Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 34 / 6

Let $P_{s,t}>0$ denote the time-$s$ price of a zero-coupon bond paying one at maturity $t$, with $P_{t,t}=1$. Information available at $s$ is the sigma-algebra ${\mathcal{F}}_s$ in the market filtration. For a curve differentiable in maturity, the instantaneous forward rate and its diagonal short rate areThe last identity follows by integrating the maturity derivative and using $P_{s,s}=1$. The continuous-time bank account reinvesting at the short rate is $B_s=\exp ({\int }_0^s{R}_{u}du)$, and its reciprocal is the discount factor. Modelling forward rates means specifying a random maturity-indexed curve with an evolution consistent with the same pricing measure for every bond. The expectation below is under that money-market risk-neutral measure, not an unspecified physical measure.

For fixed maturity $t$, if $P_{s,t}/B_s$ is a true martingale on $0\le s\le t$, its terminal value is $1/B_t$. ConsequentlyConversely, this representation gives $P_{s,t}/B_s={\mathbb{E}}_Q[B_t^{-1}\mid {\mathcal{F}}_s]$, a conditional-expectation martingale by the tower property. For example, for $r<s$, conditioning the right side on ${\mathcal{F}}_r$ returns ${\mathbb{E}}_Q[B_t^{-1}\mid {\mathcal{F}}_r]$. These claims assume ${\mathbb{E}}_Q{B}_t^{-1}<\infty$. A positive discounted local martingale alone need not equal the conditional expectation of its terminal value, so the word “martingale” cannot be weakened without an extra integrability argument.

In the Vasicek model, the Brownian motion $W$ and coefficients in $d{R}_s=\alpha (\beta -R_s)ds+\sigma d{W}_s$ must be understood under this pricing measure. Multiplying by $e^{\alpha s}$ and integrating gives, conditionally on ${\mathcal{F}}_s$,Assume first $\alpha \ne 0$, and put $\tau =t-s$ and $B(\tau )=(1-e^{-\alpha \tau })/\alpha$. Integrating the explicit solution over $s\le u\le t$ yieldsThe interchange of the stochastic integrals is justified by their deterministic square-integrable kernels on the finite triangular domain; equivalently approximate the kernels by step functions, interchange finite sums, and pass to the limit using the Itô isometry. The last integral conditionally has a centered normal distribution independent of ${\mathcal{F}}_s$, with varianceA deterministic Brownian integral is Gaussian because its step approximations are sums of independent normal increments; their variances converge by the Itô isometry. Completing the square in the normal density proves $\mathbb{E}{e}^{-Y}=e^{-m+v/2}$ for a normal variable of mean $m$ and variance $v$. Hence the zero-coupon bond price is exponential-affine:An equivalent familiar form isAt $t=s$, both coefficients are zero and the bond value is one. These formulas are valid for negative as well as positive nonzero $\alpha$ on finite horizons, though a genuinely mean-reverting Vasicek model has $\alpha >0$. For $\alpha =0$, direct integration gives $B(\tau )=\tau$, $J(\tau )={\tau }^3/3$ andAll the exponential moments used above are finite, so the prices defined by the conditional expectation do give true discounted martingales.

To fit an initial bond curve, let its observed prices $p_0(t)$ be positive, satisfy $p_0(0)=1$, and have a sufficiently smooth instantaneous forward rate $f_0(t)=-d\log p_0(t)/dt$. Set $R_0=f_0(0)$; this compatibility is forced by the infinitesimal-maturity bond price. Replace $\beta$ by a deterministic function $\theta (t)$, with $\alpha \ne 0$, so thatIts mean $m(t)={\mathbb{E}}_Q{R}_t$ satisfies $m^{\prime }=\alpha (\theta -m)$, $m(0)=R_0$. The variance of the integrated rate remains ${\sigma }^{2}J(t)$, since changing deterministic drift does not change the centered Gaussian process. ThereforeThe required mean is consequently $m(t)=f_0(t)+{\sigma }^{2}B(t{)}^2/2$. Solve its mean equation for $\theta$:Combining the last two terms gives the explicit Hull-White model calibration in the question's parameter convention:With $R_0=f_0(0)$, the specified mean solves the same linear differential equation and initial condition as the model mean, so it is the actual mean. The resulting model forward curve equals $f_0$; integrating and using $P_{0,0}=p_0(0)=1$ proves $P_{0,t}=p_0(t)$ for every maturity. If the drift is instead written $\tilde{\theta }(t)-\alpha R_t$, its calibration parameter is $\tilde{\theta }=\alpha \theta$, not $\theta$ itself.

The printed allowance of arbitrary constants needs a qualification for this final claim. If $\alpha =0$, replacing $\beta$ by $\theta (t)$ changes no coefficient at all: $R_t=R_0+\sigma W_t$ regardless of $\theta$. Its initial curve is fixed at $P_{0,t}=\exp (-R_{0}t+{\sigma }^2{t}^3/6)$. With $\sigma \ne 0$, the compatible smooth observed curve $p_0(t)=e^{-R_{0}t}$ is already a counterexample to the unconditional fitting claim. Thus arbitrary compatible smooth positive initial curves can be fitted by the stated replacement when $\alpha \ne 0$; it cannot fit arbitrary curves when $\alpha =0$. In that degenerate case one can instead introduce an independent deterministic drift $k(t)$ in $d{R}_t=k(t)dt+\sigma d{W}_t$, choosing $k(t)=f_0^{\prime }(t)+{\sigma }^{2}t$ to recover the desired initial curve. Likewise an externally fixed $R_0\ne f_0(0)$ is incompatible with exact fitting, and an arbitrary nonsmooth set of prices need not define the differentiable forward curve assumed in the model.