Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 48 / 4

An instantaneous forward rate curve determines the unit-face-value zero-coupon bond through $f(t,T)=-{\partial }_T\log P(t,T)$ and $P(t,t)=1$. Integrating in maturity givesThe diagonal is the short rate, $\boxed{r_t=f(t,t)}$, and the continuous-time bank account has $B_t=\exp ({\int }_0^t{r}_{s}ds)$ when $B_0=1$. Therefore the discounted bond price martingale has valueUnder the risk-neutral measure of the Heath-Jarrow-Morton model, sufficiently integrable discounted bond prices are true martingales; without that integrability their stochastic dynamics initially establish a local martingale.

Since $d(B_t^{-1})=-r_t{B}_t^{-1}dt$, multiplying the given bond dynamics by $B_t^{-1}$ cancels the drift and gives $d{Z}_t=Z_t\Sigma (t,T)d{W}_t^Q$. For an integrable discounted terminal payoff, the risk-neutral value process is the discounted-payoff martingaleFor an attainable claim this also follows directly from its self-financing discounted wealth being a true martingale, with terminal value $B_T^{-1}X$. In a Brownian market, the Brownian martingale representation theorem provides $d{M}_t={\psi }_{t}d{W}_t^Q$; a traded bond with nonzero diffusion exposure can hedge it by holding ${\psi }_t/(Z_t{\Sigma }_t)$ bond units, with the remainder in the bank account. Thus, under the usual attainability and integrability assumptions, this construction gives the replicated price as well as the risk-neutral pricing formula.

Because $B_t$ is ${\mathcal{F}}_t$-measurable,In a general Heath-Jarrow-Morton model, writing this value as $V(t,r_t)$ requires an additional state-reduction assumption. The conditional expectation is valid without asserting that the short rate alone is the full state.

Assume the forward field is sufficiently differentiable in maturity to differentiate its deterministic and stochastic integrals. Moving along the diagonal adds the maturity derivative to the fixed-maturity dynamics:The Heath-Jarrow-Morton model no-arbitrage drift is ${\alpha }_Q(t,T)=\sigma (t,T){\int }_t^T\sigma (t,u)du$, so ${\alpha }_Q(t,t)=0$. Equivalently, from the integrated forward equation,That drift can depend on the entire previous forward curve or driving path, rather than on the current scalar rate.

The short rate need not be Markov. For a concrete One Brownian factor does not imply a Markov short rate example, take the deterministic volatility $\sigma (t,T)=T-t$. Its admissible risk-neutral drift is $(T-t{)}^3/2$, andSubtracting the deterministic part leaves $I_t={\int }_0^t{W}_{s}ds$. The past of $I$ reveals its derivative $W_t$, and $\mathbb{E}[I_{t+h}-I_t\mid {\mathcal{F}}_t^I]=h{W}_t$. Yet $W_t$ is not determined by $I_t$: the Gaussian covariance calculation givesConsequently the conditional law of the future depends on more than $I_t$, proving failure of the Markov property for the scalar short rate. The enlarged pair $(I_t,W_t)$ is Markov. Special maturity-volatility structures, such as those producing a Vasicek-type short-rate diffusion, can instead admit a scalar Markov realization.