Let $(W_t)_t$ a Brownian motion. We know that for $0<s <t$ we have $$W_{t} - W_s$$ is $N(0, t-s)$.
But I've al seen that $W_s - W_t$ is distributed again as $N(0,t-s)$. Why?
Is this because for a normal random variable $X$, $-X$ has the same distribution (since the density function is symmetric)? SO I have that $-(W_t-W_s)$ is distributed ad $W_t - W_s$
Or am I still missing something?