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What does it mean (definition) for two stochastic processes to be independent? like two independent Brownian motion $B_1(t), B_2(t)$. I come across this when I saw a solution of a problem says if $B_1(t), B_2(t)$ are independent, then $(dB_1t)(dB_2t)=0$. How do we prove this?

This is something I dont understand in part of solution from my homework problem, and here is the original problem:

Let $X(t)=B_1(t)B_2(t)$, where $B_1(t)$ and $B_2(t)$ are two independent Brownian motions, check if $X(t)$ is a martingale and find its martingale representation.

Solution: Since $\Bbb E[X(t)]=\Bbb E[B_1(t)]\Bbb E[B_2(t)]=0=X(0)$ so it is a martingale.

By Ito's product rule, $$dB_1(t)dB_2(t)=B_1(t)dB_2(t)+B_2(t)dB_1(t)+dB_1(t)dB_2(t)$$

since $B_1(t),B_2(t)$ are independent so we take $dB_1(t)dB_2(t)=0$

...

Dylan Zhu
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  • OK, I see there is a definition in this question: http://math.stackexchange.com/questions/22360/independent-stochastic-processes-and-independent-random-vectors – Dylan Zhu May 08 '14 at 19:12
  • Please use proper notations: $(dB_1)(dB_2)$ does not exist (and please correct the typo in the title). – Did May 08 '14 at 21:50
  • Second try: please provide a definition of $(dB_1t)(dB_2t)$. – Did May 09 '14 at 21:38

1 Answers1

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First of all, note that $\mathbb{E}(X_t)=0$ does in general not imply that $(X_t)_{t \geq 0}$ is a martingale. In order to prove that $(X_t)_{t \geq 0}$ is a martingale, i.e. satisfies

$$\mathbb{E}(X_t \mid \mathcal{F}_s ) = X_s, \qquad s \leq t, \tag{1}$$

you have to use the independence of the processes as well as the independence of the increments of a Brownian motion.

Turnining to your original question: Itô's formula states that

$$d(B_1(t) B_2(t)) = B_1(t) \, dB_2(t) + B_2(t) \, dB_1(t) + d\langle B_1,B_2 \rangle_t$$

where $\langle B_1,B_2 \rangle_t$ is the quadratic covariation of the processes $(B_1(t))_{t \geq 0}$ and $(B_2(t))_{t \geq 0}$. By definition, the quadratic covariation is the unique process (with some nice properties) such that

$$B_1(t) \cdot B_2(t)- \langle B_1,B_2 \rangle_t=X_t-\langle B_1,B_2 \rangle_t$$

is a martingale. Since we have already seen in $(1)$ that $(X_t)_{t \geq 0}$ is a martinale, we get $\langle B_1,B_2\rangle_t=0$. Consequently,

$$d(B_1(t) B_2(t)) = B_1(t) \, dB_2(t) + B_2(t) \, dB_1(t).$$

Remark As already pointed out by @Did you have to be careful with the notations. You cannot simply multiply differentials of stochastic processes. See e.g. this question for some (heuristical) explanations in a similar context.

Community
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saz
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  • Many thanks @saz . the condition you gave for the martingale is certainly right, but in my exam (stochastic for finance) my tutor only ask me to show $\Bbb E[X(t)]=X(0)$, it is a necessary condition for sure, but i doubt it is not sufficient to prove $X(t)$ is martingale, is it? – Dylan Zhu May 10 '14 at 12:27
  • @DylanZhu Yes, $\mathbb{E}X_t=\mathbb{E}X_0$ is a necessary but not sufficient condition for being a martingale. (And thanks, @Did.) – saz May 10 '14 at 14:13