Let a stock price be a geometric Brownian motion
$$
d S(t)=\alpha S(t) d t+\sigma S(t) d W(t),
$$
and let $r$ denote the interest rate. We define the market price of risk to be
$$
\theta=\frac{\alpha-r}{\sigma}
$$
and the state price density process to be
$$
\zeta(t)=\exp \left\{-\theta W(t)-\left(r+\frac{1}{2} \theta^2\right) t\right\} .
$$
(i) Show that
$$
d \zeta(t)=-\theta \zeta(t) d W(t)-r \zeta(t) d t .
$$
(ii) Let $X$ denote the value of an investor's portfolio when he uses a portfolio process $\Delta(t)$. From (4.5.2), we have
$$
d X(t)=r X(t) d t+\Delta(t)(\alpha-r) S(t) d t+\Delta(t) \sigma S(t) d W(t) .
$$
Show that $\zeta(t) X(t)$ is a martingale. (Hint: Show that the differential $d(\zeta(t) X(t))$ has no $d t$ term.)
(iii) Let $T>0$ be a fixed terminal time. Show that if an investor wants to begin with some initial capital $X(0)$ and invest in order to have portfolio value $V(T)$ at time $T$, where $V(T)$ is a given $\mathcal{F}(T)$-measurable random variable, then he must begin with initial capital
$$
X(0)=\mathbb{E}[\zeta(T) V(T)] .
$$
In other words, the present value at time zero of the random payment $V(T)$ at time $T$ is $\mathbb{E}[\zeta(T) V(T)]$. This justifies calling $\zeta(t)$ the state price density process.
ons to conclude that