If $B(t)$ is standard $1$d Brownian motion, the why does $t \cdot B(1/t)$ have Gaussian distributin?

Question

I think the title is self-explanatory. In fact, this is only part of proving that $t B(1/t)$ is standard $1$d Brownian motion, but I understand the other steps. And in similar questions, the process being Gaussian is said to be trivial.

Answer 1

If $\{B(t)\}_{t\geq 0}$ is a standard Brownian motion, then for each fixed real number $s>0$, $B(s)$ is a Gaussian random variable with mean zero and variance $s$.

In particular, setting $s=\frac{1}{t}$ for some $t>0$ shows that $B(\frac{1}{t})$ is normally distributed with mean zero and variance $\frac{1}{t}$, hence $tB(\frac{1}{t})$ is a Gaussian with mean zero and variance $\frac{t^2}{t}=t$.