I read the following paragraph. Let $B_t, \ t \in [0, \infty)$ be a standard linear Brownian motion. For each $q > 4$, define the following sequence of sets.
$$ \Omega_k := \left\{\omega \in \Omega: \min_{q^k \leq t \leq q^{k+1}} B_t(\omega) - B_{q^k}(\omega)\geq -\sqrt{q^k}\right\}. $$
Then it states that by scaling invariance, it is true that $\mathbb P(\Omega_k)$ is independent of $k$. I could not see how this result is derived. Could anyone explain to me, please? Thank you!
For the Brownian motions $$W_t := B_{q^k+t}-B_{q^k} \qquad \quad \tilde{W}_t := \frac{1}{\sqrt{q^{k+1}-q^k}} W_{t (q^{k+1}-q^k)} $$ we have $$\begin{align*} \mathbb{P}(\Omega_k) &= \mathbb{P} \left( \min_{t \leq q^{k+1}-q^k} W_t \geq - \sqrt{q^k} \right) = \mathbb{P} \left( \min_{t \leq 1} W_{t (q^{k+1}-q^k)} \geq - \sqrt{q^k} \right) \\ &= \mathbb{P}\left( \sqrt{q^{k+1}-q^k} \min_{t \leq 1} \tilde{W}_t \geq - \sqrt{q^k} \right) \\ &= \mathbb{P} \left( \sqrt{q-1} \min_{t \leq 1} \tilde{W}_t \geq -1 \right). \end{align*}$$