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I came across the following exercise in Stochastic Calculus:

Let $B=(B_t)_{t\geq0}$ be a standard Brownian motion. Let also $M$ be the following process:

$M_t=B^4_t-6t(B^2_t-\dfrac{t}{3})$ for $t\geq0$

Prove that $M=(M_t)_{t\geq0}$ is a martingale and if we set $\sigma=inf\{t\geq0 : |B_t|=\sqrt{3}\}$, compute $\mathbb{E}[M_{\sigma}]$ and $\mathbb{E}[\sigma^2]$.

It was easy to prove the martingale part. Can we use the Optional Sampling Theorem for $\sigma$ and if so, how can we calculate $\mathbb{E}[\sigma^2]$?

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Are you sure that $\frac{t}{3}$ (in the definiton of $M_t$) is correct? I think it should be $\frac{t}{2}$... –  saz Nov 20 '12 at 20:17

1 Answer 1

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Let $\tau := \inf\{t \geq 0; B_t \in (-a,b)^c\}$ where $a,b>0$. Using Wald's identities one can show that $\mathbb{E}\tau = a \cdot b$. Moreover $\mathbb{E}\tau^n<\infty$ for all $n \geq 1$ (Proof (Exercise 5.16(a))). Choose $a=b=\sqrt{3}$, then $\tau=\sigma$. Thus $\sigma \in L^2$, $\sigma \in L^1$, $\mathbb{E}\sigma = \sqrt{3} \cdot \sqrt{3}=3$. We obtain

$$M_{\sigma}(w) = 3^2 - 6\sigma(w) \cdot \left(3-\frac{\sigma(w)}{2} \right) \\ \Rightarrow \mathbb{E}M_\sigma = 9 - 18 \underbrace{\mathbb{E}\sigma}_{\sqrt{3} \cdot \sqrt{3}} + 3 \mathbb{E}(\sigma^2) = 9-18 \cdot 3 +3\mathbb{E}(\sigma^2)$$

On the other hand

$$\mathbb{E}(M_{t \wedge \sigma}) = \mathbb{E}M_0 = 0$$

by optinal sampling theorem (applied to the stopping time $\sigma \wedge t$). Since

$$|M_{t \wedge \sigma}| \leq 3^2 + 6\sigma \cdot \left(3+\frac{\sigma}{2}\right) \in L^1$$

we have (by dominated convergence) $\mathbb{E}M_\sigma = 0$. This implies

$$\mathbb{E}\sigma^2 =\frac{45}{3}=15$$

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Yes, you are absolutely right! It's $M_t := B_t^4-6t \cdot \left(B_t^2-\frac{t}{2} \right)$!$\\$ Would it suffice to say that: $P[-\sqrt{3}<B_t<\sqrt{3}]=P[-\dfrac{\sqrt{3}}{\sqrt{t}}<\dfrac{B_t}{\sqrt{t}}<\‌​dfrac{\sqrt{3}}{\sqrt{t}}]=2\Phi(\dfrac{\sqrt{3}}{\sqrt{t}})-1$ whis is zero as $t\rightarrow\infty$ and so $\sigma$ is almost surely bounded? –  Nick Papadopoulos Nov 21 '12 at 10:05
    
No, $\mathbb{P}[|B_t|<\sqrt{3}] \to 0$ as $t \to \infty$ implies only $\mathbb{P}[\sigma<\infty]=1$. As far as I can see $\sigma$ is not necessarily a.s. bounded. –  saz Nov 21 '12 at 10:43
    
If so, how can we use the Optional Sampling Theorem? Shoudn't we have that $\mathbb{E}[\sigma]<\infty$ in order to use it? Sorry for my perhaps naive question but I am a novice and I am trying to clear things out in my head. –  Nick Papadopoulos Nov 21 '12 at 11:02
    
I applied the Optional Sampling Theorem to the stopping time $\sigma \wedge t$ (which is bounded (by $t$)) - this step we could do without knowing $\sigma \in L^1$. But you are right, we need $\sigma \in L^2$, $\sigma \in L^1$ (for the rest of the proof). Here is a proof (Exercise 5.16) (they show $\mathbb{E}\tau^n<\infty$ for all $n \geq 1$ where $\tau := \inf\{t \geq 0; B_t \in (-a,b)^c\}$ ... choose $a=b=\sqrt{3}$, then $\tau=\sigma$ since $B$ has continuous paths.) –  saz Nov 21 '12 at 11:28
    
Really nice proof! And it covers every case of a stopping time of this kind! Thank you very much! If we had the following: $M_t=B^3_t-B^2_t-3t(B_t-\dfrac{1}{3})$ and we want to calculate $\mathbb{E}[M_\sigma]$ (now, I know how to do it) and $\mathbb{E}[\sigma B_\sigma]$? How will we treat $B^3_\sigma$? –  Nick Papadopoulos Nov 21 '12 at 12:59

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