \bigg(\sup_{t \in [0,h]}|B_t+x-x_0|\geq \delta \bigg)$$
These probabilities are actually equal. By translation invariance, i.e. $\{B_t-B_0,t\ge 0\}$ is independent of $B_0$ and has the same law as a BM starting at zero, we have
\begin{align} \mathsf{P}^x\left(\sup_{t \in [0,h]}|B_t-x_0|\geq \delta \right) &=\mathsf{P}^0 \left(\sup_{t \in [0,h]}|B_t+x-x_0|\geq \delta \right) \\ &= \mathsf{P}^{x-x_0} \left(\sup_{t \in [0,h]}|B_t|\geq \delta \right). \end{align}