The Heath–Jarrow–Morton Framework

Abstract

Interest rate modelling can also be performed by starting from the dynamics of the instantaneous forward rate. As we shall see the dynamics of all other quantities of interest can then be derived from it. This approach has its origin in Ho and Lee (J Finance XLI:1011–1029, 1986) but was most clearly articulated in Heath et al. (Econometrica 60(1):77–105, 1992a), to which we shall subsequently refer as Heath–Jarrow–Morton. In this framework, the condition of no riskless arbitrage results in the drift coefficient of the forward rate dynamics being expressed in terms of the forward rate volatility function. The major weakness in implementing the Heath–Jarrow–Morton approach is that the spot rate dynamics are usually path dependent (non-Markovian). We consider a class of functional forms of the forward rate volatility that allow the model to be reduced to a finite dimensional Markovian system of stochastic differential equations. This class contains some important models considered in the literature.

Appendices

Appendix

Appendix 25.1 Proof of Proposition 25.1

Recall that r(t) satisfies the stochastic integral equation (25.23) and f(t, τ) satisfies the stochastic integral equation (25.26) with T set equal to τ. We assume the forward rate volatility specifications

$$\displaystyle{ \sigma \left (v,t,\omega (v)\right ) =\bar{\sigma } e^{-\lambda (t-v)}g\left (r(v),f(v,\tau )\right ) }$$

and set

$$\displaystyle\begin{array}{rcl} \sigma ^{{\ast}}\left (v,t,\omega (v)\right )& =& \sigma \left (v,t,\omega (v)\right )\int _{ v}^{t}\sigma \left (v,s,\omega (v)\right )\mathit{ds} {}\\ & =& \bar{\sigma }^{2}e^{-\lambda (t-v)}g\left (r(v),f(v,\tau )\right )\int _{ v}^{t}e^{-\lambda (s-v)}g\left (r(v),f(v,\tau )\right )\mathit{ds} {}\\ & =& \bar{\sigma }^{2}g^{2}\left (r(v),f(v,\tau )\right )e^{-\lambda (t-v)}\left (\frac{1 - e^{-\lambda (t-v)}} {\lambda } \right ). {}\\ \end{array}$$

Note that the first integral term in Eq. (25.23) can be written

$$\displaystyle\begin{array}{rcl} \int _{0}^{t}\sigma ^{{\ast}}\left (v,t,\omega (v)\right )\mathit{dv}& =& \bar{\sigma }^{2}\int _{ 0}^{t}g^{2}\left (r(v),f(v,\tau )\right )e^{-\lambda (t-v)}\frac{\left (1 - e^{-\lambda (t-v)}\right )} {\lambda } \mathit{dv} {}\\ & =& \frac{e^{-\lambda t}\bar{\sigma }^{2}} {\lambda } \int _{0}^{t}g^{2}(r(v),f(v,\tau ))e^{\lambda v}\mathit{dv} {}\\ & & \qquad -\frac{e^{-2\lambda t}\bar{\sigma }^{2}} {\lambda } \int _{0}^{t}g^{2}(r(v),f(v,\tau ))e^{2\lambda v}\mathit{dv} {}\\ & \equiv & \frac{e^{-\lambda t}} {\lambda } \,I(t;\lambda ) -\frac{e^{-2\lambda t}} {\lambda } \,I(t;2\lambda ). {}\\ \end{array}$$

Next note that the second integral in Eq. (25.23) may be written as

$$\displaystyle\begin{array}{rcl} \int _{0}^{t}\sigma \left (v,t,\omega (v)\right )d\tilde{W}(v)& =& \bar{\sigma }\int _{ 0}^{t}e^{-\lambda (t-v)}g\left (r(v),f(v,\tau )\right )d\tilde{W}(v) {}\\ & =& \bar{\sigma }e^{-\lambda t}\int _{ 0}^{t}g\left (r(v),f(v,\tau )\right )e^{\lambda v}d\tilde{W}(v) {}\\ & \equiv & e^{-\lambda t}\,J(t;\lambda ). {}\\ \end{array}$$

Similarly the first integral term in Eq. (25.26) can be written

$$\displaystyle\begin{array}{rcl} \int _{0}^{t}\sigma ^{{\ast}}\left (v,\tau,\omega (v)\right )\mathit{dv}& =& \bar{\sigma }^{2}\int _{ 0}^{t}g^{2}\left (r(v),f(v,\tau )\right )e^{-\lambda (\tau -v)}\frac{\left (1 - e^{-\lambda (\tau -v)}\right )} {\lambda } \mathit{dv} {}\\ & =& \frac{e^{-\lambda \tau }} {\lambda } \,I(t;\lambda ) -\frac{e^{-2\lambda \tau }} {\lambda } \,I(t;2\lambda ). {}\\ \end{array}$$

The second integral term in Eq. (25.26) may be similarly treated, so that

$$\displaystyle\begin{array}{rcl} \int _{0}^{t}\sigma \left (v,\tau,\omega (v)\right )d\tilde{W}(v)& =& \bar{\sigma }\int _{ 0}^{t}e^{-\lambda (\tau -v)}g\left (r(v),f(v,\tau )\right )d\tilde{W}(v) {}\\ & =& \bar{\sigma }e^{-\lambda \tau }\int _{ 0}^{t}e^{\lambda v}g\left (r(v),f(v,\tau )\right )d\tilde{W}(v) {}\\ & \equiv & e^{-\lambda \tau }J(t;\lambda ). {}\\ \end{array}$$

We may thus write the stochastic integral equations for r(t) and f(t, τ) in terms of the integrals I(t; λ), I(t; 2λ) and J(t; λ) as

$$\displaystyle{ r(t) = f(0,t) + \frac{e^{-\lambda t}} {\lambda } I(t;\lambda ) -\frac{e^{-2\lambda t}} {\lambda } I(t;2\lambda ) + e^{-\lambda t}J(t;\lambda ), }$$

(25.124)

$$\displaystyle{ f(t,\tau ) = f(0,\tau ) + \frac{e^{-\lambda \tau }} {\lambda } I(t;\lambda ) -\frac{e^{-2\lambda \tau }} {\lambda } I(t;2\lambda ) + e^{-\lambda \tau }J(t;\lambda ). }$$

(25.125)

We note that Eqs. (25.124) and (25.125) can be re-expressed as

$$\displaystyle\begin{array}{rcl} r(t) - f(0,t) + \frac{e^{-2\lambda t}} {\lambda } I(t;2\lambda )& =& e^{-\lambda t}\left [\frac{I(t;\lambda )} {\lambda } + J(t;\lambda )\right ], {}\\ f(t,\tau ) - f(0,\tau ) + \frac{e^{-2\lambda \tau }} {\lambda } I(t;2\lambda )& =& e^{-\lambda \tau }\left [\frac{I(t;\lambda )} {\lambda } + J(t;\lambda )\right ]. {}\\ \end{array}$$

We may combine the above equations to express I(t; 2λ) as a function of r(t) and f(t, τ), i.e.,

$$\displaystyle{ I(t;2\lambda ) = \frac{\lambda e^{\lambda \tau }} {e^{-\lambda t} - e^{-\lambda \tau }}\left [f(t,\tau ) - f(0,\tau )\right ] - \frac{\lambda e^{\lambda t}} {e^{-\lambda t} - e^{-\lambda \tau }}\left [r(t) - f(0,t)\right ] }$$

(25.126)

Finally we note that

$$\displaystyle\begin{array}{rcl} \psi (t)& =& \int _{0}^{t}\sigma ^{2}\left (v,t,\omega (v)\right )\mathit{dv} =\bar{\sigma } ^{2}\int _{ 0}^{t}e^{-2\lambda (t-v)}g^{2}\left (r(v),f(v,\tau )\right )\mathit{dv} {}\\ & =& \bar{\sigma }^{2}e^{-2\lambda t}\int _{ 0}^{t}e^{2\lambda v}g^{2}\left (r(v),f(v,\tau )\right )\mathit{dv} = e^{-2\lambda t}I(t;2\lambda ). {}\\ \end{array}$$

Thus we finally have

$$\displaystyle{ \psi (t) =\lambda \alpha (t,\tau )\left [r(t) - f(0,t)\right ] -\lambda e^{-\lambda (t-\tau )}\alpha (t,\tau )\left [f(t,\tau ) - f(0,\tau )\right ], }$$

(25.127)

where we set

$$\displaystyle{ \alpha (t,\tau ) \equiv \frac{e^{-\lambda t}} {e^{-\lambda \tau }- e^{-\lambda t}}. }$$

Appendix 25.2 Proof of Proposition 25.2

It is readily verified that the manipulations that led to Eq. (25.125) of Appendix 25.1 are equally valid for t set to a general maturity T. Thus (25.126) holds for t set to T, i.e.,

$$\displaystyle\begin{array}{rcl} I(t;2\lambda )& =& \frac{\lambda e^{\lambda T}} {e^{-\lambda t}-e^{-\lambda T}}\left [f(t,T)-f(0,T)\right ]- \frac{\lambda e^{\lambda t}} {e^{-\lambda t}-e^{-\lambda T}}\left [r(t)-f(0,t)\right ] {}\\ & =& e^{2\lambda t}\psi (t). {}\\ \end{array}$$

Substituting the expression for ψ(t) we find that

$$\displaystyle\begin{array}{rcl} I(t;2\lambda )& =& \lambda e^{2\lambda t}\left (\alpha (t,\tau )[r(t) - f(0,t)] - e^{-\lambda (t-\tau )}\alpha (t,\tau )[f(t,\tau ) - f(0,\tau )]\right ) {}\\ & =& \lambda \left ( \frac{e^{\lambda T}} {e^{-\lambda t} - e^{-\lambda T}}[f(t,T) - f(0,T)] - \frac{e^{\lambda t}} {e^{-\lambda t} - e^{-\lambda T}}[r(t) - f(0,t)]\right ). {}\\ \end{array}$$

On rearranging

$$\displaystyle\begin{array}{rcl} \frac{e^{\lambda T}} {e^{-\lambda t} - e^{-\lambda T}}[f(t,T) - f(0,T)]& =& \frac{e^{\lambda t}} {e^{-\lambda t} - e^{-\lambda T}}[r(t) - f(0,t)] {}\\ & & +e^{2\lambda t}\alpha (t,\tau )[r(t) - f(0,t)] {}\\ & & -e^{2\lambda t}e^{-\lambda (t-\tau )}\alpha (t,\tau )[f(t,\tau ) - f(0,\tau )], {}\\ \end{array}$$

from which

$$\displaystyle\begin{array}{rcl} f(t,T) - f(0,T)& =& [r(t) - f(0,t)]\left ( \frac{e^{\lambda t}} {e^{\lambda T}} + \frac{e^{2\lambda t}(e^{-\lambda t} - e^{-\lambda T})} {e^{\lambda T}} \alpha (t,\tau )\right ) \\ & & -\frac{e^{2\lambda t}e^{-\lambda (t-T)}} {e^{\lambda T}} \alpha (t,\tau )(e^{-\lambda t} - e^{-\lambda T})[f(t,\tau ) - f(0,\tau )].{}\end{array}$$

(25.128)

Consider the following:

• (i)
  $$\displaystyle\begin{array}{rcl} & & \frac{e^{\lambda t} + e^{2\lambda t}(e^{-\lambda t} - e^{-\lambda T})} {e^{\lambda T}} \alpha (t,\tau ) = \frac{e^{\lambda t}} {e^{\lambda T}} + \frac{e^{2\lambda t}(e^{-\lambda t} - e^{-\lambda T})e^{-\lambda t}} {e^{\lambda T}(e^{-\lambda \tau }- e^{-\lambda t})} {}\\ & =& \frac{e^{\lambda t}(e^{-\lambda \tau }- e^{-\lambda t}) + e^{\lambda t}(e^{-\lambda t} - e^{-\lambda T})} {e^{\lambda T}(e^{-\lambda \tau }- e^{-\lambda t})} = \frac{e^{\lambda t}(e^{-\lambda \tau }- e^{-\lambda T})} {e^{\lambda T}(e^{-\lambda \tau }- e^{-\lambda t})} {}\\ & =& \frac{e^{2\lambda t}} {e^{2\lambda T}} \frac{e^{-\lambda t}} {e^{-\lambda \tau }- e^{-\lambda t}} \frac{e^{-\lambda \tau }- e^{-\lambda T}} {e^{-\lambda T}} = e^{-2\lambda (T-t)} \frac{\alpha (t,\tau )} {\alpha (T,\tau )} {}\\ \end{array}$$
• (ii)
  $$\displaystyle\begin{array}{rcl} & & e^{2\lambda t}e^{-\lambda (t-\tau )}\alpha (t,\tau )\frac{(e^{-\lambda t} - e^{-\lambda T})} {e^{\lambda T}} = e^{2\lambda t-\lambda t+\lambda \tau }\alpha (t,\tau )\frac{(e^{-\lambda t} - e^{-\lambda T})} {e^{2\lambda T}e^{-\lambda T}} {}\\ & =& \frac{e^{\lambda t}e^{\lambda \tau }e^{-\lambda t}(e^{-\lambda t} - e^{-\lambda T})} {e^{2\lambda T}e^{-\lambda T}(e^{-\lambda \tau }- e^{-\lambda t})} = \frac{e^{2\lambda \tau }} {e^{2\lambda T}} \frac{e^{-\lambda \tau }} {-(e^{-\lambda t} - e^{-\lambda \tau })} \frac{(e^{-\lambda t} - e^{-\lambda T})} {e^{-\lambda T}} {}\\ & =& -e^{-2\lambda (T-\tau )} \frac{\alpha (\tau,t)} {\alpha (T,t)}. {}\\ \end{array}$$

  Hence Eq. (25.128) can be rewritten

  $$\displaystyle\begin{array}{rcl} f(t,T) - f(0,T)& =& e^{-2\lambda (T-t)} \frac{\alpha (t,\tau )} {\alpha (T,\tau )}[r(t) - f(0,t)] {}\\ & -& e^{-2\lambda (T-\tau )} \frac{\alpha (\tau,t)} {\alpha (T,t)}[f(t,\tau ) - f(0,\tau )] {}\\ \end{array}$$

  where

  $$\displaystyle{ \alpha (\theta _{1},\theta _{2}) \equiv \frac{e^{-\lambda \theta _{1}}} {e^{-\lambda \theta _{2}} - e^{-\lambda \theta _{1}}}. }$$

  We have thus proved Proposition 25.2.

Appendix 25.3 Details of the Infinitesimal Generator $\mathcal{K}$

We recall the following result from Sect. 5.4 concerning the infinitesimal generator of an n dimensional Ito process. In our application we set

$$\displaystyle\begin{array}{rcl} X_{1}& \equiv & f(t,\tau ), {}\\ a_{1}& \equiv & \sigma ^{2}(t,\tau,\omega (t))\frac{(e^{\lambda (\tau -t)} - 1)} {\lambda }, {}\\ \sigma _{11}& \equiv & \sigma _{1} \equiv \sigma (t,\tau,\omega (t)), {}\\ X_{2}& \equiv & r(t), {}\\ a_{2}& \equiv & f_{2}(0,t) +\lambda f(0,t) +\psi (t) -\lambda r(t), {}\\ \sigma _{21}& \equiv & \sigma _{r} \equiv \sigma (t,t,\omega (t)). {}\\ \end{array}$$

Thus the matrix S assumes the form

$$\displaystyle{ \left [\begin{array}{*{10}c} \sigma _{1}^{2} & \sigma _{1}\sigma _{r} \\ \sigma _{1}\sigma _{r}&\sigma _{r}^{2} \end{array} \right ]. }$$

Using the foregoing expression for S the expression for the operator $\mathcal{K}$ in Eq. (25.67) is readily derived.

Appendix 25.4 Proof of Proposition 25.3

Using the relationship

$$\displaystyle{P(t,T) =\exp \left (-\int _{t}^{T}f(t,s)\mathit{ds}\right )}$$

and Eq. (25.26) for the forward rate f(t, s) we obtain for the bond price the expression

$$\displaystyle{P(t,T)=\frac{P(0,T)} {P(0,t)} \exp \left [-\left (\int _{t}^{T}\!\int _{ 0}^{t}\sigma ^{{\ast}}(v,s,\cdot )\mathit{dv}\mathit{ds}+\int _{ t}^{T}\!\int _{ 0}^{t}\sigma (v,s,\cdot )d\tilde{W}(v)\mathit{ds}\right )\right ],}$$

where

$$\displaystyle\begin{array}{rcl} \sigma (v,T,\cdot )& =& \bar{\sigma }e^{-\lambda (T-v)}g(r(v),f(v,\tau )) {}\\ \sigma ^{{\ast}}(v,T,\cdot )& =& \sigma (v,T,\cdot )\int _{ v}^{T}\sigma (v,s,\cdot )\mathit{ds} {}\\ & =& \bar{\sigma }^{2}g^{2}(r(v),f(v,\tau ))e^{-\lambda (T-v)}\int _{ v}^{T}e^{-\lambda (s-v)}\mathit{ds}. {}\\ \end{array}$$

Set

$$\displaystyle\begin{array}{rcl} I& =& \int _{t}^{T}\int _{ 0}^{t}\sigma ^{{\ast}}(v,s,\cdot )\mathit{dv}\mathit{ds} +\int _{ t}^{T}\int _{ 0}^{t}\sigma (v,s,\cdot )d\tilde{W}(v)\mathit{ds} {}\\ & \equiv & I_{1} + I_{2} {}\\ & =& \int _{0}^{t}\int _{ t}^{T}\sigma ^{{\ast}}(v,s,\cdot )\mathit{ds}\mathit{dv} +\int _{ 0}^{t}\int _{ t}^{T}\sigma (v,s,\cdot )\mathit{ds}d\tilde{W}(v), {}\\ \end{array}$$

where we have interchanged the order of integration to obtain the last equality. Next note that

$$\displaystyle\begin{array}{rcl} \int _{t}^{T}\sigma ^{{\ast}}(v,s,\cdot )\mathit{ds}& =& \sigma \left (r(v),f(v,\tau )\right )\int _{ t}^{T}e^{-\lambda (s-v)}\int _{ v}^{s}e^{-\lambda (y-v)}\sigma \left (r(v),f(v,\tau )\right )\mathit{dy}\mathit{ds} {}\\ & =& \sigma \left (r(v),f(v,\tau )\right )\int _{t}^{T}e^{-\lambda (s-v)}\left \{\int _{ v}^{t}e^{-\lambda (y-v)}\sigma \left (r(v),f(v,\tau )\right )\mathit{dy}\right. {}\\ & & \!\qquad \left.+\int _{t}^{s}e^{-\lambda (y-v)}\sigma \left (r(v),f(v,\tau )\right )\mathit{dy}\right \}\mathit{ds} {}\\ & =& \bar{\sigma }^{2}g^{2}(r(v),f(v,\tau ))\int _{ t}^{T}e^{-\lambda (s-v)}\mathit{ds}\int _{ v}^{t}e^{-\lambda (y-v)}\mathit{dy} {}\\ & & \!\qquad +\bar{\sigma } ^{2}g^{2}(r(v),f(v,\tau ))\int _{ t}^{T}e^{-\lambda (s-v)}\int _{ t}^{s}e^{-\lambda (y-v)}\mathit{dy}\mathit{ds} {}\\ & =& \bar{\sigma }^{2}g^{2}(r(v),f(v,\tau ))e^{-\lambda (t-v)}\left (\int _{ t}^{T}e^{-\lambda (s-t)}\mathit{ds}\right )\int _{ v}^{t}e^{-\lambda (y-v)}\mathit{dy} {}\\ & & \!\qquad +\bar{\sigma } ^{2}g^{2}(r(v),f(v,\tau ))e^{-2\lambda (t-v)}\int _{ t}^{T}e^{-\lambda (s-t)}\int _{ t}^{s}e^{-\lambda (y-t)}\mathit{dy}\mathit{ds} {}\\ & =& \sigma ^{{\ast}}(v,t,\cdot )\beta (t,T) +\sigma ^{2}(v,t,\cdot )\alpha (t,T), {}\\ \end{array}$$

where

$$\displaystyle\begin{array}{rcl} \beta (t,T)& =& \int _{t}^{T}e^{-\lambda (s-t)}\mathit{ds} = \frac{1} {\lambda } \left (1 - e^{-\lambda (T-t)}\right ), {}\\ \alpha (t,T)& =& \int _{t}^{T}e^{-\lambda (s-t)}\int _{ t}^{s}e^{-\lambda (y-t)}\mathit{dy}\mathit{ds} = \frac{1} {2}\beta ^{2}(t,T), {}\\ \end{array}$$

i.e. we have shown that

$$\displaystyle{\int _{t}^{T}\sigma ^{{\ast}}(v,s,\cdot )\mathit{ds} =\beta (t,T)\sigma ^{{\ast}}(v,t,\cdot ) + \frac{1} {2}\beta ^{2}(t,T)\sigma ^{2}(v,t,\cdot ).}$$

Next consider

$$\displaystyle\begin{array}{rcl} \int _{t}^{T}\sigma (v,s,\cdot )\mathit{ds}& =& \int _{ t}^{T}e^{-\lambda (s-v)}\sigma \left (r(v),f(v,\tau )\right )\mathit{ds} {}\\ & =& \sigma \left (r(v),f(v,\tau )\right )e^{-\lambda (t-v)}\left (\int _{ t}^{T}e^{-\lambda (s-t)}\mathit{ds}\right ), {}\\ \end{array}$$

i.e. we have shown that

$$\displaystyle{\int _{t}^{T}\sigma (v,s,\cdot )\mathit{ds} =\sigma (v,t,\cdot )\beta (t,T).}$$

Returning to the expressions for I ₁, I ₂ we can now write

$$\displaystyle{I_{1} =\int _{ 0}^{t}\left [\beta (t,T)\sigma ^{{\ast}}(v,t,\cdot ) + \frac{1} {2}\beta ^{2}(t,T)\sigma ^{2}(v,t,\cdot )\right ]\mathit{dv},}$$

and

$$\displaystyle{I_{2} =\int _{ 0}^{t}\beta (t,T)\sigma (v,t,\cdot )d\tilde{W}(v),}$$

so that

$$\displaystyle\begin{array}{rcl} I& =& \frac{1} {2}\beta ^{2}(t,T)\int _{ 0}^{t}\sigma ^{2}(v,t,\cdot )\mathit{dv} {}\\ & & +\beta (t,T)\left [\int _{0}^{t}\sigma ^{{\ast}}(v,t,\cdot )\mathit{dv}\,+\,\int _{ 0}^{t}\sigma (v,t,\cdot )d\tilde{W}(v)\right ]. {}\\ \end{array}$$

However we note from Eq. (25.23), for the instantaneous spot rate r(t), that

$$\displaystyle{\int _{0}^{t}\sigma ^{{\ast}}(v,t,\cdot )\mathit{dv}+\int _{ 0}^{t}\sigma (v,t,\cdot )d\tilde{W}(v) = r(t) - f(0,t).}$$

Hence

$$\displaystyle{I = \frac{1} {2}\beta ^{2}(t,T)\int _{ 0}^{t}\sigma ^{2}(v,t,\cdot )\mathit{dv} +\beta (t,T)\left [r(t) - f(0,t)\right ].}$$

Recalling the definition of the subsidiary stochastic variable ψ(t) we can finally write

$$\displaystyle{I = \frac{1} {2}\beta ^{2}(t,T)\psi (t) +\beta (t,T)\left [r(t) - f(0,t)\right ].}$$

Hence the expression for the bond price may be written as in Proposition 25.3.

Problems

Problem 25.1

Show that the Hull–White model can be obtained within the Heath–Jarrow–Morton framework by setting

$$\displaystyle{\sigma (t,T) =\bar{\sigma } e^{-k(T-t)},}$$

where $\bar{\sigma },k$ are constants.

Problem 25.2

The Heath–Jarrow–Morton model takes as its starting point a stochastic differential equation for the instantaneous forward rate of the form

$$\displaystyle{ df(t,T) =\alpha (t,T)\mathit{dt} +\sigma (t,T)\mathit{dW }(t), }$$

and from this determines the stochastic dynamics of the instantaneous spot rate spot rate r(t) and pure discount bond price P(t, T).

Suppose instead we take as the starting point a stochastic differential equation for P(t, T) of the form

$$\displaystyle{ \frac{\mathit{dP}(t,T)} {P(t,T)} =\beta (t,T)\mathit{dt} +\delta (t,T)\mathit{dW }(t). }$$

Determine the corresponding stochastic dynamics for r(t) and f(t, T).

Express in terms of β(t, T) and δ(t, T) the Heath–Jarrow–Morton drift restriction that guarantees no riskless arbitrage opportunities between bonds of different maturities.

Problem 25.3

In Sect. 23.6 we considered the volatility function

$$\displaystyle{ \sigma (t,T) =\bar{\sigma } e^{-\lambda (T-t)} }$$

and showed how this allowed the system dynamics to be Markovianised.

Now consider the volatility function

$$\displaystyle{ \sigma (t,T) = [\sigma _{0} +\sigma _{1}(T - t)]e^{-\lambda (T-t)}. }$$

Show the system dynamics can be Markovianised in this case. In particular obtain the stochastic differential equations for the bond price and the instantaneous spot interest rate.

Hint: You will need to obtain a linked stochastic differential equation system for

$$\displaystyle{ Z_{1}(t) =\int _{ 0}^{t}(t - v)e^{-\lambda (t-v)}\mathit{dW }(v), }$$

and

$$\displaystyle{ Z_{0}(t) =\int _{ 0}^{t}e^{-\lambda (t-v)}\mathit{dW }(v). }$$

Problem 25.4

The Ho–Lee model is obtained within the Heath–Jarrow–Morton framework by setting

$$\displaystyle{\sigma (t,T) =\bar{\sigma },}$$

where $\bar{\sigma }$ is a constant. Show that

$$\displaystyle{f(t,T) = r(t) + f(0,T) - f(0,t) +\bar{\sigma } ^{2}t(T - t).}$$

By obtaining the dynamics for r(t) under the risk neutral measure, show also that

$$\displaystyle{r(t) = f(0,t) + \frac{1} {2}\sigma ^{2}t^{2} +\bar{\sigma } \tilde{W}(t).}$$

Hence, show that for this model the bond price is given by

$$\displaystyle{P(t,T) =\exp [-a(t,T) - (T - t)r(t)],}$$

where

$$\displaystyle{a(t,T) =\ln \frac{P(0,t)} {P(0,T)} - (T - t)f(0,t) + \frac{1} {2}\bar{\sigma }^{2}t(T - t)^{2}.}$$

Problem 25.5

Computational Problem—Consider the Heath–Jarrow–Morton model with the volatility function

$$\displaystyle{\sigma (t,T) =\sigma _{0}e^{-\lambda (T-t)}.}$$

We know in this case that the dynamics for the instantaneous spot rate are given by [Eq. (25.54)].

Take σ ₀ = 0. 02 and λ = 0. 6. Assume also that the initial forward curve is given by

$$\displaystyle{f(0,T) = 0.08 - 0.03e^{-1.5T}.}$$

Consider the bond pricing formula [Eq. (25.29)]. Write a program to calculate the bond price by simulating the stochastic differential equation for r(t) from 0 to t and performing the $\tilde{\mathbb{E}}_{t}$ operation by simulating a large number of paths from t to T. This will give the bond price conditional on the value of r(t) that has been obtained.

You can check the accuracy of your algorithm (and hence choose appropriate Δ t and number of paths) by using the fact that when t = 0 we have the exact solution

$$\displaystyle{P(0,T) =\exp \left (-\int _{0}^{T}f(0,s)\mathit{ds}\right ).}$$

Use this to check the accuracy for T = 0. 5, 1. 0, 1. 5 and 2. 0.

Then use the simulation procedure to calculate P(0. 5, 1. 0), P(0. 5, 1. 5) and P(0. 5, 2. 0).

Note that the evaluation of ρ(t, T) will be conditional on the interest rate r(t). Obtain r(t) by simulating from 0 to t and be sure to specify the value of r(t) that you are using.

Check the accuracy of these approximations by using the exact bond-pricing formula (here you need to refer to Sect. 23.4.2, but use the θ(t) that arises in the Heath–Jarrow–Morton model).