The question lies in the title. I am trying to show that $Z(t)$ is a Martingale. The martingale property associated with the filtration I have computed..
I tried using the Hölder's Inequality to prove it, but then I arrive with having to compute the fourth moment of the Wiener process, so I think I have done something wrong.
Any help is appreciated!
Consider the function $f(x,t)=tx^2$. By Ito's lemma you have $$ df (W_t,t)= \left(\frac{\partial f}{\partial t}(W_t,t)+\frac{1}{2}\frac{\partial^2 f}{\partial x^2}(W_t,t)\right)dt +\frac{\partial f}{\partial x}(W_t,t) dW_t $$ The second addendum vanish in mean because Brownian motion are martigales so you only need to estimae the first addendum. Remember that $\mathbb{E}\left[W_t^2\right]=t$. So by applying Fubini you obtain
$$ \mathbb{E}\left[W_t^2 t\right]=2 \int_0^t s ds=t^2 $$ so it is not a martingala
EDIT: if there is a typo and the process is $W_t^2-t$ you can prove it by using the definition of Brownian motion. In particular, the fact that $W_t-W_s$ is a Gaussian random variable with law $\mathcal{N}(0,t-s)$. Because $W_0=0$ this imply: $$ \mathbb{E}\left[W_t^2\right]=\mathbb{E}\left[W_t^2-2 W_0W_t+W_0\right]=\mathbb{E}\left[(W_t-W_0)^2\right]=t $$