T-Forward Measure

When pricing a derivative that pays at a fixed future time \(T\), the standard risk-neutral expectation \(\mathbb{E}^{\mathbb{Q}}[e^{-\int_t^T r(s)\,ds}V(T)\,|\,\mathcal{F}(t)]\) involves both the payoff \(V(T)\) and the stochastic discount factor. The \(T\)-forward measure \(\mathbb{Q}^T\) eliminates the discount factor by using the zero-coupon bond \(P(t,T)\) as numeraire, replacing the expectation with \(P(t,T)\,\mathbb{E}^{T}[V(T)\,|\,\mathcal{F}(t)]\). This section defines the \(T\)-forward measure, derives the change of measure from \(\mathbb{Q}\) to \(\mathbb{Q}^T\), and establishes the forward rate martingale property.

Numeraire and Measure Change

The general change-of-numeraire theorem states that if \(N(t)\) is a positive, tradable numeraire process, then the measure \(\mathbb{Q}^N\) defined by

\[ \frac{d\mathbb{Q}^N}{d\mathbb{Q}}\Bigg|_{\mathcal{F}(t)} = \frac{N(t)/N(0)}{M(t)/M(0)} \]

makes the ratio \(X(t)/N(t)\) a martingale for any tradable asset \(X(t)\). Here \(M(t) = \exp(\int_0^t r(s)\,ds)\) is the money market account.

Definition: T-Forward Measure

The \(T\)-forward measure \(\mathbb{Q}^T\) is the probability measure obtained by choosing \(N(t) = P(t,T)\) as numeraire. Its Radon-Nikodym derivative with respect to \(\mathbb{Q}\) is

\[ \frac{d\mathbb{Q}^T}{d\mathbb{Q}}\Bigg|_{\mathcal{F}(t)} = \frac{P(t,T)/P(0,T)}{M(t)/M(0)} = \frac{P(t,T)}{P(0,T)\,M(t)} \]

Radon-Nikodym Derivative in the Hull-White Model

In the Hull-White model, the zero-coupon bond dynamics under \(\mathbb{Q}\) are

\[ \frac{dP(t,T)}{P(t,T)} = r(t)\,dt + \sigma_P(t,T)\,dW^{\mathbb{Q}}(t) \]

where the bond volatility is

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

with \(B(t,T) = -\frac{1}{\lambda}(1 - e^{-\lambda(T-t)})\) using the sign convention from the named functions. Taking the logarithm of the Radon-Nikodym derivative:

\[\begin{array}{lllll} \displaystyle \log\frac{d\mathbb{Q}^T}{d\mathbb{Q}}\Bigg|_{\mathcal{F}(t)} &=&\displaystyle \log P(t,T) - \log P(0,T) - \int_0^t r(s)\,ds \\[6pt] &=&\displaystyle -\frac{1}{2}\int_0^t \sigma_P(s,T)^2\,ds + \int_0^t \sigma_P(s,T)\,dW^{\mathbb{Q}}(s) \end{array}\]

Proof

From the bond price dynamics under \(\mathbb{Q}\):

\[\begin{array}{lllll} \displaystyle d\log P(t,T) &=&\displaystyle \left(r(t) - \frac{1}{2}\sigma_P^2(t,T)\right)dt + \sigma_P(t,T)\,dW^{\mathbb{Q}}(t) \end{array}\]

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

\[\begin{array}{lllll} \displaystyle \log P(t,T) - \log P(0,T) &=&\displaystyle \int_0^t r(s)\,ds - \frac{1}{2}\int_0^t \sigma_P^2(s,T)\,ds + \int_0^t \sigma_P(s,T)\,dW^{\mathbb{Q}}(s) \end{array}\]

Subtracting \(\int_0^t r(s)\,ds\) yields the stated expression. \(\square\)

Girsanov Transformation

By Girsanov's theorem, the process

\[ dW^{T}(t) = dW^{\mathbb{Q}}(t) - \sigma_P(t,T)\,dt \]

is a standard Brownian motion under \(\mathbb{Q}^T\). Equivalently,

\[ dW^{\mathbb{Q}}(t) = dW^{T}(t) + \sigma_P(t,T)\,dt \]

Theorem: Girsanov Change from Q to T-Forward

Under the \(T\)-forward measure \(\mathbb{Q}^T\), the Brownian motion is

\[ dW^{T}(t) = dW^{\mathbb{Q}}(t) - \sigma_P(t,T)\,dt \]

where \(\sigma_P(t,T) = -\frac{\sigma}{\lambda}(1 - e^{-\lambda(T-t)})\) is the Hull-White bond volatility.

Pricing Formula Under the T-Forward Measure

The key advantage of the \(T\)-forward measure is that it absorbs the stochastic discount factor into the measure change. For any \(\mathcal{F}(T)\)-measurable payoff \(V(T)\):

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

Proof

Using the abstract Bayes formula for conditional expectations under a change of measure:

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

This follows from the identity \(\frac{M(t)}{M(T)} = P(t,T)\frac{d\mathbb{Q}^T}{d\mathbb{Q}}\Big|_{\mathcal{F}(T)} / \frac{d\mathbb{Q}^T}{d\mathbb{Q}}\Big|_{\mathcal{F}(t)}\) and the tower property of conditional expectation. \(\square\)

This formula is particularly powerful for options on zero-coupon bonds, where \(V(T) = \max(P(T,S) - K, 0)\) and the expectation under \(\mathbb{Q}^T\) involves only the distribution of \(r(T)\) without the complicating discount factor.

Forward Rate as a Martingale

A central property of the \(T\)-forward measure is that the instantaneous forward rate \(f(t,T)\) is a martingale.

Theorem: Forward Rate Martingale Property

Under \(\mathbb{Q}^T\), the instantaneous forward rate satisfies

\[ df(t,T) = \sigma(t,T)\,dW^{T}(t) \]

where \(\sigma(t,T) = \sigma\,e^{-\lambda(T-t)}\) is the Hull-White forward rate volatility. In particular, \(f(t,T)\) is a \(\mathbb{Q}^T\)-martingale.

Proof

Under \(\mathbb{Q}\), the HJM drift condition gives

\[ df(t,T) = \sigma(t,T)\!\left(\int_t^T \sigma(t,u)\,du\right)dt + \sigma(t,T)\,dW^{\mathbb{Q}}(t) \]

Since \(\int_t^T \sigma(t,u)\,du = -\sigma_P(t,T)\), substituting \(dW^{\mathbb{Q}}(t) = dW^{T}(t) + \sigma_P(t,T)\,dt\):

\[\begin{array}{lllll} \displaystyle df(t,T) &=&\displaystyle -\sigma(t,T)\sigma_P(t,T)\,dt + \sigma(t,T)\bigl(dW^{T}(t) + \sigma_P(t,T)\,dt\bigr) \\[4pt] &=&\displaystyle \sigma(t,T)\,dW^{T}(t) \end{array}\]

The drift terms cancel exactly, confirming the martingale property. \(\square\)

Forward Price Martingale

More generally, the forward price of any tradable asset \(X(t)\) relative to the \(T\)-maturity bond is a \(\mathbb{Q}^T\)-martingale:

\[ \mathbb{E}^{T}\!\left[\frac{X(T)}{P(T,T)}\,\Big|\,\mathcal{F}(t)\right] = \frac{X(t)}{P(t,T)} \]

Since \(P(T,T) = 1\), this simplifies to

\[ \mathbb{E}^{T}\!\left[X(T)\,\Big|\,\mathcal{F}(t)\right] = \frac{X(t)}{P(t,T)} \]

This result underpins the pricing of all European-style interest rate derivatives in the Hull-White framework.

Numerical Example

Consider pricing a European call option on a zero-coupon bond. The payoff at \(T = 5\) is \(\max(P(5,10) - K, 0)\) with \(K = 0.75\). Under the \(T\)-forward measure, the price at \(t = 0\) is

\[ V(0) = P(0,5)\,\mathbb{E}^{5}\!\left[\max(P(5,10) - 0.75,\, 0)\,\Big|\,\mathcal{F}(0)\right] \]

The advantage is clear: instead of computing \(\mathbb{E}^{\mathbb{Q}}[e^{-\int_0^5 r(s)\,ds}\max(P(5,10) - 0.75, 0)]\) -- which involves the joint distribution of the discount factor and the payoff -- we need only the distribution of \(P(5,10)\) under \(\mathbb{Q}^5\). In the Hull-White model, \(P(5,10) = e^{A(5,10) + B(5,10)r(5)}\) and \(r(5)\) is Gaussian under \(\mathbb{Q}^5\), leading directly to the Black-Scholes-type closed-form formula derived in the zero-coupon bond options section.

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Summary

The \(T\)-forward measure \(\mathbb{Q}^T\) uses the zero-coupon bond \(P(t,T)\) as numeraire and is obtained from \(\mathbb{Q}\) via Girsanov's theorem with drift adjustment \(\sigma_P(t,T)\). Under \(\mathbb{Q}^T\), the discount factor is absorbed into the measure change, simplifying derivative pricing to \(P(t,T)\,\mathbb{E}^T[V(T)]\). The instantaneous forward rate \(f(t,T)\) becomes a \(\mathbb{Q}^T\)-martingale, and the Gaussian structure of the Hull-White model is preserved under this measure change.

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Exercises

Exercise 1. State the general change-of-numeraire theorem. If the numeraire is \(N(t) = P(t,T)\) and the base measure is \(\mathbb{Q}\) with numeraire \(M(t) = e^{\int_0^t r(s)\,ds}\), write down the Radon-Nikodym derivative \(\frac{d\mathbb{Q}^T}{d\mathbb{Q}}\big|_{\mathcal{F}(t)}\) and verify that \(P(t,S)/P(t,T)\) is a \(\mathbb{Q}^T\)-martingale for any \(S\).

Solution to Exercise 1

General change-of-numeraire theorem: Let \(N(t)\) be a positive, tradable numeraire and \(M(t) = e^{\int_0^t r(s)\,ds}\) the money market account. The measure \(\mathbb{Q}^N\) defined by

\[ \frac{d\mathbb{Q}^N}{d\mathbb{Q}}\Bigg|_{\mathcal{F}(t)} = \frac{N(t)/N(0)}{M(t)/M(0)} = \frac{N(t)}{N(0)\,M(t)} \]

makes the ratio \(X(t)/N(t)\) a \(\mathbb{Q}^N\)-martingale for every tradable asset \(X(t)\).

Radon-Nikodym derivative for the T-forward measure: Setting \(N(t) = P(t,T)\):

\[ \frac{d\mathbb{Q}^T}{d\mathbb{Q}}\Bigg|_{\mathcal{F}(t)} = \frac{P(t,T)}{P(0,T)\,M(t)} \]

Verification that \(P(t,S)/P(t,T)\) is a \(\mathbb{Q}^T\)-martingale: By the change-of-numeraire theorem, for any tradable asset \(X(t)\), the ratio \(X(t)/P(t,T)\) is a \(\mathbb{Q}^T\)-martingale. Taking \(X(t) = P(t,S)\) (which is a tradable asset — it is the price of the \(S\)-maturity zero-coupon bond):

\[ \mathbb{E}^T\!\left[\frac{P(u,S)}{P(u,T)}\,\Big|\,\mathcal{F}(t)\right] = \frac{P(t,S)}{P(t,T)} \qquad \text{for all } t \leq u \]

This confirms that \(P(t,S)/P(t,T)\) is a \(\mathbb{Q}^T\)-martingale. In particular, at \(u = T\) where \(P(T,T) = 1\):

\[ \mathbb{E}^T\!\left[P(T,S)\,\Big|\,\mathcal{F}(t)\right] = \frac{P(t,S)}{P(t,T)} \]

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Exercise 2. Verify the Girsanov transformation \(dW^T(t) = dW^{\mathbb{Q}}(t) - \sigma_P(t,T)\,dt\) by substituting \(\sigma_P(t,T) = -\frac{\sigma}{\lambda}(1 - e^{-\lambda(T-t)})\) and computing the drift adjustment explicitly for \(\lambda = 0.05\) and \(\sigma = 0.01\) at \(t = 3\) and \(T = 10\).

Solution to Exercise 2

The Girsanov transformation is \(dW^T(t) = dW^{\mathbb{Q}}(t) - \sigma_P(t,T)\,dt\) where

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

For \(\lambda = 0.05\), \(\sigma = 0.01\), \(t = 3\), \(T = 10\):

\[ \sigma_P(3, 10) = -\frac{0.01}{0.05}\left(1 - e^{-0.05 \times 7}\right) = -0.2\left(1 - e^{-0.35}\right) \]

Computing \(e^{-0.35} \approx 0.7047\):

\[ \sigma_P(3, 10) = -0.2 \times (1 - 0.7047) = -0.2 \times 0.2953 = -0.05906 \]

The drift adjustment in the Girsanov transformation is therefore \(-\sigma_P(3, 10) = 0.05906\). This means

\[ dW^T(t)\big|_{t=3} = dW^{\mathbb{Q}}(t)\big|_{t=3} + 0.05906\,dt \]

or equivalently \(dW^{\mathbb{Q}}(t)\big|_{t=3} = dW^T(t)\big|_{t=3} - 0.05906\,dt\). The drift adjustment is positive (subtracting a negative quantity), reflecting the fact that under the \(T\)-forward measure, the Brownian motion has an upward drift relative to the \(\mathbb{Q}\)-Brownian motion. Since the bond volatility \(\sigma_P\) is negative (bond prices move inversely to rates), this adjustment shifts probability toward lower interest rate paths, consistent with using the bond as numeraire.

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Exercise 3. Show that the forward rate \(f(t,T)\) has zero drift under \(\mathbb{Q}^T\) by starting from the HJM drift condition under \(\mathbb{Q}\) and applying the Girsanov transformation. Explain why the exact cancellation of drift terms is a consequence of the HJM no-arbitrage condition.

Solution to Exercise 3

Under \(\mathbb{Q}\), the HJM drift condition gives the forward rate dynamics:

\[ df(t,T) = \sigma(t,T)\!\left(\int_t^T \sigma(t,u)\,du\right)dt + \sigma(t,T)\,dW^{\mathbb{Q}}(t) \]

The key identity is \(\int_t^T \sigma(t,u)\,du = -\sigma_P(t,T)\), which follows from \(\sigma_P(t,T) = -\int_t^T \sigma(t,u)\,du\) (the bond volatility is minus the integral of forward rate volatilities). Therefore:

\[ df(t,T) = -\sigma(t,T)\,\sigma_P(t,T)\,dt + \sigma(t,T)\,dW^{\mathbb{Q}}(t) \]

Now apply the Girsanov transformation \(dW^{\mathbb{Q}}(t) = dW^T(t) + \sigma_P(t,T)\,dt\):

\[\begin{array}{lllll} \displaystyle df(t,T) &=&\displaystyle -\sigma(t,T)\,\sigma_P(t,T)\,dt + \sigma(t,T)\!\left(dW^T(t) + \sigma_P(t,T)\,dt\right) \\[6pt] &=&\displaystyle -\sigma(t,T)\,\sigma_P(t,T)\,dt + \sigma(t,T)\,\sigma_P(t,T)\,dt + \sigma(t,T)\,dW^T(t) \\[6pt] &=&\displaystyle \sigma(t,T)\,dW^T(t) \end{array}\]

The drift terms cancel exactly, confirming that \(f(t,T)\) is driftless under \(\mathbb{Q}^T\).

Why the cancellation occurs: The HJM no-arbitrage condition under \(\mathbb{Q}\) is precisely the statement that the drift of \(f(t,T)\) equals \(\sigma(t,T)\int_t^T \sigma(t,u)\,du\). This drift is exactly what is needed to cancel against the Girsanov adjustment \(\sigma(t,T)\sigma_P(t,T)\) when changing to \(\mathbb{Q}^T\). In other words, the HJM drift condition was designed (via the no-arbitrage requirement) so that the forward rate becomes a martingale under the measure that uses \(P(t,T)\) as numeraire. This is not a coincidence but a fundamental consequence of the absence of arbitrage.

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Exercise 4. Prove that the forward price \(X(t)/P(t,T)\) is a \(\mathbb{Q}^T\)-martingale using the abstract Bayes formula. Explain why this martingale property makes the \(T\)-forward measure the natural choice for pricing European derivatives with payoff at time \(T\).

Solution to Exercise 4

Proof using the abstract Bayes formula: Let \(X(t)\) be a tradable asset. Under \(\mathbb{Q}\), the discounted price \(X(t)/M(t)\) is a \(\mathbb{Q}\)-martingale:

\[ \frac{X(t)}{M(t)} = \mathbb{E}^{\mathbb{Q}}\!\left[\frac{X(T)}{M(T)}\,\Big|\,\mathcal{F}(t)\right] \]

The abstract Bayes formula for conditional expectations states that for any random variable \(Y\) and Radon-Nikodym derivative \(L_T = \frac{d\mathbb{Q}^T}{d\mathbb{Q}}\big|_{\mathcal{F}(T)}\):

\[ \mathbb{E}^T\!\left[Y\,|\,\mathcal{F}(t)\right] = \frac{\mathbb{E}^{\mathbb{Q}}\!\left[L_T\,Y\,|\,\mathcal{F}(t)\right]}{L_t} \]

Setting \(Y = X(T)/P(T,T) = X(T)\) and using \(L_t = \frac{P(t,T)}{P(0,T)M(t)}\):

\[\begin{array}{lllll} \displaystyle \mathbb{E}^T\!\left[X(T)\,|\,\mathcal{F}(t)\right] &=&\displaystyle \frac{\mathbb{E}^{\mathbb{Q}}\!\left[\frac{P(T,T)}{P(0,T)M(T)} X(T)\,\Big|\,\mathcal{F}(t)\right]}{\frac{P(t,T)}{P(0,T)M(t)}} \\[10pt] &=&\displaystyle \frac{M(t)}{P(t,T)}\,\mathbb{E}^{\mathbb{Q}}\!\left[\frac{X(T)}{M(T)}\,\Big|\,\mathcal{F}(t)\right] \\[10pt] &=&\displaystyle \frac{M(t)}{P(t,T)} \cdot \frac{X(t)}{M(t)} = \frac{X(t)}{P(t,T)} \end{array}\]

This confirms \(X(t)/P(t,T)\) is a \(\mathbb{Q}^T\)-martingale.

Why this is the natural choice for European derivatives: A European derivative pays \(V(T)\) at time \(T\). Its price is \(V(t) = P(t,T)\,\mathbb{E}^T[V(T)\,|\,\mathcal{F}(t)]\). Under \(\mathbb{Q}^T\), the stochastic discount factor \(e^{-\int_t^T r(s)\,ds}\) is absorbed into the measure change, so one only needs the distribution of \(V(T)\) under \(\mathbb{Q}^T\) — not the joint distribution of the payoff and the discount factor. This separation dramatically simplifies computations, especially when the payoff depends on the same interest rate driving the discount factor.

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Exercise 5. In the numerical example, the option price is \(V(0) = P(0,5)\,\mathbb{E}^5[\max(P(5,10) - 0.75, 0)]\). Explain why computing this expectation under \(\mathbb{Q}^5\) is simpler than computing \(\mathbb{E}^{\mathbb{Q}}[e^{-\int_0^5 r(s)\,ds}\max(P(5,10) - 0.75, 0)]\). What correlation structure makes the \(\mathbb{Q}\)-expectation more difficult?

Solution to Exercise 5

Under \(\mathbb{Q}\), the expectation \(\mathbb{E}^{\mathbb{Q}}[e^{-\int_0^5 r(s)\,ds}\max(P(5,10) - 0.75, 0)]\) involves two correlated stochastic quantities:

1. The stochastic discount factor \(e^{-\int_0^5 r(s)\,ds}\), which depends on the entire path of \(r(s)\) for \(s \in [0, 5]\).
2. The bond price \(P(5,10) = e^{A(5,10) + B(5,10)r(5)}\), which depends on \(r(5)\).

These are correlated because \(r(5)\) is determined by the path \(\{r(s)\}_{s \in [0,5]}\), and the discount factor is a functional of the same path. The payoff is large when \(P(5,10)\) is high, i.e., when \(r(5)\) is low. But when rates are low, the discount factor \(e^{-\int_0^5 r(s)\,ds}\) is close to 1 (less discounting). So the discount factor and the payoff are positively correlated, making the \(\mathbb{Q}\)-expectation harder to evaluate because one must account for this dependence.

Under \(\mathbb{Q}^5\), the expectation \(\mathbb{E}^5[\max(P(5,10) - 0.75, 0)]\) involves only the distribution of \(r(5)\) under \(\mathbb{Q}^5\), which is Gaussian. Since \(P(5,10) = e^{A(5,10) + B(5,10)r(5)}\) is log-linear in the Gaussian variable \(r(5)\), the payoff is the maximum of a lognormal variable minus a constant. This is a standard Black-Scholes-type integral with a known closed-form solution. The correlation structure is completely absorbed into the measure change, eliminating the need to handle the joint distribution.

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Exercise 6. Consider two different \(T\)-forward measures, \(\mathbb{Q}^{T_1}\) and \(\mathbb{Q}^{T_2}\) with \(T_1 < T_2\). Derive the Radon-Nikodym derivative \(\frac{d\mathbb{Q}^{T_2}}{d\mathbb{Q}^{T_1}}\big|_{\mathcal{F}(t)}\) and the corresponding Girsanov transformation between the two measures.

Solution to Exercise 6

Radon-Nikodym derivative between two forward measures: Starting from their respective definitions relative to \(\mathbb{Q}\):

\[ \frac{d\mathbb{Q}^{T_i}}{d\mathbb{Q}}\Bigg|_{\mathcal{F}(t)} = \frac{P(t,T_i)}{P(0,T_i)\,M(t)}, \qquad i = 1, 2 \]

The Radon-Nikodym derivative from \(\mathbb{Q}^{T_1}\) to \(\mathbb{Q}^{T_2}\) is:

\[ \frac{d\mathbb{Q}^{T_2}}{d\mathbb{Q}^{T_1}}\Bigg|_{\mathcal{F}(t)} = \frac{d\mathbb{Q}^{T_2}/d\mathbb{Q}\big|_{\mathcal{F}(t)}}{d\mathbb{Q}^{T_1}/d\mathbb{Q}\big|_{\mathcal{F}(t)}} = \frac{P(t,T_2)/P(0,T_2)}{P(t,T_1)/P(0,T_1)} = \frac{P(t,T_2)\,P(0,T_1)}{P(t,T_1)\,P(0,T_2)} \]

Girsanov transformation: Under \(\mathbb{Q}^{T_1}\), the Brownian motion is \(dW^{T_1}(t) = dW^{\mathbb{Q}}(t) - \sigma_P(t,T_1)\,dt\). Under \(\mathbb{Q}^{T_2}\), it is \(dW^{T_2}(t) = dW^{\mathbb{Q}}(t) - \sigma_P(t,T_2)\,dt\). Therefore the relationship between the two forward measure Brownian motions is:

\[ dW^{T_2}(t) = dW^{T_1}(t) - \left(\sigma_P(t,T_2) - \sigma_P(t,T_1)\right)dt \]

In the Hull-White model, the drift adjustment is:

\[ \sigma_P(t,T_2) - \sigma_P(t,T_1) = -\frac{\sigma}{\lambda}\left(e^{-\lambda(T_1-t)} - e^{-\lambda(T_2-t)}\right) \]

This is the Girsanov kernel for moving between the two forward measures, and it is deterministic, preserving the Gaussian framework.

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Exercise 7. The forward LIBOR rate \(L(t, T, T+\delta)\) defined by \(1 + \delta L(t,T,T+\delta) = P(t,T)/P(t,T+\delta)\) is a martingale under \(\mathbb{Q}^{T+\delta}\). Verify this claim for the Hull-White model and explain how this connects to the pricing of caplets.

Solution to Exercise 7

The forward LIBOR rate is defined by \(1 + \delta L(t,T,T+\delta) = P(t,T)/P(t,T+\delta)\), so

\[ L(t,T,T+\delta) = \frac{1}{\delta}\!\left(\frac{P(t,T)}{P(t,T+\delta)} - 1\right) \]

Martingale property: By the change-of-numeraire theorem, using \(P(t,T+\delta)\) as numeraire, the ratio \(P(t,T)/P(t,T+\delta)\) is a \(\mathbb{Q}^{T+\delta}\)-martingale. Since \(L(t,T,T+\delta) = \frac{1}{\delta}(P(t,T)/P(t,T+\delta) - 1)\) is an affine transformation of a martingale (with constants \(1/\delta\) and \(-1/\delta\)), it is also a \(\mathbb{Q}^{T+\delta}\)-martingale:

\[ \mathbb{E}^{T+\delta}\!\left[L(u,T,T+\delta)\,\Big|\,\mathcal{F}(t)\right] = L(t,T,T+\delta) \qquad \text{for all } t \leq u \leq T \]

Verification in the Hull-White model: In the Hull-White model, \(P(t,T)/P(t,T+\delta)\) can be written as \(\exp(A^* + B^* r(t))\) for appropriate deterministic functions \(A^*\) and \(B^*\). Under \(\mathbb{Q}^{T+\delta}\), \(r(t)\) is Gaussian with a drift adjusted by \(\sigma_P(t, T+\delta)\). The ratio \(P(t,T)/P(t,T+\delta)\) is the discounted value of \(P(T,T)/P(T,T+\delta) = 1/P(T,T+\delta)\) using \(P(t,T+\delta)\) as numeraire, so it is indeed a martingale.

Connection to caplet pricing: A caplet pays \(\delta\max(L(T,T,T+\delta) - K, 0)\) at time \(T+\delta\). Its price at time \(t\) is:

\[ \text{Caplet}(t) = \delta\,P(t,T+\delta)\,\mathbb{E}^{T+\delta}\!\left[\max(L(T,T,T+\delta) - K, 0)\,\Big|\,\mathcal{F}(t)\right] \]

Since \(L(t,T,T+\delta)\) is a \(\mathbb{Q}^{T+\delta}\)-martingale and (in the Hull-White model) is a function of the Gaussian variable \(r(T)\), this expectation can be evaluated in closed form, yielding a Black-type formula for the caplet price. The martingale property ensures that the forward LIBOR rate is the "correct" underlying for the option pricing formula, with \(L(t,T,T+\delta)\) serving as the forward price of the caplet payoff.