$$Z(t)=(\frac{S(t)}{H})^p$$where $S$ has a standard Black-scholes Dynamics for a stock, $H$ is a postive constant and $p =1 - \frac{2r}{\sigma^2}$. How can I show that $Z(t)/Z(0)$ is a postive Q-martingale with mean 1?
The questions doesn't specify anything about the filtration beside that we are in the standard Black-SCholes model. My concern here is how to treat $S_0$. Will $S_0$ be a process or simply a constant?
Assuming standard BS dynamics for $S_t$, you have
$\frac{Z_t}{Z_0}\equiv(\frac{S_t}{S_0})^p = \exp{((rt-\frac{\sigma^2}{2}t+\sigma W_t)(1-\frac{2r}{\sigma^2}))}$
Now, this is a lognormally distributed r.v. and the Gaussian inside the exponential has mean and variance
$m=2rt - \frac{\sigma^2}{2}t -\frac{2r^2}{\sigma^2}t$
$v = (\sigma-\frac{2r}{\sigma})^2 t$
Applying the standard formula for the expectation of a lognormal r.v. $E(e^X)=e^{m_X+\frac{v_X}{2}}$, you will find after a little arithmetic that
$E(\frac{Z_t}{Z_0}) = 1$
As a starting point: for price dynamics $dS(t) = rS(t)dt + \sigma S(t) dW^\mathbb{Q}(t)$, to show that $Z(t)/Z(0)$ is a positive mean 1 Q-martingale, use Itô's formula to get:
$dZ(t)=p \sigma Z(t)dW^\mathbb{Q}(t)$
