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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 $t\geq0$.

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2 Answers 2

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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

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Suggestion: The processes $M-B=(M_t-B_t)_{t\geqslant0}$ and $|B|=(|B_t|)_{t\geqslant0}$ are equal in law. –  Did Dec 17 '12 at 13:16
    
If we apply Ito's formula to $f(|B_t|,t)$, we have: $$f(|B_t|,t)=F(|B_0|,0)+\int_0^t12s|B_s|\text{d}s+\int_0^t4|B_s|^3-12s|B_s| \text{d} |B_s|+\\ +\frac{1}{2}\int_0^t12|B_s|^2-12s\text{d}{\left<|B|,|B|\right>}_s$$ How can we proceed? –  Nick Papadopoulos Dec 17 '12 at 18:03
    
@Nick : Apply Itô not to $|B_t|$ but to $B_t^2$. I edit my post accordingly. –  TheBridge Dec 17 '12 at 21:15

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.

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@ saz : I am not sure to understand how you can switch expectation with max operator in your proof, could you please give some details ? Best regards –  TheBridge Dec 17 '12 at 14:26
    
sorry I still don't get it, take 2 positive random variables, we usually don't have $E[X_1\vee X_2]=E[X_1]\vee E[ X_2]$ this is what I dont get in your proof even with the argument of monotone convergence. Best regards. –  TheBridge Dec 17 '12 at 15:12
    
@TheBridge I thought that it would be simply monotone convergence for conditional expectations - but I see your point. I'll check whether I can fix it somehow... –  saz Dec 17 '12 at 15:39
    
@TheBridge I corrected it. Hopefully it's fine now. –  saz Dec 17 '12 at 16:01
    
I am convinced now. ps : the fact that processes $(M_t-B_t)$ and $|B_t|$ have the same law as also comes from reflection principle I think. –  TheBridge Dec 17 '12 at 16:18

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