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 $s<t$ 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.