Little help is needed Can I use geometric Brownian motion here?
The question I get:
$Let z=(z_t)$ be a one-dimensional standard Brownian motion and define the process $y = ( y_t )$ by $y_t = z_t^2 − t$ .
Show that $y_t = 2t\int^t_0 z_udz_u$. Hint: Notice that $y_t = g(z_t, t)$ where the function $g(x, t) =x^2−t$. Then apply Itˆo’s formula.
how that $E[y_t] = 0$ and that $var[y_t] = 2t^2$.
$z$ is a one-dimensional standard Brownian motion - it is another name for the Wiener process. By the definition of Wiener process w have that $z_t - z_0$ is normally distributed with mean $0$ and variance $t$.
Applying the Ito's lemma to $y=g(t,x)=x^2-t$ one obtains
$$dg(t,z)=g_t(t,z)dt+g_z(t,z)dz+\frac{1}{2} g_{zz}(t,z)dt$$
so we have $dy=2zdz$, and if we rewrite this as a stochastic integral we have the desired result.
By definition $\mathrm{Var}(z_t)=E[z_t^2]-(E[z_t])^2=t$, while the martingale property of the Wiener process implies $E[z_t]=z_0=0$, consequently $E[z_t^2-t]=E[y_t]=0$.
Further calculation gives $\mathrm{Var}(z_t^2-t)=E[(z_t^2-t)^2]-(E[z_t^2-t])^2=E[z_t^4]-2tE[z_t^2]+t^2.$ Since $E[z_t^4]$ is the 4th central moment of random variable with normal distribution so it is equal $3\mathrm{Var}(z_t)^2=3t^2$.
Pulling everything together: $\mathrm{Var}(z_t^2-t) = 3t^2 - 2t^2 + t^2 = 2t^2.$