Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\mathbb{E}[\sigma^2]$?

share|cite|improve this question
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
up vote 0 down vote accepted

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

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.