Checking if $B_t^3 $ and $3tB_t$ are martingales?

Question

$$\mathbb{E}[ B_t^3 - 3tB_t + 3B_t | \mathcal{F}_s]$$

$$\mathbb{E}[B_t^3 | \mathcal{F}_s] - 3\mathbb{E}[t B_t | \mathcal{F}_s\}$$

$$\mathbb{E}[(B_t^3 - B_s^3 + B_s^3) | \mathcal{F}_s] + [ not \space sure \space about \space this ]$$

Using $(a-b)(a^2+ab+b^2)$ for cube expansions, with $a = B_t - B_s, b = B_s$

... I have tried to simplify this expression, but nothing fruitful yielded.

I got $$ B_t^3 - B_t^2B_s + B_tB_s - 2B_sB_t^2 + 2B_tB_s^3 - 2B_s^3$$

Which must be a mistake... but is there perhaps a better method to approach this?

Also, for the $$-3 \mathbb{E}[t B_t | \mathcal{F}_s] $$ where can I proceed from here? My instinct tells me $t$ is $\mathcal{F}$ - measurable, although I am not 100% confident.

Answer 1

\begin{align} E[B_t^3|\mathcal{F}_s]&=E[(B_t-B_s+B_s)^3|\mathcal{F}_s]\\ &=E[(B_t-B_s)^3+3(B_t-B_s)^2B_s+3(B_t-B_s)B_s^2+B_s^3|\mathcal{F}_s]\\ &=E[3(B_t-B_s)^2B_s|\mathcal{F}_s]+B_s^3\neq B_s^3\\ E[3tB_t|\mathcal{F}_s]&=E[3t(B_t-B_s+B_s)|\mathcal{F}_s]\\ &=E[3t(B_t-B_s)|\mathcal{F}_s]+E[3tB_s|\mathcal{F}_s]\\ &=3tB_s \neq 3sB_s \end{align}

---

EDIT: \begin{align} E[3(B_t-B_s)^2B_s|\mathcal{F}_s]&=3B_sE[(B_t-B_s)^2|\mathcal{F}_s]\\ =3B_s(t-s) \end{align}