Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 39 / 1

For the autoregressive process of order one, the formula at $t=0$ reads $r_0=r_0$, with an empty sum. If it holds at time $t$, substitution into the recursion givesThus mathematical induction provesNo restriction such as $|\beta |<1$ is needed for this finite-time identity.

Put $n=T-t$ and define the finite geometric seriesStarting the autoregressive process of order one at time $t$ gives $r_{t+k}={\beta }^k{r}_t+{\sum }_{j=1}^k{\beta }^{k-j}{\xi }_{t+j}$. ConsequentlyThe future innovations are independent and identically distributed random variables, independent of ${\mathcal{F}}_t$. Factorizing their exponential conditional expectation and using the cumulant-generating function $K$ yieldsThus the exponential-affine bond pricing coefficients areThey vanish at maturity, giving $P(T,T)=1$. Everywhere finiteness of $K$ ensures that each zero-coupon bond price is positive and finite, including when $\beta$ is zero, negative, or one.

Let $q(T)$ denote the original zero-coupon bond price $P(0,T)$ computed above. A deterministic shift $h_t$ of the short rate, with $h_0=0$, gives ${\hat{r}}_t=r_t+h_t$. Subtracting the two recursions shows that the required added drift isSince the shift is nonrandom, the new discount factor differs by $\exp (-{\sum }_{s=1}^T{h}_s)$, and hence $\hat{P}(0,T)=q(T)\exp (-{\sum }_{s=1}^T{h}_s)$.

For deterministic calibration of an autoregressive short rate, defineAll these quantities are finite. The telescoping sum ${\sum }_{s=1}^T{h}_s=d_T$ gives $\hat{P}(0,T)=p(T)$. Explicitly,These constants fit every maturity simultaneously. Each successive maturity fixes $h_T$ uniquely, so the deterministic calibration is also unique with the prescribed $r_0$.