Optional Stopping Theorem for martingales

Question

I'm having a bit of trouble understanding/follow the following I found in a proof:

For basic BM, the probability of hitting $a$ before $-b$

E(B(τ )) = E(B(0)) = 0 implies that $aP_a$ − b(1 − $P_a$) = 0;

Why is that implied?

Also....

"Here we use optional stopping on the the MG $B^2(t)$ − t yielding E($B^2$(τ )) = E(τ), or E(τ) = $a^2P_a$ + $b^2(1 − P_a)$ = ab"

I follow E($B^2$(τ )) = E(τ) from the basic definition of BM, but not the part after that.

Answer 1

$B(\tau)$ takes the value $a$ with some probability $P_a$ and $-b$ with probability $1-P_a.$ It can't take any other value since we've defined $\tau$ to be the moment it hits $a$ or $-b.$ $B(\tau)$ must be either $a$ or $-b.$

The expected value of this two-valued distribution is just $E(B(\tau)) = aP_a-b(1-P_a).$

Your second statement has a similar answer.