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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?

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To see that $W_{2,t}$ is a martingale wrt $\mathcal{F}^{W_2}_s:=\sigma(\{W_{2,h},h\leq s\})$ for $s<t$ 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.

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  • $\begingroup$ $$E[W_{2,t}|\mathcal{F}_s^{W_1}]=E[W_{2,s}|\mathcal{F}_s^{W_1}]=E[W_{2,s}]=0$$ Why is the time index s in the second step,i.e. $W_{2,s}$ $\endgroup$
    – Sarah
    Sep 12, 2021 at 10:15
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    $\begingroup$ @Sarah $$E[W_{2,t}|\mathcal{F}^{W_1}_s]=E[(W_{2,t}-W_{2,s})+W_{2,s}|\mathcal{F}^{W_1}_s]=E[W_{2,s}|\mathcal{F}^{W_1}_s]$$ $\endgroup$
    – Snoop
    Sep 12, 2021 at 10:18
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    $\begingroup$ 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? $\endgroup$
    – Snoop
    Sep 12, 2021 at 10:36
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    $\begingroup$ Yes I think so too. Thank you for the clear view on this. $\endgroup$
    – Sarah
    Sep 12, 2021 at 10:40
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    $\begingroup$ @Sarah you're welcome. you should post another question for the case with $B$ $\endgroup$
    – Snoop
    Sep 12, 2021 at 10:56

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