Skip to content

Bond Price Derivation via PDE

The expectation-based derivation of the Hull-White bond price exploits the Gaussian structure of the short rate integral. An independent route to the same formula starts from the Feynman-Kac theorem, which converts the pricing problem into a partial differential equation. Substituting an exponential-affine ansatz reduces the PDE to a system of ordinary differential equations -- a linear ODE for \(B(\tau)\) and a quadrature for \(A(t,T)\) -- that can be solved explicitly. This section carries out the PDE derivation from start to finish, filling in every algebraic step.

Prerequisites

  • Hull-White SDE: \(dr(t) = [\theta(t) - ar(t)]\,dt + \sigma\,dW^{\mathbb{Q}}(t)\)
  • Feynman-Kac theorem connecting conditional expectations to PDEs
  • Bond price formula from the expectation derivation (sibling section) for cross-verification

Learning Objectives

By the end of this section, you will be able to:

  1. Derive the bond pricing PDE from the Feynman-Kac theorem
  2. Substitute the exponential-affine ansatz and separate variables
  3. Solve the Riccati ODE for \(B(\tau)\) and the quadrature for \(A(t,T)\)
  4. Express the complete bond price in terms of market observables
  5. Verify consistency with the expectation-based derivation

From Feynman-Kac to the Bond Pricing PDE

The zero-coupon bond price satisfies the risk-neutral pricing identity

\[ P(t,T) = \mathbb{E}^{\mathbb{Q}}\!\left[e^{-\int_t^T r(s)\,ds}\,\Big|\,\mathcal{F}(t)\right] \]

By the Feynman-Kac theorem, any function \(P(t,r)\) satisfying this conditional expectation also satisfies a backward PDE with the short rate dynamics providing the drift and diffusion coefficients.

Theorem: Hull-White Bond Pricing PDE

Under the Hull-White model, the zero-coupon bond price \(P(t,T)\) satisfies

\[ \frac{\partial P}{\partial t} + \bigl[\theta(t) - ar\bigr]\frac{\partial P}{\partial r} + \frac{1}{2}\sigma^2\frac{\partial^2 P}{\partial r^2} - rP = 0 \]

with terminal condition \(P(T,T) = 1\).

Derivation

The Feynman-Kac theorem states that if \(X_t\) satisfies \(dX_t = \mu(t,X_t)\,dt + \sigma(t,X_t)\,dW_t\), then

\[ u(t,x) = \mathbb{E}\!\left[e^{-\int_t^T c(s,X_s)\,ds}\,g(X_T)\,\Big|\,X_t = x\right] \]

solves \(\frac{\partial u}{\partial t} + \mu\frac{\partial u}{\partial x} + \frac{1}{2}\sigma^2\frac{\partial^2 u}{\partial x^2} - c(t,x)\,u = 0\) with \(u(T,x) = g(x)\).

For the zero-coupon bond: \(X_t = r(t)\), \(\mu(t,r) = \theta(t) - ar\), \(\sigma(t,r) = \sigma\) (constant), \(c(t,r) = r\), and \(g(r) = 1\). Substitution gives the stated PDE. The terminal condition \(P(T,T) = 1\) reflects the $1 payoff at maturity. \(\square\)


The Exponential-Affine Ansatz

The PDE is linear in \(r\) (the drift \(\theta(t) - ar\) is affine and the diffusion \(\sigma\) is constant), suggesting that the solution may be exponential-affine in \(r\).

Ansatz

Assume the bond price has the form

\[ P(t,T) = \exp\!\bigl(A(t,T) + B(t,T)\,r\bigr) \]

where \(A(t,T)\) and \(B(t,T)\) are deterministic functions to be determined, subject to the terminal conditions \(A(T,T) = 0\) and \(B(T,T) = 0\).

The terminal conditions follow from \(P(T,T) = e^{A(T,T) + B(T,T)r} = 1\) for all \(r\), which requires both \(A(T,T) = 0\) and \(B(T,T) = 0\).


Substitution into the PDE

Compute the required partial derivatives of \(P = e^{A + Br}\):

\[ \frac{\partial P}{\partial t} = \left(\frac{\partial A}{\partial t} + \frac{\partial B}{\partial t}\,r\right)P \]
\[ \frac{\partial P}{\partial r} = B\,P \]
\[ \frac{\partial^2 P}{\partial r^2} = B^2\,P \]

Substitute into the PDE and divide by \(P > 0\):

\[ \frac{\partial A}{\partial t} + \frac{\partial B}{\partial t}\,r + [\theta(t) - ar]\,B + \frac{1}{2}\sigma^2 B^2 - r = 0 \]

Collect terms by powers of \(r\):

\[ \underbrace{\left(\frac{\partial B}{\partial t} - aB - 1\right)}_{\text{coefficient of }r}\,r + \underbrace{\left(\frac{\partial A}{\partial t} + \theta(t)\,B + \frac{1}{2}\sigma^2 B^2\right)}_{\text{constant term}} = 0 \]

Since this must hold for all values of \(r\), both the coefficient of \(r\) and the constant term must vanish independently.


The ODE for B

Setting the coefficient of \(r\) to zero yields a first-order linear ODE.

Theorem: ODE for \(B(t,T)\)

\[ \frac{\partial B}{\partial t}(t,T) = aB(t,T) + 1, \qquad B(T,T) = 0 \]

Change to time-to-maturity. Let \(\tau = T - t\) and write \(\tilde{B}(\tau) = B(t,T)\). Since \(\frac{\partial B}{\partial t} = -\frac{d\tilde{B}}{d\tau}\), the ODE becomes

\[ \frac{d\tilde{B}}{d\tau} = -a\tilde{B} - 1, \qquad \tilde{B}(0) = 0 \]

Theorem: Solution for \(B(\tau)\)

\[ B(\tau) = -\frac{1 - e^{-a\tau}}{a} \]
Proof

The ODE \(\tilde{B}' + a\tilde{B} = -1\) is solved by the integrating factor \(\mu(\tau) = e^{a\tau}\):

\[ \frac{d}{d\tau}\!\left(e^{a\tau}\tilde{B}\right) = e^{a\tau}\!\left(\tilde{B}' + a\tilde{B}\right) = -e^{a\tau} \]

Integrating from \(0\) to \(\tau\):

\[ e^{a\tau}\tilde{B}(\tau) - \tilde{B}(0) = -\int_0^{\tau} e^{as}\,ds = -\frac{e^{a\tau} - 1}{a} \]

Using \(\tilde{B}(0) = 0\):

\[ \tilde{B}(\tau) = -\frac{e^{a\tau} - 1}{a}\,e^{-a\tau} = -\frac{1 - e^{-a\tau}}{a} \]

\(\square\)

Properties of \(B(\tau)\):

  • \(B(0) = 0\): the terminal condition is satisfied
  • \(B(\tau) < 0\) for \(\tau > 0\): bond prices decrease when rates increase
  • \(B(\tau) \to -1/a\) as \(\tau \to \infty\): the sensitivity saturates due to mean reversion
  • \(B(\tau) \approx -\tau\) for \(a\tau \ll 1\): recovers the zero-mean-reversion (Ho-Lee) limit

The magnitude \(|B(\tau)| = (1 - e^{-a\tau})/a\) is the duration-like function used in the positive-\(B\) convention where the bond price is written \(P = e^{A - |B|\,r}\).


The ODE for A

Setting the constant term to zero:

Theorem: ODE for \(A(t,T)\)

\[ \frac{\partial A}{\partial t}(t,T) = -\theta(t)\,B(t,T) - \frac{1}{2}\sigma^2\,B(t,T)^2, \qquad A(T,T) = 0 \]

In the \(\tau\) variable:

\[ \frac{d\tilde{A}}{d\tau} = \theta(T - \tau)\,B(\tau) + \frac{1}{2}\sigma^2\,B(\tau)^2, \qquad \tilde{A}(0) = 0 \]

Since \(B(\tau)\) is already known, this is a quadrature (direct integration), not a differential equation in the usual sense.

Theorem: Solution for \(A\) via Quadrature

\[ A(t,T) = \int_t^T \theta(u)\,B(u,T)\,du + \frac{1}{2}\sigma^2 \int_t^T B(u,T)^2\,du \]
Proof

Integrating the ODE for \(A\) from \(t\) to \(T\) (equivalently, from \(\tau = T - t\) down to \(0\)):

\[ A(T,T) - A(t,T) = -\int_t^T \theta(u)\,B(u,T)\,du - \frac{1}{2}\sigma^2\int_t^T B(u,T)^2\,du \]

Since \(A(T,T) = 0\):

\[ A(t,T) = \int_t^T \theta(u)\,B(u,T)\,du + \frac{1}{2}\sigma^2\int_t^T B(u,T)^2\,du \]

\(\square\)


Evaluating the Integrals

The \(B^2\) Integral

The second integral can be computed in closed form.

Proposition: Integral of \(B^2\)

\[ \frac{1}{2}\sigma^2\int_t^T B(u,T)^2\,du = \frac{\sigma^2}{2a^2}\left[\tau - \frac{2}{a}(1 - e^{-a\tau}) + \frac{1}{2a}(1 - e^{-2a\tau})\right] \]

where \(\tau = T - t\).

Proof

Substituting \(v = T - u\) and using \(B(u,T) = -(1 - e^{-av})/a\):

\[ \int_t^T B(u,T)^2\,du = \frac{1}{a^2}\int_0^{\tau}(1 - e^{-av})^2\,dv \]

Expanding and integrating term by term:

\[ = \frac{1}{a^2}\int_0^{\tau}\!\left(1 - 2e^{-av} + e^{-2av}\right)dv = \frac{1}{a^2}\left[\tau + \frac{2}{a}(e^{-a\tau} - 1) - \frac{1}{2a}(e^{-2a\tau} - 1)\right] \]
\[ = \frac{1}{a^2}\left[\tau - \frac{2}{a}(1 - e^{-a\tau}) + \frac{1}{2a}(1 - e^{-2a\tau})\right] \]

\(\square\)

The theta Integral and the Market Curve

The integral \(\int_t^T \theta(u) B(u,T)\,du\) involves the time-dependent drift \(\theta(t)\), which encodes the market yield curve. Using

\[ \theta(t) = \frac{\partial f^M(0,t)}{\partial t} + a\,f^M(0,t) + \frac{\sigma^2}{2a}(1 - e^{-2at}) \]

and performing integration by parts (details in the sibling section on Riccati equations), the \(\theta\)-integral combines with the \(B^2\)-integral to give a formula involving only market observables.

Theorem: \(A(t,T)\) in Terms of the Market Curve

\[ A(t,T) = \ln\frac{P^M(0,T)}{P^M(0,t)} + B(t,T)\,f^M(0,t) + \frac{\sigma^2}{4a}\,B(t,T)^2\left(1 - e^{-2at}\right) \]

where \(B(t,T) = -(1 - e^{-a\tau})/a\) and \(f^M(0,t) = -\partial_t \ln P^M(0,t)\).

Proof Sketch

The key step is eliminating \(\theta(u)\) in favor of \(P^M\). Starting from \(\theta(u) = f'(0,u) + af(0,u) + \frac{\sigma^2}{2a}(1 - e^{-2au})\) and integrating \(\theta(u) B(u,T)\) from \(t\) to \(T\):

The \(f'(0,u)B(u,T)\) term is integrated by parts. The \(af(0,u)B(u,T)\) term combines with the boundary term from integration by parts. After collecting all contributions with the \(B^2\)-integral, one obtains

\[ A(t,T) = -\int_t^T f^M(0,u)\,du + \int_0^t f^M(0,u)\,du + B(t,T) f^M(0,t) + \frac{\sigma^2}{4a} B(t,T)^2(1 - e^{-2at}) \]

Using \(\int_0^T f^M(0,u)\,du = -\ln P^M(0,T)\) and \(\int_0^t f^M(0,u)\,du = -\ln P^M(0,t)\), this simplifies to the stated formula. \(\square\)


The Complete Bond Price

Combining \(A(t,T)\) and \(B(t,T)\):

Corollary: Hull-White Bond Price (PDE Derivation)

The zero-coupon bond price is

\[ P(t,T) = \frac{P^M(0,T)}{P^M(0,t)}\exp\!\left(B(t,T)\bigl[f^M(0,t) - r(t)\bigr] + \frac{\sigma^2}{4a}\,B(t,T)^2(1 - e^{-2at})\right) \]

where \(B(t,T) = -(1 - e^{-a(T-t)})/a\).

Verification at \(t = 0\)

At \(t = 0\): \(P^M(0,0) = 1\), \(f^M(0,0) = r(0)\), and \(1 - e^0 = 0\). Therefore

\[ P(0,T) = P^M(0,T)\exp\!\bigl(B(0,T)(r(0) - r(0)) + 0\bigr) = P^M(0,T) \]

confirming exact calibration to the initial curve. \(\square\)


Equivalence with the Expectation Derivation

The bond price derived via the PDE method is identical to the one obtained by computing \(\mathbb{E}^{\mathbb{Q}}[e^{-\int_t^T r(s)\,ds} \mid \mathcal{F}(t)]\) directly (see the sibling section). Both routes exploit the same structural feature -- the affine dependence of the Hull-White drift on \(r\) -- but through different mathematical machinery:

Method Key tool Role of Gaussian structure
Expectation MGF of \(\mathcal{N}(\mu, \sigma^2)\): \(\mathbb{E}[e^{-X}] = e^{-\mu + \sigma^2/2}\) Evaluates the exponential expectation directly
PDE Variable separation via affine ansatz Reduces the PDE to ODEs that can be solved explicitly

The agreement between the two derivations serves as a consistency check: any error in either calculation would produce a discrepancy.


Numerical Example

Consider a Hull-White model with \(a = 0.05\), \(\sigma = 0.01\), flat forward curve \(f^M(0,t) = 0.03\), and current short rate \(r(2) = 0.035\) at time \(t = 2\). Compute the price of a zero-coupon bond maturing at \(T = 12\) (\(\tau = 10\) years).

Step 1: Compute \(B(2, 12)\).

\[ B(2,12) = -\frac{1 - e^{-0.05 \times 10}}{0.05} = -\frac{1 - e^{-0.5}}{0.05} = -\frac{1 - 0.6065}{0.05} \approx -7.869 \]

Step 2: Compute \(A(2, 12)\).

With a flat forward curve: \(P^M(0,T) = e^{-0.03T}\).

\[ \ln\frac{P^M(0,12)}{P^M(0,2)} = -0.03(12) + 0.03(2) = -0.30 \]
\[ B(2,12)\,f^M(0,2) = (-7.869)(0.03) = -0.2361 \]
\[ \frac{\sigma^2}{4a}\,B^2(1 - e^{-2at}) = \frac{0.0001}{0.2}(61.92)(1 - e^{-0.2}) = 0.0005 \times 61.92 \times 0.1813 \approx 0.00561 \]
\[ A(2,12) = -0.30 + (-0.2361) + 0.00561 \approx -0.5305 \]

Step 3: Compute the bond price.

\[ P(2,12) = e^{A + Br} = e^{-0.5305 + (-7.869)(0.035)} = e^{-0.5305 - 0.2754} = e^{-0.8059} \approx 0.4466 \]

For comparison, the market discount factor is \(P^M(0,12)/P^M(0,2) = e^{-0.30} \approx 0.7408/1.0 = 0.7408\). The model price is lower because \(r(2) = 0.035 > f^M(0,2) = 0.03\): the short rate is above the forward rate, depressing bond prices.


Summary

The PDE derivation of the Hull-White bond price proceeds by applying the Feynman-Kac theorem to obtain the backward pricing equation, substituting the exponential-affine ansatz \(P = e^{A + Br}\), and separating the resulting equation into a linear ODE for \(B(\tau)\) with solution \(B(\tau) = -(1 - e^{-a\tau})/a\) and a quadrature for \(A(t,T)\) that reduces to \(A(t,T) = \ln(P^M(0,T)/P^M(0,t)) + B(t,T)f^M(0,t) + \frac{\sigma^2}{4a}B(t,T)^2(1-e^{-2at})\) when \(\theta(t)\) is eliminated via the market curve. The result is identical to the expectation-based derivation, confirming the internal consistency of the Hull-White framework.


Exercises

Exercise 1. State the Feynman-Kac theorem precisely. Identify the functions \(\mu(t,x)\), \(\sigma(t,x)\), \(c(t,x)\), and \(g(x)\) that correspond to the Hull-White zero-coupon bond pricing problem.

Solution to Exercise 1

The Feynman-Kac theorem states: Let \(X_t\) satisfy the SDE \(dX_t = \mu(t,X_t)\,dt + \sigma(t,X_t)\,dW_t\), and define

\[ u(t,x) = \mathbb{E}\left[e^{-\int_t^T c(s,X_s)ds}\,g(X_T)\,\Big|\,X_t = x\right] \]

Then, under suitable regularity conditions (Lipschitz drift and diffusion, polynomial growth of \(g\)), \(u\) solves the PDE:

\[ \frac{\partial u}{\partial t} + \mu(t,x)\frac{\partial u}{\partial x} + \frac{1}{2}\sigma^2(t,x)\frac{\partial^2 u}{\partial x^2} - c(t,x)\,u = 0 \]

with terminal condition \(u(T,x) = g(x)\).

For the Hull-White zero-coupon bond pricing problem, the identifications are:

  • State process: \(X_t = r(t)\), the short rate
  • Drift: \(\mu(t,r) = \theta(t) - ar\), the Hull-White mean-reverting drift
  • Diffusion: \(\sigma(t,r) = \sigma\) (constant volatility)
  • Discount rate: \(c(t,r) = r\), since the discount factor is \(e^{-\int_t^T r(s)ds}\)
  • Terminal payoff: \(g(r) = 1\), since the bond pays one unit at maturity

Substituting these into the Feynman-Kac PDE yields:

\[ \frac{\partial P}{\partial t} + (\theta(t) - ar)\frac{\partial P}{\partial r} + \frac{1}{2}\sigma^2\frac{\partial^2 P}{\partial r^2} - rP = 0, \qquad P(T,r) = 1 \]

Exercise 2. The ODE for \(B(\tau)\) is \(\tilde{B}' + a\tilde{B} = -1\) with \(\tilde{B}(0) = 0\). Solve this ODE using the method of variation of parameters instead of the integrating factor. Verify you obtain the same solution \(B(\tau) = -(1 - e^{-a\tau})/a\).

Solution to Exercise 2

The ODE is \(\tilde{B}' + a\tilde{B} = -1\) with \(\tilde{B}(0) = 0\).

Using variation of parameters: First solve the homogeneous equation \(\tilde{B}'_h + a\tilde{B}_h = 0\), which gives \(\tilde{B}_h(\tau) = Ce^{-a\tau}\).

Assume the particular solution has the form \(\tilde{B}_p(\tau) = C(\tau)e^{-a\tau}\). Substituting:

\[ C'(\tau)e^{-a\tau} - aC(\tau)e^{-a\tau} + aC(\tau)e^{-a\tau} = -1 \]
\[ C'(\tau)e^{-a\tau} = -1 \quad\Longrightarrow\quad C'(\tau) = -e^{a\tau} \]

Integrating:

\[ C(\tau) = -\frac{e^{a\tau}}{a} + K \]

The general solution is:

\[ \tilde{B}(\tau) = C(\tau)e^{-a\tau} = \left(-\frac{e^{a\tau}}{a} + K\right)e^{-a\tau} = -\frac{1}{a} + Ke^{-a\tau} \]

Applying the initial condition \(\tilde{B}(0) = 0\):

\[ 0 = -\frac{1}{a} + K \quad\Longrightarrow\quad K = \frac{1}{a} \]

Therefore:

\[ \tilde{B}(\tau) = -\frac{1}{a} + \frac{1}{a}e^{-a\tau} = -\frac{1 - e^{-a\tau}}{a} \]

This matches the integrating factor solution.


Exercise 3. Show that \(B(\tau) \approx -\tau + \frac{a\tau^2}{2}\) for small \(a\tau\) by Taylor expanding \(e^{-a\tau}\). Use this to estimate the error in approximating \(B(\tau) \approx -\tau\) (the Ho-Lee limit) when \(a = 0.05\) and \(\tau = 10\).

Solution to Exercise 3

Taylor expand \(e^{-a\tau} = 1 - a\tau + \frac{(a\tau)^2}{2} - \frac{(a\tau)^3}{6} + \cdots\) :

\[ 1 - e^{-a\tau} = a\tau - \frac{(a\tau)^2}{2} + \frac{(a\tau)^3}{6} - \cdots \]
\[ B(\tau) = -\frac{1-e^{-a\tau}}{a} = -\tau + \frac{a\tau^2}{2} - \frac{a^2\tau^3}{6} + \cdots \]

For small \(a\tau\), keeping the first two terms: \(B(\tau) \approx -\tau + \frac{a\tau^2}{2}\).

The Ho-Lee approximation is \(B(\tau) \approx -\tau\), so the error is approximately \(\frac{a\tau^2}{2}\).

Numerical estimate for \(a = 0.05\), \(\tau = 10\):

\[ \text{Error} \approx \frac{0.05 \times 100}{2} = 2.5 \]

Exact value: \(B(10) = -\frac{1-e^{-0.5}}{0.05} = -\frac{0.3935}{0.05} = -7.869\). The Ho-Lee approximation gives \(B(10) \approx -10\). The error is \(|-10 - (-7.869)| = 2.131\), which is close to the estimate of 2.5. The quadratic correction \(\frac{a\tau^2}{2} = 2.5\) slightly overestimates the true error because higher-order terms (negative) partially cancel.

The relative error is \(\frac{2.131}{7.869} \approx 27\%\), indicating that the Ho-Lee approximation is poor for \(a\tau = 0.5\).


Exercise 4. Explain why the \(A\)-equation is a quadrature rather than a genuine ODE. What structural property of the Hull-White model makes this simplification possible, and how would the situation change for a model with state-dependent volatility (e.g., CIR)?

Solution to Exercise 4

The \(A\)-equation is a quadrature rather than a genuine ODE because the right-hand side depends only on \(\tau\) (through the known function \(B(\tau)\)) and on \(t\) (through \(\theta(T-\tau)\)), but not on \(A\) itself:

\[ \frac{d\tilde{A}}{d\tau} = \theta(T-\tau)\,B(\tau) + \frac{1}{2}\sigma^2 B(\tau)^2 \]

The function \(A\) does not appear on the right-hand side, so the equation can be solved by direct integration (quadrature) once \(B(\tau)\) is known. There is no feedback from \(A\) to its own derivative.

Structural property: This simplification occurs because the Hull-White model has constant (state-independent) volatility \(\sigma\). The \(B^2\) term in the PDE substitution comes from \(\frac{1}{2}\sigma^2 B^2\), which is independent of \(r\), and hence independent of \(A\).

CIR model comparison: In the CIR model, \(\sigma(r) = \sigma\sqrt{r}\), so \(\frac{1}{2}\sigma^2 r \cdot B^2\) contributes a term proportional to \(rB^2\) when substituting the ansatz. This \(r\)-dependent term mixes with the coefficient of \(r\) in the separation, modifying the \(B\)-equation to a genuine (nonlinear) Riccati ODE:

\[ B'(\tau) = -aB - 1 + \frac{1}{2}\sigma^2 B^2 \]

The \(B^2\) term makes the ODE nonlinear, and the \(A\)-equation also becomes more complex. However, \(A\) is still determined by quadrature once \(B\) is known, because even in CIR the ansatz \(P = e^{A+Br}\) separates cleanly into \(r\)-dependent and \(r\)-independent parts.


Exercise 5. Using the numerical example with \(a = 0.05\), \(\sigma = 0.01\), \(f^M(0,t) = 0.03\), and \(r(2) = 0.035\), compute \(P(2, 12)\) step by step. Then recompute for \(r(2) = 0.025\) and comment on the sensitivity of the bond price to the current short rate.

Solution to Exercise 5

Given: \(a = 0.05\), \(\sigma = 0.01\), \(f^M(0,t) = 0.03\), \(\tau = 10\).

Case 1: \(r(2) = 0.035\) (already computed in the text).

\[ B(2,12) = -\frac{1-e^{-0.5}}{0.05} \approx -7.869 \]
\[ \ln\frac{P^M(0,12)}{P^M(0,2)} = -0.03(12-0) + 0.03(2) = -0.36 + 0.06 = -0.30 \]
\[ B \cdot f^M(0,2) = -7.869 \times 0.03 = -0.2361 \]
\[ \frac{\sigma^2}{4a}B^2(1-e^{-2at}) = \frac{0.0001}{0.2} \times 61.92 \times (1-e^{-0.2}) = 0.0005 \times 61.92 \times 0.1813 \approx 0.00561 \]
\[ A(2,12) = -0.30 + (-0.2361) + 0.00561 = -0.5305 \]
\[ P(2,12) = e^{-0.5305 + (-7.869)(0.035)} = e^{-0.5305 - 0.2754} = e^{-0.8059} \approx 0.4466 \]

Case 2: \(r(2) = 0.025\).

\(A(2,12)\) is unchanged at \(-0.5305\) (it does not depend on \(r\)).

\[ P(2,12) = e^{-0.5305 + (-7.869)(0.025)} = e^{-0.5305 - 0.1967} = e^{-0.7272} \approx 0.4833 \]

Comparison: When \(r(2)\) drops from 0.035 to 0.025 (a decrease of 100 basis points), the bond price increases from 0.4466 to 0.4833, a change of \(+0.0367\) or approximately \(+8.2\%\). The sensitivity is \(\frac{\Delta P}{\Delta r} \approx \frac{0.0367}{-0.01} = -3.67\), which is related to \(B(2,12) \times P \approx -7.869 \times 0.465 \approx -3.66\), confirming the affine structure.


Exercise 6. Verify the consistency check at \(t = 0\): show that \(P(0,T) = P^M(0,T)\) by substituting \(t = 0\) into the complete bond price formula. Which terms simplify or vanish, and why?

Solution to Exercise 6

At \(t = 0\), substitute into the complete bond price formula:

\[ P(0,T) = \frac{P^M(0,T)}{P^M(0,0)}\exp\left(B(0,T)[f^M(0,0) - r(0)] + \frac{\sigma^2}{4a}B(0,T)^2(1-e^{-2a\cdot 0})\right) \]

Terms that simplify or vanish:

  1. \(P^M(0,0) = 1\): the price of a bond maturing immediately is 1.

  2. \(f^M(0,0) = r(0)\): the instantaneous forward rate at time zero equals the current short rate, so \(f^M(0,0) - r(0) = 0\).

  3. \(1 - e^{-2a \cdot 0} = 1 - 1 = 0\): the convexity correction vanishes.

Therefore:

\[ P(0,T) = \frac{P^M(0,T)}{1}\exp(0 + 0) = P^M(0,T) \]

The model exactly reproduces the market discount curve at \(t = 0\). This verification confirms that the PDE derivation is consistent with the calibration requirement: the time-dependent drift \(\theta(t)\) has been chosen precisely so that the \(A(t,T)\) function absorbs the market term structure at the initial time.


Exercise 7. Compare the PDE and expectation derivations from a computational perspective. For a model where closed-form solutions are not available, which approach (numerical PDE vs. Monte Carlo) would you prefer for computing bond prices? Discuss the trade-offs in terms of accuracy, computational cost, and dimensionality.

Solution to Exercise 7

Numerical PDE (finite differences):

  • Accuracy: Deterministic error that decreases systematically with grid refinement. For smooth solutions (as in Hull-White), second-order schemes like Crank-Nicolson converge as \(O(\Delta r^2 + \Delta t^2)\).
  • Cost: Scales as \(O(N_r \times N_t)\) where \(N_r\) is the number of spatial grid points and \(N_t\) is the number of time steps. Very efficient for one-factor models.
  • Dimensionality: Cost grows exponentially with the number of factors (curse of dimensionality). A two-factor model (e.g., Heston) requires a 2D grid; three factors become prohibitively expensive.
  • Output: Produces the bond price for all values of \(r\) simultaneously (the entire price surface), which is useful for computing Greeks.

Monte Carlo (expectation approach):

  • Accuracy: Statistical error decreasing as \(O(1/\sqrt{N})\) where \(N\) is the number of paths. Convergence is slow but dimension-independent.
  • Cost: Scales as \(O(N \times N_t \times d)\) where \(d\) is the number of factors. Adding dimensions increases cost linearly, not exponentially.
  • Dimensionality: Monte Carlo is the preferred method for high-dimensional problems (multi-factor models, path-dependent payoffs).
  • Output: Produces the bond price at a single point \((t, r(t))\) per simulation. Computing the full price surface requires separate simulations.

Recommendation: For single-factor models like Hull-White, the PDE approach is strongly preferred: it is faster, more accurate, and provides the full price surface. For multi-factor models (two or more state variables) or path-dependent payoffs, Monte Carlo becomes the practical choice despite its slower convergence, because the PDE grid becomes infeasible in high dimensions.