Let $ B(t)(t\geq 0) $ be the standard Brownian motion and $ M(t)=\max_{0\leq s\leq t}{B(s)} $. If we define $ X(t)=M(t)-B(t) $ as a new stochastic process, how can I show that $ X(t) $ has the same distribution as $ B(t) $? I have tried to use the property of Brownian motion to calculate $ P(M(t)>x,B(t)>y) $, but I cannot figure out this probability when $ x>y $. Can anyone give some hint or thoughts about it?