If $B_t$ is a standard brownian motion process, is $B_t^2 - \frac{t}{2}$ a martingale w.r.t. brownian motion?

Question

If I have that $B_t$ is a standard brownian motion process, is $B_t^2 - \frac{t}{2}$ a martingale w.r.t. brownian motion? I know that $B_t^2 - t$ is but can't see it for the latter.

Answer 1

No, it's not. Just calculate the conditional expectation. It is a submartingale.

Answer 2

\begin{align} E(B_t^2-\frac t2 \Big|B_s)&=E((B_t-B_s+B_s)^2-\frac t2 \Big|B_s)\\ &=E((B_t-B_s)^2+B_s^2+2B_s(B_t-B_s)-\frac t2 \Big|B_s)\\ &=t-s+B_s^2-0-\frac t2 \\ &=B_s^2-\frac s2+\frac {t-s}2 \\ & \geq B_s^2-\frac s2 \end{align}