Show that this process is a martingale Let $B_t$ be a Brownian motion and $M_t=\max_{0\leq s\leq t}B_s$. Show that: $$(M_t-B_t)^4-6t(M_t-B_t)^2+3t^2$$
is a martingale for $t\geq0$. 
 A: I would go this way using the fact  that (see my comment below) :  
$(M_t-B_t)$ and $|B_t|$ are processes equal in law. 
I would use Itô's lemma with $f(x,t)=x^2-6t.x+3t^2$ and show (if true) that there is no drift.  
The delicate part is then to justify the fact that if you have 2 processes equal in law and if one of them is a martingale then the other one is also a martingale.
Comment (edited thank's to did's comment):
For the fact that  $(M_t-B_t)$ and $|B_t|$ are equal in law you can have a look at Karatzas and Shreve's book for a proof. 
Best regards
A: Observe that
$$(M_t-B_t)^4 = \left(\max_{0 \leq r \leq t} (B_r-B_t)\right)^4 = \max_{0 \leq r \leq t} ((B_r-B_t)^4) \\ = \max_{0 \leq r \leq s} ((B_r-B_s)^4) + \max_{0 \leq r \leq t-s} \left(((B_{s+r}-B_s)-(B_t-B_s))\right)^4$$
Moreover,
$$\max_{0 \leq r \leq t-s} ((B_{s+r}-B_s)-(B_t-B_s))^4 = \max_{0 \leq r \leq t-s} (W_r-W_{t-s})^4 = (M_{t-s}^\ast-W_{t-s})^4$$
where $W_{r} := B_{s+r}-B_s$ (is again Brownian motion), $M_t^\ast := \max_{0 \leq r \leq t} W_r$. From reflection principle we know that
$$M_{t-s}^\ast - W_{t-s} \sim \sqrt{\frac{2}{\pi \cdot (t-s)}} \cdot \exp \left(- \frac{x^2}{2(t-s)} \right) \cdot 1_{(0,\infty)}(x)$$
hence $$\mathbb{E}((M_{t-s}^\ast-W_{t-s})^4)=3(t-s)^2$$
Thus
$$\mathbb{E}((M_t-B_t)^4|\mathcal{F}_s) = (M_s-B_s)^4 + \mathbb{E} \left( \max_{0 \leq r \leq t-s} (B_{s+r}-B_t)^4 |\mathcal{F}_s \right) \\ =(M_s-B_s)^4 + \underbrace{\mathbb{E} \left( \max_{0 \leq r \leq t-s} (B_{s+r}-B_t)^4 \right)}_{\mathbb{E}((M_{t-s}^\ast-W_{t-s})^4)} =(M_s-B_s)^4 +3 \cdot (t-s)^2$$
Similar proof works for $(M_t-B_t)^2$. Adding it all up finishs the proof.
