Bond prices

Consider the general Gaussian real-world n-factor model under objective measure

$$P:\quad \mathrm{d}\xi = (A\xi+a)\,\mathrm{d}t + \Sigma\,\ma... ...{d}W\, , A\in\mathbb{R}^{n\times n}, a\in\mathbb{R}^n \quad .$$ (1)

Via Girsanov transform to risk-neutral measure $P^\star$ with $\mathrm{d}W = \lambda(\xi) \mathrm{d}t+ \mathrm{d}W^\star$ and $\lambda(\xi)$ affine:

$$\lambda(\xi)=\Sigma^-(\Lambda\xi+\lambda^1)\, , \Lambda\in\mathbb{R}^{n\times n}\quad .$$ (2)

Introducing $A^\star=A+\Lambda$ and $a^\star=a+\lambda^1$ leads to risk-neutral dynamics

$$P^\star:\quad \mathrm{d}\xi = (A^\star\xi+a^\star)\,\mathrm{d}t + \Sigma\,\mathrm{d}W^\star \quad ,$$ (3)

The price $p(\tau,T)$ of a T-Bond (i.e. a zero-bond of maturity $T$) at time $\tau$ then satisfies the PDE

$$-r p + \frac{\partial p}{\partial\tau} + \left(\frac{\parti... ...ac{\partial^2 p}{\partial \xi\partial \xi}\right) \Sigma = 0.$$ (4)

Setting $t=T-\tau$ the ansatz $p(\tau,T) = p(t) = \exp\left(-u(t)^\prime\xi + v(t)\right)$ leads to the following two ODEs

$$\frac{\partial u}{\partial t} - {A^\star}^\prime u = (1,0,0)^\prime$$ (5)

$$\frac{\partial v}{\partial t} = -u^\prime a^\star + {1\over 2} u^\prime V u \qquad ,$$ (6)

with initial conditions $u(0) = 0$ and $v(0)=0$. The function $u(t)$ has a natural interpretation as duration with respect to the rates $\xi$, since $\partial\ln(P)/\partial\xi = u$. The solutions to the two ODEs are ¹

$$u = \left(e^{t {A^\star}^\prime} -1\right) {{A^\star}^\prime}^{-1} e_1$$ (7)

$$v = \int_0^t ds\left[-{a^\star}^\prime u + {1\over 2} u^\prime V u\right]\quad .$$ (8)

The asymptotics of $u$ for $t\to 0$ will be useful:

$$u = \left(t\,\mathrm{\bf 1}+{t^2\over 2}\,{A^\star}^\prime\right)\,e_1 + o(t^3) \quad .$$ (9)

We turn to the calculation of $u$ and $v$ via diagonalization of ${A^\star}^\prime$. To avoid clutter the star$^\star$ will be dropped for the remainder of this subsection, i.e. we write $A$ for $A^\star$, etc.

Assuming $A^\prime$ can be diagonalised $A^\prime = Q D Q^{-1}$ with diagonal $D$ the solution may be rewritten

$$\begin{eqnarray*} u &=& Q\left(e^{t D} -1\right) b\\ v &=& \int_0^t ds\left[a... ...\over 2} b^\prime e^{s D} Q^\prime V Q e^{s D} b\right] \quad , \end{eqnarray*}$$

where the following abbrevation was introduced:

$$b := D^- Q^- e_1\quad .$$ (10)

Denote the eigenvalues of $A^\prime$ by $\nu=(\nu_1,\nu_2,\dots,\nu_n)$. Then $D=\rm {diag}(\nu)$ and $e^{s D}=\rm {diag}(e^{s \nu_i})$ and

$$\left(e^{s D} Q^\prime V Q e^{s D}\right)_{ij} = \left(Q^\prime V Q\right)_{ij} e^{s(\nu_i+\nu_j)}$$ (11)

which makes integration towards $v$ straightforward:

$$\begin{eqnarray*} v = && t\left[a^\prime Q b + {1\over 2} b^\prime Q^\prime V Q... ..._{ij} \frac{e^{t(\nu_i+\nu_j)}-1}{\nu_i+\nu_j}\right) b \quad . \end{eqnarray*}$$

This can be written in another form,

$$\begin{eqnarray*} v = && a^\prime Q\left(t-\left(e^{tD}-1\right)D^-\right)b \\ ... ..._j)}-1}{\nu_i+\nu_j}-\frac{e^{t\nu_j}-1}{\nu_j}\right]\right) b \end{eqnarray*}$$

or, using the result for $y(\infty)$ from section below,

$$\begin{eqnarray*} v = && -t\,y(\infty) + a^\prime Q\left(-\left(e^{tD}-1\right)... ...\nu_i+\nu_j}-\frac{e^{t\nu_j}-1}{\nu_j}\right]\right) b \quad . \end{eqnarray*}$$