# martingale representation of two independent Brownian motion

Let $$W^1_t$$ and $$W^2_t$$ be two independent standard Brownian motions. Then $$W^2_t$$ is a martingale with respect to the its own filtration but not adapted to the filtration generated by $$W^1_t$$. I want to proof this. I have to show: There cannot be an Ito integrand $$H$$ such that

$$W_t^2=\int_0^t H dW^1_s$$

Applying Ito's formula on $$W_t^1W_t^2$$ yields:

$$W_t^1W_t^2=\int_0^t W_s^2 dW^1_s +\int_0^t W_s^1 dW^2_s$$

How can I make my argument from here on rigious?

To see that $$W_{2,t}$$ is a martingale wrt $$\mathcal{F}^{W_2}_s:=\sigma(\{W_{2,h},h\leq s\})$$ for $$s we see that, since $$W_{2,t}-W_{2,s}\sim\mathcal{N}(0,t-s)$$ is independent from $$\mathcal{F}^{W_2}_s$$ and $$W_{2,s}$$ is $$\mathcal{F}^{W_2}_s$$-measurable so it can be 'taken out' from the conditional expectation we get $$E[W_{2,t}|\mathcal{F}^{W_2}_s]=E[(W_{2,t}-W_{2,s})+W_{2,s}|\mathcal{F}^{W_2}_s]=W_{2,s}E[1|\mathcal{F}^{W_2}_s]=W_{2,s}$$ To see that $$W_{2,t}$$ is not a $$\mathcal{F}_t^{W_1}$$-martingale, since $$W_{2,s}$$ is independent from $$W_{1,s}$$ we get $$E[W_{2,t}|\mathcal{F}_s^{W_1}]=E[W_{2,s}|\mathcal{F}_s^{W_1}]=E[W_{2,s}]=0$$ which is not a.s. equal to $$W_{2,s}$$. Indeed for $$A \in \mathcal{F}_s^{W_1}$$ \begin{aligned}E[E[W_{2,s}|\mathcal{F}_s^{W_1}]\mathbb{I}_A]&=E[W_{2,s}\mathbb{I}_A]=\\&=E[W_{2,s}]E[\mathbb{I}_A]=\\ &=E[E[W_{2,s}]\mathbb{I}_A]\end{aligned} so $$E[W_{2,s}|\mathcal{F}_s^{W_1}]=E[W_{2,s}]=0$$ a.s.
• $$E[W_{2,t}|\mathcal{F}_s^{W_1}]=E[W_{2,s}|\mathcal{F}_s^{W_1}]=E[W_{2,s}]=0$$ Why is the time index s in the second step,i.e. $W_{2,s}$ Sep 12, 2021 at 10:15
• @Sarah $$E[W_{2,t}|\mathcal{F}^{W_1}_s]=E[(W_{2,t}-W_{2,s})+W_{2,s}|\mathcal{F}^{W_1}_s]=E[W_{2,s}|\mathcal{F}^{W_1}_s]$$ Sep 12, 2021 at 10:18
• I think we could argue like this: $$W_{2,t}\textrm{ is a }\mathcal{F}_t^{W_1}\textrm{-martingale} \implies E[W_{2,t}|\mathcal{F}_s^{W_1}]=W_{2,s}\textrm{ a.s.}$$ $$E[W_{2,t}|\mathcal{F}_s^{W_1}]\neq W_{2,s}\textrm{ a.s.} \implies W_{2,t}\textrm{ is not a }\mathcal{F}_t^{W_1}\textrm{-martingale}$$ what do you think? Sep 12, 2021 at 10:36
• @Sarah you're welcome. you should post another question for the case with $B$ Sep 12, 2021 at 10:56