TITLE: Showing that the square of Brownian motion, minus time, is a martingale
QUESTION [9 upvotes]: What exactly are we supposed to do to show what they have given is a martingale.
If I try to follow through I'm getting a bit confused. In the third to last line I don't understand how they have simplified the first two terms to get $0$ and $t-s$ respectively.

REPLY [2 votes]: For a Wiener process you have that
$$(W_t - W_s) \sim N(0, t-s)$$
Therefore the first term in the third to last line becomes $0$.
and the second term becomes $t-s$.