Is $\frac{W^{2}(t)}{t}$ a martingale w.r.t. the usual filtration? $t>0$ and $W(t)$ is the Wiener process.
What I have so far: Define $Z(t)=\frac{W^2(t)}{t}$.
By Ito we get
$dZ(t)=\frac{-W^2(t)}{t^2}dt+\frac{2W(t)}{t}+\frac{1}{t}dt$
Now we find
$Z(t)=\int_{0}^{t}\frac{s-W^2(s)}{s^2}ds+2\int_{0}^{t}\frac{W(s)}{s}dW(s)$
Taking conditional expectations gives
$\mathbb{E}[Z(t)|F_{u}]=\mathbb{E}[\int_{0}^{u}\frac{s-W^2(s)}{s^2}ds|F_{u}]+\mathbb{E}[\int_{u}^{t}\frac{s-W^2(s)}{s^2}ds|F_{u}]+2\mathbb{E}[\int_{0}^t\frac{W(s)}{s}dW(s)|F_{u}]$ $=\int_{0}^{u}\frac{s-W^2(s)}{s^2}ds+2\int_0^u\frac{W(s)}{s}dW(s)+\mathbb{E}[\int_{u}^{t}\frac{s-W^2(s)}{s^2}ds|F_{u}]$
$=Z(u)+\mathbb{E}[\int_{u}^{t}\frac{s-W^2(s)}{s^2}ds|F_{u}]$
So if $\mathbb{E}[\int_{u}^{t}\frac{s-W^2(s)}{s^2}ds|F_{u}]=0$ then $Z(t)$ is a martingale.
I know that $\mathbb{E}[W^2(s)]=s$. Can I just say $\mathbb{E}[\int_{s}^{t}\frac{s-W^2(s)}{s^2}ds|F_{u}]=\int_{u}^{t}\frac{s-\mathbb{E}[W^2(s)|F_{u}]}{s^2}ds$?? And then does $\mathbb{E}[W^2(s)]=s$ imply $\mathbb{E}[W^2(s)|F_{u}]=s$??
Is Z a martingale?