Standard Brownian motion and stopping time

Question

Let be $B$ standard Brownian motion and let $S \leq T$ two stopping times with $E(T) < \infty $ and $E(S) < \infty$. Prove that hold $$ E[(B_T – B_S)^2] = E[B_T^2 – B_S^2] = E(T-S).$$

Please help me solve this.

Answer 1

For the first equality, expand the square $(B_T - B_S)^2$ and use properties of Brownian motion to obtain $E[B_T^2 - B_S^2]$.

By expanding the square, $$E[(B_T - B_S)^2] - E[(B_T^2 - B_S^2)]= 2 E[B_S^2] - 2 E[B_S B_T].$$ Then note $E[B_S B_T] = E[B_S E[B_T \mid S]] = E[B_S^2]$.

For the second equality, show $E[B_T]=E[T]$ for any stopping time $T$.

$E[B_T^2] = E[E[B_T^2 \mid T]] = E[T]$.