I am reading John Hull's "Options, Futures, and Other Derivatives". I am currently in Ch. 31 on the HJM Model. Hull makes a statement which a need an explanation for. First, some notation. Let $P(t,T)$ be the price at time $t$ of a zero-coupon bond with principal $\$1$ maturing at time $T$. Let $r(t)$ be the short term risk free interest rate at time. dz(t) is a Wiener process a.k.a z(t) is a Brownian motion. And $v(t,T,\Omega_t)$ is the volatity based on $t$, $T$, and $\Omega_t$ which a vector of pertinent information. Now here is the statement I need an explanation for:
"We use the risk-neutral world. A zero-coupon bond is a traded security providing no income. Its return in the risk-neutral world must therefore be $r$. This means its stochastic process has the form $dP(t,T)=r(t)P(t,T)dt+v(t,T,\Omega_t)P(t,T)dz(t)$."
I don't understand how we get that stochastic process. If $r$ was constant as was the volaltilty, then we would have Geometric Brownian Motion and I would follow the result for the Expected value I think. But since $r$ and $v$ are not constant, I am lost. Thanks for the help!
In the risk neutral world, assets discounted at the risk free rate are martingales - that means they are neither expected to increase or decrease in value over a time horizon. This is only possible if it drifts according to $r(t)P(t,T)dt$. To see what this means, take the expectation of the present value of the bond at time $t$, and you will find it's expectation to be just $P(0,T)$.
$$ \begin{align*} d\left(e^{-\int_0^t r(t) dt} P(t,T)\right) &= e^{-\int_0^t r(t) dt} dP(t,T) + P(t,T)d\left(e^{-\int_0^t r(t) dt}\right) \\ &= e^{-\int_0^t r(t) dt} r(t)P(t,T) dt + e^{-\int_0^t r(t) dt}v(t,T,\Omega_t)P(t,T) dz(t) - e^{-\int_0^t r(t) dt}r(t)P(t,T)dt \\ &= e^{-\int_0^t r(t) dt}v(t,T,\Omega_t)P(t,T) dz(t) \\ \Longrightarrow e^{-\int_0^t r(t) dt}P(t,T) &= P(0,T)+ \int_0^t e^{-\int_0^s r(u) du}v(s,T,\Omega_s)P(s,T) dz(s) \end{align*} $$
Taking expectations shows that in fact the discounted value of the bond at time $t$ is exactly the same as its value at time 0. viz.,
$$ \mathbb{E}\left[e^{-\int_0^t r(t) dt}P(t,T)\right]=P(0,T) $$