Show that $(B_t)$ and $(tB_{1_t})$ has the same distribution where ($B_t)_t$ is a brownian motion

Question

Let $(B_t)_{t\geq 0}$ a brownian motion s.t. $B_0=0$. I want to show that $B_t$ and $tB_{1/t}$ has the same law.

$$p\{tB_{1/t}\leq x\}\underset{u=1/t}{=}p\{\frac{1}{u}B_u\leq x\}=p\{B_u\leq ux\}$$ but how can $$p\{B_u\leq ux\}=p\{B_u\leq x\}$$ and thus get $$p\{tB_t\leq x\}=p\{B_t\leq x\}\ \ ?$$

Answer 1

Since $B_t\sim N(0, \,t)$ and $tB_{1/t}\sim tN(0,\,1/t)\sim N(0,\,t^2\times 1/t) \sim N(0,\, t)$, they have the same distribution.