Bond Price via Expectation

Under the risk-neutral measure \(\mathbb{Q}\), the price of a zero-coupon bond equals the expected discounted payoff. In the Hull-White model the short rate \(r(t)\) is Gaussian, so the integral \(\int_t^T r(s)\,ds\) is also Gaussian. This section derives the closed-form bond price \(P(t,T) = A(t,T)e^{-B(t,T)r(t)}\) by computing the expectation of the exponential of a Gaussian random variable directly, without appealing to PDE methods.

Risk-Neutral Pricing Formula

The fundamental pricing identity for a zero-coupon bond paying one unit at maturity \(T\) states that its time-\(t\) price equals the conditional expectation of the stochastic discount factor under \(\mathbb{Q}\).

Definition: Zero-Coupon Bond Price

Under the risk-neutral measure \(\mathbb{Q}\), the zero-coupon bond price is

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

where \(r(s)\) is the short rate process and \(\mathcal{F}(t)\) is the filtration at time \(t\).

The key observation is that when \(r(s)\) follows the Hull-White process

\[ dr(s) = \lambda\bigl(\theta^{\mathbb{Q}}(s) - r(s)\bigr)\,ds + \sigma\,dW^{\mathbb{Q}}(s) \]

the integral \(\int_t^T r(s)\,ds\) is a Gaussian random variable conditional on \(\mathcal{F}(t)\). This Gaussian structure allows us to evaluate the expectation in closed form using the moment generating function of the normal distribution.

Conditional Distribution of the Integral

Recall from the short rate solution that for \(s \ge t\),

\[ r(s) = r(t)e^{-\lambda(s-t)} + \alpha(s) - \alpha(t)e^{-\lambda(s-t)} + \sigma\int_t^s e^{-\lambda(s-u)}\,dW^{\mathbb{Q}}(u) \]

where

\[ \alpha(s) = f^M(0,s) + \frac{\sigma^2}{2\lambda^2}\left(1 - e^{-\lambda s}\right)^2 \]

Integrating \(r(s)\) from \(t\) to \(T\) and interchanging the order of integration with the stochastic integral, the random variable \(I(t,T) := \int_t^T r(s)\,ds\) conditional on \(\mathcal{F}(t)\) is normally distributed.

Proposition: Conditional Gaussian Integral

The integrated short rate \(I(t,T) = \int_t^T r(s)\,ds\) conditional on \(\mathcal{F}(t)\) satisfies

\[ I(t,T)\,\Big|\,\mathcal{F}(t) \;\sim\; \mathcal{N}\!\left(M(t,T),\; V(t,T)\right) \]

where the conditional mean and variance are given below.

Conditional mean. Separating the deterministic and stochastic parts of \(r(s)\) and integrating from \(t\) to \(T\):

\[ M(t,T) = B(t,T)\bigl[r(t) - \alpha(t)\bigr] + \int_t^T \alpha(s)\,ds \]

where \(B(t,T) = \frac{1}{\lambda}\left[1 - e^{-\lambda(T-t)}\right]\) collects the contribution of the current short rate \(r(t)\).

Derivation of Conditional Mean

Integrating the short rate solution over \([t,T]\):

\[\begin{array}{lllll} \displaystyle \int_t^T r(s)\,ds &=&\displaystyle \bigl[r(t) - \alpha(t)\bigr]\int_t^T e^{-\lambda(s-t)}\,ds + \int_t^T \alpha(s)\,ds + \sigma\int_t^T\!\int_t^s e^{-\lambda(s-u)}\,dW^{\mathbb{Q}}(u)\,ds \end{array}\]

The first integral evaluates to

\[ \int_t^T e^{-\lambda(s-t)}\,ds = \frac{1}{\lambda}\left[1 - e^{-\lambda(T-t)}\right] = B(t,T) \]

Taking the conditional expectation, the stochastic integral vanishes and we obtain \(M(t,T)\).

Conditional variance. The variance arises entirely from the stochastic integral term. Using the Fubini-type interchange of integration order:

\[ V(t,T) = \frac{\sigma^2}{\lambda^2}\left[T - t + \frac{2}{\lambda}e^{-\lambda(T-t)} - \frac{1}{2\lambda}e^{-2\lambda(T-t)} - \frac{3}{2\lambda}\right] \]

Derivation of Conditional Variance

Starting from the double integral in the stochastic part:

\[\begin{array}{lllll} \displaystyle \sigma\int_t^T\!\int_t^s e^{-\lambda(s-u)}\,dW^{\mathbb{Q}}(u)\,ds &=&\displaystyle \sigma\int_t^T\!\left(\int_u^T e^{-\lambda(s-u)}\,ds\right)dW^{\mathbb{Q}}(u) \end{array}\]

The inner integral equals \(B(u,T) = \frac{1}{\lambda}\left[1 - e^{-\lambda(T-u)}\right]\), so

\[\begin{array}{lllll} \displaystyle V(t,T) &=&\displaystyle \sigma^2\int_t^T B(u,T)^2\,du \;=\; \frac{\sigma^2}{\lambda^2}\int_t^T \left[1 - e^{-\lambda(T-u)}\right]^2 du \end{array}\]

Expanding the square and integrating term by term with the substitution \(v = T - u\):

\[\begin{array}{lllll} \displaystyle V(t,T) &=&\displaystyle \frac{\sigma^2}{\lambda^2}\left[(T-t) - \frac{2}{\lambda}\left(1 - e^{-\lambda(T-t)}\right) + \frac{1}{2\lambda}\left(1 - e^{-2\lambda(T-t)}\right)\right] \end{array}\]

which simplifies to the stated formula.

Gaussian Moment Generating Function

For any Gaussian random variable \(X \sim \mathcal{N}(\mu, \sigma^2)\), the moment generating function evaluated at \(u = -1\) gives

\[ \mathbb{E}\!\left[e^{-X}\right] = e^{-\mu + \frac{1}{2}\sigma^2} \]

This identity is the central tool for converting the conditional Gaussian distribution of \(I(t,T)\) into a closed-form bond price.

Bond Price Formula

Applying the moment generating function identity to \(I(t,T) \sim \mathcal{N}(M(t,T), V(t,T))\) yields the Hull-White affine bond pricing formula.

Theorem: Hull-White Bond Price via Expectation

The zero-coupon bond price in the Hull-White model is

\[ P(t,T) = A(t,T)\,e^{-B(t,T)\,r(t)} \]

where

\[\begin{array}{lllll} \displaystyle B(t,T) &=&\displaystyle \frac{1}{\lambda}\left[1 - e^{-\lambda(T-t)}\right] \\[8pt] \displaystyle A(t,T) &=&\displaystyle \frac{P^M(0,T)}{P^M(0,t)}\exp\!\left\{B(t,T)f^M(0,t) - \frac{\sigma^2}{4\lambda}B(t,T)^2\left(1 - e^{-2\lambda t}\right)\right\} \end{array}\]

Proof

Starting from the pricing formula:

\[\begin{array}{lllll} \displaystyle P(t,T) &=&\displaystyle \mathbb{E}^{\mathbb{Q}}\!\left[e^{-I(t,T)}\,\Big|\,\mathcal{F}(t)\right] \;=\; e^{-M(t,T) + \frac{1}{2}V(t,T)} \end{array}\]

Substituting the expressions for \(M(t,T)\) and \(V(t,T)\) and collecting terms:

\[\begin{array}{lllll} \displaystyle P(t,T) &=&\displaystyle \exp\!\left\{-B(t,T)\bigl[r(t) - \alpha(t)\bigr] - \int_t^T \alpha(s)\,ds + \frac{1}{2}V(t,T)\right\} \end{array}\]

The terms not involving \(r(t)\) combine into \(\log A(t,T)\), and the \(r(t)\)-dependent part gives \(-B(t,T)r(t)\). Using the relation

\[ \alpha(t) = f^M(0,t) + \frac{\sigma^2}{2\lambda^2}\left(1 - e^{-\lambda t}\right)^2 \]

and the market bond price identity \(P^M(0,T) = \exp\!\left(-\int_0^T f^M(0,s)\,ds\right)\), the deterministic factor simplifies to

\[\begin{array}{lllll} \displaystyle A(t,T) &=&\displaystyle \frac{P^M(0,T)}{P^M(0,t)}\exp\!\left\{B(t,T)f^M(0,t) - \frac{\sigma^2}{4\lambda}B(t,T)^2\left(1 - e^{-2\lambda t}\right)\right\} \end{array}\]

which completes the derivation. \(\square\)

Interpretation of the Formula

The bond price formula \(P(t,T) = A(t,T)e^{-B(t,T)r(t)}\) has the affine structure characteristic of Gaussian short rate models. Two features deserve emphasis:

• Factor \(B(t,T)\): This function measures the sensitivity of the bond price to changes in the current short rate \(r(t)\). For small mean reversion \(\lambda \to 0\), we recover \(B(t,T) \to T - t\), the duration of the bond. For large \(\lambda\), the exponential decay dampens the effect of \(r(t)\) on long-dated bonds.

• Factor \(A(t,T)\): This deterministic function encodes the initial market term structure through \(P^M(0,T)/P^M(0,t)\) and the instantaneous forward rate \(f^M(0,t)\). The correction term involving \(\sigma^2\) accounts for the convexity adjustment due to the volatility of the short rate.

Numerical Example

Consider a Hull-White model with parameters \(\lambda = 0.05\), \(\sigma = 0.01\), and a flat initial forward rate curve \(f^M(0,t) = 0.03\) for all \(t\). We compute the 10-year zero-coupon bond price at time \(t = 0\).

With \(r(0) = f^M(0,0) = 0.03\), the function values are:

\[\begin{array}{lllll} \displaystyle B(0,10) &=&\displaystyle \frac{1}{0.05}\left[1 - e^{-0.05 \times 10}\right] \;=\; 20\left(1 - e^{-0.5}\right) \;\approx\; 7.8694 \end{array}\]

Since \(P^M(0,T) = e^{-0.03T}\) for a flat curve:

\[\begin{array}{lllll} \displaystyle A(0,10) &=&\displaystyle e^{-0.03 \times 10}\exp\!\left\{7.8694 \times 0.03 - \frac{0.01^2}{4 \times 0.05}\times 7.8694^2 \times 0\right\} \;=\; e^{-0.3}\,e^{0.2361} \end{array}\]

At \(t=0\) the exponential correction term involving \((1 - e^{-2\lambda t})\) vanishes, giving

\[ P(0,10) = A(0,10)\,e^{-B(0,10) \times 0.03} = e^{-0.03 \times 10} \approx 0.7408 \]

This recovers the market discount factor, confirming the model's consistency with the initial term structure at \(t = 0\).

Connection to PDE Derivation

The expectation-based derivation and the PDE-based derivation yield identical bond price formulas. The PDE approach substitutes the affine ansatz \(P(t,T) = e^{A(\tau) + B(\tau)r(t)}\) into the fundamental PDE and solves the resulting Riccati ODEs for \(A\) and \(B\). The expectation approach instead evaluates the Gaussian integral directly. Both methods exploit the affine (linear-in-\(r\)) structure of the Hull-White drift, which ensures that all relevant conditional distributions remain Gaussian.

---

Summary

This section derived the Hull-White zero-coupon bond price by computing \(\mathbb{E}^{\mathbb{Q}}[e^{-\int_t^T r(s)\,ds}\,|\,\mathcal{F}(t)]\) directly. The Gaussian property of the OU-type short rate process implies that the integrated short rate is conditionally normal, and the moment generating function of the normal distribution delivers the closed-form result \(P(t,T) = A(t,T)e^{-B(t,T)r(t)}\). The function \(A(t,T)\) encodes the initial market term structure and a volatility convexity correction, while \(B(t,T)\) captures the mean-reversion-adjusted duration.

---

Exercises

Exercise 1. Verify the conditional variance formula by computing \(\sigma^2 \int_t^T B(u,T)^2\,du\) directly. Expand the square \((1 - e^{-\lambda(T-u)})^2\), integrate each term, and confirm the result matches \(V(t,T)\).

Solution to Exercise 1

We need to compute \(V(t,T) = \sigma^2 \int_t^T B(u,T)^2\,du\) where \(B(u,T) = \frac{1}{\lambda}(1 - e^{-\lambda(T-u)})\).

Substituting \(v = T - u\) (so \(du = -dv\), and the limits change from \(u = t\) to \(v = \tau = T-t\) and from \(u = T\) to \(v = 0\)):

\[ V(t,T) = \frac{\sigma^2}{\lambda^2}\int_0^{\tau}(1 - e^{-\lambda v})^2\,dv \]

Expand the square:

\[ (1 - e^{-\lambda v})^2 = 1 - 2e^{-\lambda v} + e^{-2\lambda v} \]

Integrate term by term:

\[ \int_0^{\tau} 1\,dv = \tau \]

\[ \int_0^{\tau} 2e^{-\lambda v}\,dv = 2\left[-\frac{1}{\lambda}e^{-\lambda v}\right]_0^{\tau} = \frac{2}{\lambda}(1 - e^{-\lambda\tau}) \]

\[ \int_0^{\tau} e^{-2\lambda v}\,dv = \left[-\frac{1}{2\lambda}e^{-2\lambda v}\right]_0^{\tau} = \frac{1}{2\lambda}(1 - e^{-2\lambda\tau}) \]

Combining:

\[ V(t,T) = \frac{\sigma^2}{\lambda^2}\left[\tau - \frac{2}{\lambda}(1 - e^{-\lambda\tau}) + \frac{1}{2\lambda}(1 - e^{-2\lambda\tau})\right] \]

Rewriting by distributing the negative signs:

\[ = \frac{\sigma^2}{\lambda^2}\left[\tau + \frac{2}{\lambda}e^{-\lambda\tau} - \frac{1}{2\lambda}e^{-2\lambda\tau} - \frac{2}{\lambda} + \frac{1}{2\lambda}\right] \]

\[ = \frac{\sigma^2}{\lambda^2}\left[\tau + \frac{2}{\lambda}e^{-\lambda\tau} - \frac{1}{2\lambda}e^{-2\lambda\tau} - \frac{3}{2\lambda}\right] \]

This confirms the stated formula for \(V(t,T)\).

---

Exercise 2. Show that \(V(t,T) \to \frac{\sigma^2}{\lambda^2}(T-t)\) in the limit \(\lambda \to 0\) by applying L'Hopital's rule or Taylor expanding the exponentials. Interpret this limiting variance in terms of the integrated Brownian motion.

Solution to Exercise 2

Taylor expand each exponential in \(V(t,T)\) around \(\lambda = 0\) using \(\tau = T - t\):

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

\[ e^{-2\lambda\tau} = 1 - 2\lambda\tau + 2(\lambda\tau)^2 - \frac{4(\lambda\tau)^3}{3} + \cdots \]

Substituting into \(V(t,T) = \frac{\sigma^2}{\lambda^2}\left[\tau + \frac{2}{\lambda}e^{-\lambda\tau} - \frac{1}{2\lambda}e^{-2\lambda\tau} - \frac{3}{2\lambda}\right]\):

\[ \frac{2}{\lambda}e^{-\lambda\tau} = \frac{2}{\lambda} - 2\tau + \lambda\tau^2 - \frac{\lambda^2\tau^3}{3} + \cdots \]

\[ \frac{1}{2\lambda}e^{-2\lambda\tau} = \frac{1}{2\lambda} - \tau + \lambda\tau^2 - \frac{2\lambda^2\tau^3}{3} + \cdots \]

Combining:

\[ \tau + \frac{2}{\lambda} - 2\tau + \lambda\tau^2 - \frac{1}{2\lambda} + \tau - \lambda\tau^2 - \frac{3}{2\lambda} + O(\lambda^2) \]

\[ = \tau - 2\tau + \tau + \frac{2}{\lambda} - \frac{1}{2\lambda} - \frac{3}{2\lambda} + \lambda\tau^2 - \lambda\tau^2 + O(\lambda^2) \]

The constant terms cancel: \(\frac{2}{\lambda} - \frac{1}{2\lambda} - \frac{3}{2\lambda} = 0\). The \(\tau\) terms cancel: \(\tau - 2\tau + \tau = 0\). The leading surviving term is \(\frac{\lambda^2\tau^3}{3}\) (from detailed expansion of cubic terms).

Therefore:

\[ V(t,T) = \frac{\sigma^2}{\lambda^2} \cdot \frac{\lambda^2\tau^3}{3} + O(\lambda^2) = \frac{\sigma^2\tau^3}{3} \]

Wait -- let us redo this more carefully. Actually \(V(t,T) \to \sigma^2 \tau^3/3\) for \(\lambda \to 0\), not \(\sigma^2\tau/\lambda^2\). Let us verify: in the Ho-Lee model (\(\lambda = 0\)), \(r(s) = r(t) + \theta s + \sigma W(s)\), and

\[ \text{Var}\left(\int_t^T r(s)ds\right) = \sigma^2\int_t^T\int_t^T \min(s-t,u-t)\,ds\,du = \frac{\sigma^2(T-t)^3}{3} \]

This is consistent with \(V(t,T) \to \sigma^2\tau^3/3\) as \(\lambda \to 0\).

Interpretation: In the limit \(\lambda \to 0\) there is no mean reversion, so the short rate performs a random walk. The integrated Brownian motion \(\int_t^T W(s)\,ds\) has variance proportional to \(\tau^3/3\), growing cubically with the integration horizon. Mean reversion (\(\lambda > 0\)) dampens the variance, transitioning the growth from cubic to approximately linear for large \(\tau\).

---

Exercise 3. For a Hull-White model with \(\lambda = 0.1\), \(\sigma = 0.015\), and \(f^M(0,t) = 0.04\), compute the conditional mean \(M(0, 10)\) and variance \(V(0, 10)\) of the integrated short rate \(\int_0^{10} r(s)\,ds\) given \(r(0) = 0.04\).

Solution to Exercise 3

With \(\lambda = 0.1\), \(\sigma = 0.015\), \(f^M(0,t) = 0.04\), \(r(0) = 0.04\), and \(\tau = T - t = 10\):

Conditional mean \(M(0,10)\): Using \(M(t,T) = B(t,T)[r(t) - \alpha(t)] + \int_t^T \alpha(s)\,ds\):

\[ B(0,10) = \frac{1}{0.1}(1 - e^{-1}) = 10(1 - 0.3679) = 6.321 \]

\[ \alpha(0) = f^M(0,0) + \frac{\sigma^2}{2\lambda^2}(1 - e^0)^2 = 0.04 + 0 = 0.04 \]

Since \(r(0) = \alpha(0) = 0.04\), the first term is zero: \(B(0,10)[r(0) - \alpha(0)] = 0\).

For the integral \(\int_0^{10}\alpha(s)\,ds\) with \(\alpha(s) = 0.04 + \frac{0.000225}{0.02}(1 - e^{-0.1s})^2 = 0.04 + 0.01125(1 - e^{-0.1s})^2\):

\[ \int_0^{10}\alpha(s)\,ds = 0.4 + 0.01125\int_0^{10}(1 - e^{-0.1s})^2\,ds \]

\[ \int_0^{10}(1-e^{-0.1s})^2\,ds = 10 - \frac{2}{0.1}(1-e^{-1}) + \frac{1}{0.2}(1-e^{-2}) \]

\[ = 10 - 20(0.6321) + 5(0.8647) = 10 - 12.642 + 4.3235 = 1.6815 \]

\[ M(0,10) = 0.4 + 0.01125 \times 1.6815 \approx 0.4 + 0.01892 = 0.41892 \]

Conditional variance \(V(0,10)\):

\[ V(0,10) = \frac{0.015^2}{0.01}\left[10 + \frac{2}{0.1}e^{-1} - \frac{1}{0.2}e^{-2} - \frac{3}{0.2}\right] \]

\[ = 0.0225\left[10 + 20(0.3679) - 5(0.1353) - 15\right] \]

\[ = 0.0225\left[10 + 7.358 - 0.6767 - 15\right] = 0.0225 \times 1.6813 \approx 0.03783 \]

The standard deviation is \(\sqrt{0.03783} \approx 0.1946\).

---

Exercise 4. The moment generating function identity \(\mathbb{E}[e^{-X}] = e^{-\mu + \frac{1}{2}\sigma^2}\) for \(X \sim \mathcal{N}(\mu, \sigma^2)\) is central to the derivation. Prove this identity by completing the square in the integral

\[ \int_{-\infty}^{\infty} e^{-x} \frac{1}{\sqrt{2\pi}\sigma} e^{-(x-\mu)^2/(2\sigma^2)}\,dx \]

Solution to Exercise 4

We need to evaluate:

\[ I = \int_{-\infty}^{\infty} e^{-x}\,\frac{1}{\sqrt{2\pi}\sigma}\,e^{-(x-\mu)^2/(2\sigma^2)}\,dx \]

Combine the exponents:

\[ -x - \frac{(x-\mu)^2}{2\sigma^2} = -x - \frac{x^2 - 2\mu x + \mu^2}{2\sigma^2} \]

\[ = -\frac{1}{2\sigma^2}\left[x^2 - 2\mu x + \mu^2 + 2\sigma^2 x\right] = -\frac{1}{2\sigma^2}\left[x^2 - 2(\mu - \sigma^2)x + \mu^2\right] \]

Complete the square in \(x\):

\[ x^2 - 2(\mu-\sigma^2)x + \mu^2 = \left[x - (\mu-\sigma^2)\right]^2 - (\mu-\sigma^2)^2 + \mu^2 \]

\[ = \left[x - (\mu-\sigma^2)\right]^2 + 2\mu\sigma^2 - \sigma^4 \]

Therefore the exponent becomes:

\[ -\frac{[x-(\mu-\sigma^2)]^2}{2\sigma^2} - \mu + \frac{\sigma^2}{2} \]

Substituting back:

\[ I = e^{-\mu + \sigma^2/2}\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}\sigma}\,e^{-[x-(\mu-\sigma^2)]^2/(2\sigma^2)}\,dx \]

The integral is the total probability of a \(\mathcal{N}(\mu - \sigma^2, \sigma^2)\) density, which equals 1. Therefore:

\[ \mathbb{E}[e^{-X}] = e^{-\mu + \sigma^2/2} \]

---

Exercise 5. The factor \(B(t,T)\) has the interpretation of a mean-reversion-adjusted duration. Compare \(B(0,T) = \frac{1}{\lambda}(1 - e^{-\lambda T})\) with the Macaulay duration \(T\) for a zero-coupon bond. For which values of \(\lambda\) and \(T\) does the difference exceed 10%?

Solution to Exercise 5

The Macaulay duration of a zero-coupon bond is simply \(T\) (the time to maturity). The mean-reversion-adjusted duration is \(B(0,T) = \frac{1}{\lambda}(1 - e^{-\lambda T})\).

The relative difference is:

\[ \frac{T - B(0,T)}{T} = 1 - \frac{1 - e^{-\lambda T}}{\lambda T} \]

We want this to exceed 10%, i.e.:

\[ 1 - \frac{1 - e^{-\lambda T}}{\lambda T} > 0.1 \quad\Longleftrightarrow\quad \frac{1 - e^{-\lambda T}}{\lambda T} < 0.9 \]

Let \(x = \lambda T\). The condition becomes \(\frac{1 - e^{-x}}{x} < 0.9\).

For small \(x\): \(\frac{1-e^{-x}}{x} \approx 1 - x/2 + x^2/6 - \cdots < 0.9\) when \(x/2 > 0.1\), i.e., \(x > 0.2\) approximately. Numerically solving \(\frac{1-e^{-x}}{x} = 0.9\): at \(x = 0.2\), \(\frac{1-e^{-0.2}}{0.2} = \frac{0.1813}{0.2} = 0.9066\); at \(x = 0.22\), \(\frac{1-e^{-0.22}}{0.22} = \frac{0.1975}{0.22} = 0.8977\). So \(x \approx 0.21\).

Examples where the 10% threshold is exceeded:

• \(\lambda = 0.05\), \(T = 5\): \(x = 0.25\), difference \(\approx 11.6\%\)
• \(\lambda = 0.1\), \(T = 3\): \(x = 0.3\), difference \(\approx 13.7\%\)
• \(\lambda = 0.02\), \(T = 15\): \(x = 0.3\), difference \(\approx 13.7\%\)

In general, whenever \(\lambda T \gtrsim 0.21\), the relative difference exceeds 10%. Higher mean reversion or longer maturities amplify the discrepancy between the standard duration \(T\) and the mean-reversion-adjusted duration \(B(0,T)\).

---

Exercise 6. Show that the bond price formula satisfies \(P(t,t) = 1\) for all \(t\) by verifying that \(A(t,t) = 0\) and \(B(t,t) = 0\). What does this boundary condition represent financially?

Solution to Exercise 6

At \(\tau = T - t = 0\):

\(B(t,t) = 0\): From \(B(t,T) = \frac{1}{\lambda}(1 - e^{-\lambda(T-t)})\), setting \(T = t\):

\[ B(t,t) = \frac{1}{\lambda}(1 - e^{0}) = \frac{1}{\lambda}(1 - 1) = 0 \]

\(A(t,t) = 1\): From the formula:

\[ A(t,t) = \frac{P^M(0,t)}{P^M(0,t)}\exp\left\{0 \cdot f^M(0,t) - \frac{\sigma^2}{4\lambda} \cdot 0^2 \cdot (\cdots)\right\} = 1 \cdot e^0 = 1 \]

Therefore (note \(\log A(t,t) = 0\)):

\[ P(t,t) = A(t,t)\,e^{-B(t,t)\,r(t)} = 1 \cdot e^{0} = 1 \]

Financial interpretation: The bond price \(P(t,t) = 1\) means that a zero-coupon bond at its maturity date pays exactly its face value of one unit of currency. This is the terminal (or boundary) condition: there is no discounting needed when the payment is immediate.

---

Exercise 7. Explain why the expectation-based derivation and the PDE-based derivation must produce the same bond price formula. Under what conditions could the two approaches give different results (hint: consider models where the short rate is not affine)?

Solution to Exercise 7

Why both derivations agree: The Feynman-Kac theorem establishes a rigorous equivalence between the conditional expectation \(\mathbb{E}^{\mathbb{Q}}[e^{-\int_t^T r(s)ds} \mid \mathcal{F}(t)]\) and the solution of the backward PDE \(\frac{\partial P}{\partial t} + \mu\frac{\partial P}{\partial r} + \frac{1}{2}\sigma^2\frac{\partial^2 P}{\partial r^2} - rP = 0\) with terminal condition \(P(T,T) = 1\). Under standard regularity conditions (the Hull-White model satisfies these because the drift is affine and the diffusion is bounded), the Feynman-Kac theorem guarantees that the PDE solution and the expectation yield the same function \(P(t,T)\). Therefore, any correct computation of either one must produce the same answer.

When the two approaches could diverge in practice:

1. Non-affine models: In models like the Black-Karasinski model (\(d\log r = a(\theta - \log r)dt + \sigma\,dW\)), the short rate dynamics are not affine. The exponential-affine ansatz fails, and the PDE cannot be reduced to ODEs. The expectation approach also fails to produce a closed form because \(\int_t^T r(s)ds\) is not Gaussian. Both approaches still give the same theoretical answer (by Feynman-Kac), but neither yields a closed-form solution -- both require numerical methods.

2. Numerical approximation: When solving numerically, the PDE approach (finite differences) and the expectation approach (Monte Carlo) use different approximations with different error characteristics. Finite differences introduce grid truncation and discretization error; Monte Carlo introduces statistical sampling error and time-stepping bias. These can produce different numerical values that converge to the same limit as resolution increases.

3. Regularity failures: If the Feynman-Kac conditions fail (e.g., the diffusion coefficient degenerates, or the solution grows too fast), the PDE may have multiple solutions or the expectation may not be well-defined, potentially leading to discrepancies. The Hull-White model does not suffer from these issues.