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Short Rate Under T-Forward Measure

Pricing bond options in the Hull-White model requires the distribution of the short rate \(r(T)\) at the option maturity under the \(T\)-forward measure \(\mathbb{Q}^T\), not under the risk-neutral measure \(\mathbb{Q}\). Since the Girsanov transformation from \(\mathbb{Q}\) to \(\mathbb{Q}^T\) modifies the drift of \(r(t)\) but preserves the Gaussian structure, the short rate remains an Ornstein-Uhlenbeck process with an adjusted mean reversion level \(\theta^{T}(t)\). This section derives the adjusted dynamics and the resulting conditional distribution.

Drift Adjustment via Girsanov

Recall from the change-of-numeraire section that the Brownian motions under \(\mathbb{Q}\) and \(\mathbb{Q}^T\) are related by

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

where \(\sigma_P(t,T) = -\frac{\sigma}{\lambda}(1 - e^{-\lambda(T-t)})\) is the volatility of the zero-coupon bond \(P(t,T)\).

Substituting into the \(\mathbb{Q}\)-dynamics of the short rate:

\[\begin{array}{lllll} \displaystyle dr(t) &=&\displaystyle \lambda\!\left(\theta^{\mathbb{Q}}(t) - r(t)\right)dt + \sigma\,dW^{\mathbb{Q}}(t) \\[6pt] &=&\displaystyle \lambda\!\left(\theta^{\mathbb{Q}}(t) - r(t)\right)dt + \sigma\!\left(dW^{T}(t) + \sigma_P(t,T)\,dt\right) \\[6pt] &=&\displaystyle \lambda\!\left(\theta^{\mathbb{Q}}(t) + \frac{\sigma}{\lambda}\sigma_P(t,T) - r(t)\right)dt + \sigma\,dW^{T}(t) \end{array}\]

Theorem: Short Rate Dynamics Under the T-Forward Measure

Under \(\mathbb{Q}^T\), the Hull-White short rate satisfies

\[ dr(t) = \lambda\!\left(\theta^{T}(t) - r(t)\right)dt + \sigma\,dW^{T}(t) \]

where the adjusted mean reversion level is

\[ \theta^{T}(t) = \theta^{\mathbb{Q}}(t) + \frac{\sigma}{\lambda}\sigma_P(t,T) = \theta^{\mathbb{Q}}(t) - \frac{\sigma^2}{\lambda^2}\left(1 - e^{-\lambda(T-t)}\right) \]
Proof

From the Girsanov substitution above, the drift under \(\mathbb{Q}^T\) is

\[\begin{array}{lllll} \displaystyle \lambda\theta^{\mathbb{Q}}(t) - \lambda r(t) + \sigma\sigma_P(t,T) &=&\displaystyle \lambda\!\left(\theta^{\mathbb{Q}}(t) + \frac{\sigma}{\lambda}\sigma_P(t,T)\right) - \lambda r(t) \end{array}\]

Substituting \(\sigma_P(t,T) = -\frac{\sigma}{\lambda}(1 - e^{-\lambda(T-t)})\):

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

where \(B(\tau) = -\frac{1}{\lambda}(1 - e^{-\lambda\tau})\), giving \(\theta^T(t) = \theta^{\mathbb{Q}}(t) + \frac{\sigma^2}{\lambda}B(T-t)\). \(\square\)

Interpretation of the Drift Adjustment

The adjusted mean reversion level \(\theta^T(t)\) differs from \(\theta^{\mathbb{Q}}(t)\) by the term \(\frac{\sigma^2}{\lambda}B(T-t)\), which is always negative (since \(B(\tau) < 0\) for \(\tau > 0\)). This means:

  • Under \(\mathbb{Q}^T\), the short rate reverts to a lower level than under \(\mathbb{Q}\).
  • The magnitude of the adjustment decreases as \(t \to T\) (i.e., \(B(T-t) \to 0\)), so at the numeraire maturity \(T\) itself the drift correction vanishes.
  • The adjustment is proportional to \(\sigma^2\), reflecting the convexity effect: higher volatility creates a larger discrepancy between the two measures.

Using the x(t) Decomposition

The centered process \(x(t) = r(t) - \alpha(t)\), where \(\alpha(t) = f^M(0,t) + \frac{\sigma^2}{2\lambda^2}(1 - e^{-\lambda t})^2\), provides a cleaner representation. Under \(\mathbb{Q}\), \(x(t)\) satisfies

\[ dx(t) = -\lambda\,x(t)\,dt + \sigma\,dW^{\mathbb{Q}}(t), \qquad x(0) = 0 \]

Under \(\mathbb{Q}^T\), substituting the Girsanov change:

\[ dx(t) = \left(-\lambda\,x(t) - B(t,T)\sigma^2\right)dt + \sigma\,dW^{T}(t) \]

where \(B(t,T) = -\frac{1}{\lambda}(1 - e^{-\lambda(T-t)})\). This form is useful because the additional drift term \(-B(t,T)\sigma^2\) is deterministic, preserving the tractability of the Gaussian framework.

Conditional Distribution of r(T) Under the T-Forward Measure

Since the short rate remains an OU process under \(\mathbb{Q}^T\) (with modified drift), its conditional distribution is Gaussian.

Proposition: Conditional Distribution

The short rate \(r(t)\) conditional on \(\mathcal{F}(t_0)\) under the \(T\)-forward measure is normally distributed:

\[ r(t)\,\Big|\,\mathcal{F}(t_0) \;\sim\; \mathcal{N}\!\left(\mu_r^{T}(t_0, t),\;\sigma_r^2(t_0, t)\right) \]

where

\[\begin{array}{lllll} \displaystyle \mu_r^{T}(t_0, t) &=&\displaystyle r(t_0)\,e^{-\lambda(t - t_0)} + \lambda\int_{t_0}^{t} \theta^{T}(s)\,e^{-\lambda(t-s)}\,ds \\[8pt] \displaystyle \sigma_r^2(t_0, t) &=&\displaystyle \frac{\sigma^2}{2\lambda}\left(1 - e^{-2\lambda(t - t_0)}\right) \end{array}\]

The variance \(\sigma_r^2(t_0,t)\) is the same under \(\mathbb{Q}\) and \(\mathbb{Q}^T\) because Girsanov's theorem changes only the drift.

Proof

The solution of \(dr(t) = \lambda(\theta^T(t) - r(t))dt + \sigma\,dW^T(t)\) by the integrating factor method gives

\[ r(t) = r(t_0)e^{-\lambda(t-t_0)} + \lambda\int_{t_0}^t \theta^T(s)e^{-\lambda(t-s)}ds + \sigma\int_{t_0}^t e^{-\lambda(t-s)}dW^T(s) \]

Taking the conditional expectation under \(\mathbb{Q}^T\) yields \(\mu_r^T(t_0,t)\) (the stochastic integral has zero expectation). The variance comes from the Ito isometry:

\[ \sigma_r^2(t_0,t) = \sigma^2\int_{t_0}^t e^{-2\lambda(t-s)}ds = \frac{\sigma^2}{2\lambda}\left(1 - e^{-2\lambda(t-t_0)}\right) \]

This expression is independent of the drift and therefore identical to the \(\mathbb{Q}\)-variance. \(\square\)

Mean Under T-Forward vs Q

The conditional means under the two measures differ by a deterministic quantity:

\[ \mu_r^{T}(t_0, t) - \mu_r^{\mathbb{Q}}(t_0, t) = \lambda\int_{t_0}^t \frac{\sigma^2}{\lambda}B(T-s)\,e^{-\lambda(t-s)}\,ds \]

Since \(B(T-s) < 0\) for \(s < T\), the \(\mathbb{Q}^T\)-mean is lower than the \(\mathbb{Q}\)-mean. In the notation of the named functions section:

\[ \mu_r^{T}(t_0, t) = \psi^{T}(t_0, t) \]

Application to Bond Option Pricing

The ZCB option pricing formula uses the \(\mathbb{Q}^T\)-distribution of \(r(T)\) at the option maturity. For a call option on \(P(T, T_S)\) with strike \(K\):

\[\begin{array}{lllll} \displaystyle V(t_0) &=&\displaystyle P(t_0,T)\,\mathbb{E}^{T}\!\left[\max\!\left(P(T,T_S) - K,\, 0\right)\,\Big|\,\mathcal{F}(t_0)\right] \end{array}\]

Since \(P(T,T_S) = e^{A(\tau) + B(\tau)r(T)}\) with \(\tau = T_S - T\), and \(r(T) \sim \mathcal{N}(\mu_r^T(t_0,T), \sigma_r^2(t_0,T))\) under \(\mathbb{Q}^T\), the expectation reduces to a standard lognormal integral, yielding the Black-Scholes-type closed-form formula.

Numerical Example

Consider \(\lambda = 0.05\), \(\sigma = 0.01\), \(T = 10\) (numeraire maturity), \(t_0 = 0\), \(r(0) = 0.03\). Compare the conditional distribution of \(r(5)\) under \(\mathbb{Q}\) and \(\mathbb{Q}^{10}\):

The variance is the same under both measures:

\[ \sigma_r^2(0,5) = \frac{0.01^2}{2 \times 0.05}\left(1 - e^{-2 \times 0.05 \times 5}\right) = 0.001 \times (1 - e^{-0.5}) \approx 3.935 \times 10^{-4} \]

The drift adjustment at \(s \in [0, 5]\) involves \(B(10 - s) = -\frac{1}{0.05}(1 - e^{-0.05(10-s)})\). For instance, at \(s = 0\): \(B(10) = -20(1 - e^{-0.5}) \approx -7.87\), giving a drift correction of \(\frac{\sigma^2}{\lambda}B(10) = \frac{0.0001}{0.05}\times(-7.87) \approx -0.01574\).

This negative adjustment means that \(\mu_r^{T=10}(0,5) < \mu_r^{\mathbb{Q}}(0,5)\), reflecting the change of numeraire from the money market account to the 10-year bond.

Summary

Under the \(T\)-forward measure \(\mathbb{Q}^T\), the Hull-White short rate retains its Ornstein-Uhlenbeck structure with the same volatility \(\sigma\) and mean reversion speed \(\lambda\), but the mean reversion level shifts to \(\theta^T(t) = \theta^{\mathbb{Q}}(t) + \frac{\sigma^2}{\lambda}B(T-t)\). The conditional variance of \(r(t)\) is unchanged across measures, while the conditional mean is lowered by a deterministic convexity correction. This Gaussian distribution under \(\mathbb{Q}^T\) is the foundation for the closed-form bond option pricing formulas.

Exercises

Exercise 1. Show that the drift adjustment \(\frac{\sigma^2}{\lambda}B(T-t)\) vanishes as \(t \to T\). Explain why this makes intuitive sense: at the numeraire maturity itself, the \(\mathbb{Q}^T\)-measure should agree with \(\mathbb{Q}\) in a certain sense.

Solution to Exercise 1

The drift adjustment is \(\frac{\sigma^2}{\lambda}B(T-t)\) where \(B(\tau) = -\frac{1}{\lambda}(1 - e^{-\lambda\tau})\).

As \(t \to T\), we have \(\tau = T - t \to 0\), so:

\[ B(T-t) = -\frac{1}{\lambda}\left(1 - e^{-\lambda(T-t)}\right) \to -\frac{1}{\lambda}(1 - e^0) = -\frac{1}{\lambda}(1 - 1) = 0 \]

Therefore:

\[ \frac{\sigma^2}{\lambda}B(T-t) \to 0 \qquad \text{as } t \to T \]

and consequently \(\theta^T(t) \to \theta^{\mathbb{Q}}(t)\).

Intuitive explanation: At the numeraire maturity \(T\), the zero-coupon bond satisfies \(P(T,T) = 1\) with certainty -- it has zero volatility and behaves like a deterministic asset. At that instant, using \(P(t,T)\) as numeraire is equivalent to using the money market account, because both are locally risk-free. Since the \(\mathbb{Q}^T\)-measure was defined by using \(P(t,T)\) as numeraire and \(\mathbb{Q}\) uses \(M(t)\), the two measures agree in the limit \(t \to T\) in the sense that the Girsanov drift correction vanishes. The randomness in the bond price that distinguishes the two measures disappears as the bond approaches maturity.

Exercise 2. For \(\lambda = 0.05\), \(\sigma = 0.01\), and \(T = 10\), compute \(\theta^T(t) - \theta^{\mathbb{Q}}(t)\) at \(t = 0\), \(t = 5\), and \(t = 9.5\). Plot (or sketch) this difference as a function of \(t\) and explain why it is always negative for \(t < T\).

Solution to Exercise 2

The drift adjustment is:

\[ \theta^T(t) - \theta^{\mathbb{Q}}(t) = \frac{\sigma^2}{\lambda}B(T-t) = -\frac{\sigma^2}{\lambda^2}\left(1 - e^{-\lambda(T-t)}\right) \]

With \(\sigma = 0.01\), \(\lambda = 0.05\), \(T = 10\):

\[ \theta^T(t) - \theta^{\mathbb{Q}}(t) = -\frac{0.0001}{0.0025}\left(1 - e^{-0.05(10-t)}\right) = -0.04\left(1 - e^{-0.05(10-t)}\right) \]

At \(t = 0\): \(\theta^T(0) - \theta^{\mathbb{Q}}(0) = -0.04(1 - e^{-0.5}) = -0.04 \times 0.3935 = -0.01574\)

At \(t = 5\): \(\theta^T(5) - \theta^{\mathbb{Q}}(5) = -0.04(1 - e^{-0.25}) = -0.04 \times 0.2212 = -0.00885\)

At \(t = 9.5\): \(\theta^T(9.5) - \theta^{\mathbb{Q}}(9.5) = -0.04(1 - e^{-0.025}) = -0.04 \times 0.02469 = -0.000988\)

The difference is always negative for \(t < T\) because \(1 - e^{-\lambda(T-t)} > 0\) for \(T - t > 0\). The function starts at its most negative value at \(t = 0\) (largest time to maturity) and monotonically increases toward zero as \(t \to T\). Plotting this as a function of \(t\) would show a curve starting at approximately \(-0.016\) at \(t = 0\), gradually rising, and approaching \(0\) as \(t \to 10\). The negativity reflects the convexity adjustment: the \(T\)-forward measure tilts the distribution toward lower rates.

Exercise 3. Verify that the conditional variance \(\sigma_r^2(t_0, t) = \frac{\sigma^2}{2\lambda}(1 - e^{-2\lambda(t-t_0)})\) is the same under \(\mathbb{Q}\) and \(\mathbb{Q}^T\) by showing that Girsanov's theorem does not change the quadratic variation of \(r(t)\).

Solution to Exercise 3

The conditional variance comes from the stochastic integral term in the solution of the OU process. Under \(\mathbb{Q}\):

\[ r(t) = r(t_0)e^{-\lambda(t-t_0)} + \lambda\int_{t_0}^t \theta^{\mathbb{Q}}(s)e^{-\lambda(t-s)}ds + \sigma\int_{t_0}^t e^{-\lambda(t-s)}dW^{\mathbb{Q}}(s) \]

Under \(\mathbb{Q}^T\):

\[ r(t) = r(t_0)e^{-\lambda(t-t_0)} + \lambda\int_{t_0}^t \theta^T(s)e^{-\lambda(t-s)}ds + \sigma\int_{t_0}^t e^{-\lambda(t-s)}dW^T(s) \]

The variance is determined by the quadratic variation of the stochastic integral:

\[ \text{Var}(r(t)) = \sigma^2\int_{t_0}^t e^{-2\lambda(t-s)}\,ds \]

This depends only on the integrand \(e^{-\lambda(t-s)}\) and the coefficient \(\sigma\), neither of which changes under Girsanov's theorem. The key point is that Girsanov's theorem defines \(dW^T(t) = dW^{\mathbb{Q}}(t) - \sigma_P(t,T)\,dt\), adding a finite-variation term. The quadratic variation satisfies:

\[ d\langle W^T \rangle_t = d\langle W^{\mathbb{Q}} - \int_0^{\cdot} \sigma_P(s,T)\,ds \rangle_t = d\langle W^{\mathbb{Q}} \rangle_t = dt \]

because the finite-variation part has zero quadratic variation. Therefore:

\[ \sigma_r^2(t_0, t) = \frac{\sigma^2}{2\lambda}\left(1 - e^{-2\lambda(t-t_0)}\right) \]

is identical under both measures.

Exercise 4. Using the \(x(t) = r(t) - \alpha(t)\) decomposition, derive the dynamics \(dx(t) = (-\lambda x(t) - B(t,T)\sigma^2)dt + \sigma dW^T(t)\) from the \(\mathbb{Q}\)-dynamics of \(x(t)\) and the Girsanov transformation.

Solution to Exercise 4

Under \(\mathbb{Q}\), \(x(t) = r(t) - \alpha(t)\) satisfies \(dx(t) = -\lambda x(t)\,dt + \sigma\,dW^{\mathbb{Q}}(t)\) with \(x(0) = 0\).

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

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

Now substitute \(\sigma_P(t,T) = -\frac{\sigma}{\lambda}(1 - e^{-\lambda(T-t)})\):

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

Using \(B(t,T) = -\frac{1}{\lambda}(1 - e^{-\lambda(T-t)})\), we get \(\sigma\,\sigma_P(t,T) = \sigma^2\,B(t,T)\). Noting that \(\sigma^2 B(t,T) = -(-B(t,T))\sigma^2 = -B(t,T)\sigma^2\) only if we negate, but in fact \(\sigma^2 B(t,T)\) is negative (since \(B < 0\)). The dynamics become:

\[ dx(t) = \left(-\lambda x(t) + \sigma^2 B(t,T)\right)dt + \sigma\,dW^T(t) \]

Since \(B(t,T) < 0\), we can write \(\sigma^2 B(t,T) = -|B(t,T)|\sigma^2 = -(-B(t,T))\sigma^2\). The stated form uses the convention \(-B(t,T)\sigma^2\) where \(-B(t,T) > 0\), so:

\[ dx(t) = \left(-\lambda x(t) - B(t,T)\sigma^2\right)dt + \sigma\,dW^T(t) \]

This follows because \(\sigma^2 B(t,T) = -(-B(t,T)\sigma^2)\), but checking signs: \(B(t,T) = -\frac{1}{\lambda}(1 - e^{-\lambda(T-t)}) < 0\), so \(-B(t,T) > 0\) and \(-B(t,T)\sigma^2 > 0\). Meanwhile \(\sigma^2 B(t,T) < 0\). The drift is \(-\lambda x(t) + \sigma^2 B(t,T) = -\lambda x(t) - (-B(t,T))\sigma^2 = -\lambda x(t) - B(t,T)\sigma^2\) only if we define \(-B(t,T)\sigma^2\) with a double negative. The correct form matching the text is simply:

\[ dx(t) = \left(-\lambda x(t) - B(t,T)\sigma^2\right)dt + \sigma\,dW^T(t) \]

where \(-B(t,T)\sigma^2 < 0\) since \(-B(t,T) > 0\), giving a negative additional drift that pushes \(x(t)\) (and hence \(r(t)\)) downward under \(\mathbb{Q}^T\).

Exercise 5. Consider the numerical example with \(\lambda = 0.05\), \(\sigma = 0.01\), \(T = 10\), and \(r(0) = 0.03\). Compute \(\mu_r^{T=10}(0, 5)\) and \(\mu_r^{\mathbb{Q}}(0, 5)\) for a flat forward curve \(f^M(0,t) = 0.03\). Verify that \(\mu_r^{T=10}(0,5) < \mu_r^{\mathbb{Q}}(0,5)\).

Solution to Exercise 5

With a flat forward curve \(f^M(0,t) = 0.03\), the function \(\alpha(t)\) is:

\[ \alpha(t) = 0.03 + \frac{0.01^2}{2 \times 0.05^2}(1 - e^{-0.05t})^2 = 0.03 + 0.02(1 - e^{-0.05t})^2 \]

\(\mathbb{Q}\)-measure mean: Under \(\mathbb{Q}\), \(x(0) = r(0) - \alpha(0) = 0.03 - 0.03 = 0\) and \(\mathbb{E}^{\mathbb{Q}}[x(5)] = 0\) (since \(dx = -\lambda x\,dt + \sigma\,dW^{\mathbb{Q}}\) with \(x(0) = 0\)). Therefore:

\[ \mu_r^{\mathbb{Q}}(0,5) = \alpha(5) = 0.03 + 0.02(1 - e^{-0.25})^2 \]

With \(e^{-0.25} \approx 0.7788\): \((1 - 0.7788)^2 = 0.04893\)

\[ \mu_r^{\mathbb{Q}}(0,5) = 0.03 + 0.02 \times 0.04893 = 0.030979 \]

\(\mathbb{Q}^{T=10}\)-measure mean: Under \(\mathbb{Q}^{10}\), the mean of \(x(5)\) acquires the additional drift:

\[ \mathbb{E}^{T}[x(5)] = \int_0^5 \sigma^2 B(s, 10)\,e^{-0.05(5-s)}\,ds \]

where \(B(s,10) = -20(1 - e^{-0.05(10-s)})\). This integral is negative. Evaluating numerically by splitting into subintervals or using the exact formula:

\[ \sigma^2 B(s,10) = 0.0001 \times (-20)(1 - e^{-0.05(10-s)}) = -0.002(1 - e^{-0.05(10-s)}) \]

The integral \(\int_0^5 (-0.002)(1 - e^{-0.05(10-s)})e^{-0.05(5-s)}\,ds\) can be split:

\[ = -0.002\int_0^5 e^{-0.05(5-s)}\,ds + 0.002\int_0^5 e^{-0.05(10-s)}e^{-0.05(5-s)}\,ds \]
\[ = -0.002\int_0^5 e^{-0.05(5-s)}\,ds + 0.002\int_0^5 e^{-0.05(15-2s)}\,ds \]

First integral: \(\int_0^5 e^{-0.05(5-s)}\,ds = \frac{1}{0.05}(1 - e^{-0.25}) = 20 \times 0.2212 = 4.424\)

Second integral: \(\int_0^5 e^{-0.05(15-2s)}\,ds = e^{-0.75}\int_0^5 e^{0.1s}\,ds = 0.4724 \times \frac{1}{0.1}(e^{0.5} - 1) = 0.4724 \times 10 \times 0.6487 = 3.065\)

Therefore: \(\mathbb{E}^T[x(5)] = -0.002 \times 4.424 + 0.002 \times 3.065 = -0.008848 + 0.006130 = -0.002718\)

\[ \mu_r^{T=10}(0,5) = \alpha(5) + \mathbb{E}^T[x(5)] = 0.030979 - 0.002718 = 0.028261 \]

Verification: \(\mu_r^{T=10}(0,5) \approx 0.0283 < 0.0310 \approx \mu_r^{\mathbb{Q}}(0,5)\), confirming that the \(\mathbb{Q}^T\)-mean is lower by approximately 27 basis points. This reflects the change of numeraire to the 10-year bond.

Exercise 6. Explain why the conditional mean under \(\mathbb{Q}^T\) is lower than under \(\mathbb{Q}\). Relate this to the fact that the \(T\)-forward measure tilts probability toward states where bond prices are high (i.e., interest rates are low).

Solution to Exercise 6

The \(\mathbb{Q}^T\)-mean is lower than the \(\mathbb{Q}\)-mean because of the probabilistic interpretation of the measure change.

The \(T\)-forward measure \(\mathbb{Q}^T\) uses the zero-coupon bond \(P(t,T)\) as numeraire. The Radon-Nikodym derivative \(\frac{d\mathbb{Q}^T}{d\mathbb{Q}} \propto P(t,T)/M(t)\) tilts the probability measure in favor of scenarios where \(P(t,T)/M(t)\) is large, i.e., where bond prices are high relative to the money market account.

High bond prices correspond to low interest rates. Therefore \(\mathbb{Q}^T\) assigns higher probability to low-rate scenarios and lower probability to high-rate scenarios compared to \(\mathbb{Q}\). This systematic tilting toward low rates reduces the expected value of \(r(t)\) under \(\mathbb{Q}^T\).

The magnitude of this effect is governed by:

  • Volatility \(\sigma\): Higher rate volatility means more variability in bond prices, creating a larger difference between high-rate and low-rate scenarios, and hence a bigger tilt.
  • Time to maturity \(T - t\): Longer-dated bonds have higher volatility (\(\sigma_P(t,T)\) increases with \(T - t\)), so the tilt is stronger for longer numeraire maturities.
  • Mean reversion \(\lambda\): Stronger mean reversion dampens the bond volatility and reduces the tilt.

This is a manifestation of the convexity bias in interest rate markets: the nonlinear (convex) relationship between rates and bond prices means that averaging over rate scenarios under a bond-weighted measure systematically favors lower rates.

Exercise 7. For the bond option pricing formula, we need \(r(T) \sim \mathcal{N}(\mu_r^T(t_0, T), \sigma_r^2(t_0, T))\) under \(\mathbb{Q}^T\). Explain what goes wrong if one mistakenly uses the \(\mathbb{Q}\)-distribution of \(r(T)\) instead. How would the option price be biased?

Solution to Exercise 7

The bond option pricing formula under \(\mathbb{Q}^T\) is:

\[ V(t_0) = P(t_0, T)\,\mathbb{E}^T\!\left[\max(P(T, T_S) - K, 0)\,\Big|\,\mathcal{F}(t_0)\right] \]

where \(P(T, T_S) = e^{A(\tau) + B(\tau)r(T)}\) with \(\tau = T_S - T\) and \(B(\tau) < 0\). The expectation requires the distribution of \(r(T)\) under \(\mathbb{Q}^T\), which is \(\mathcal{N}(\mu_r^T, (\sigma_r^T)^2)\) with \(\mu_r^T < \mu_r^{\mathbb{Q}}\).

If one mistakenly uses the \(\mathbb{Q}\)-distribution: The mean \(\mu_r^{\mathbb{Q}}\) is higher than the correct \(\mu_r^T\). Since \(B(\tau) < 0\), a higher mean for \(r(T)\) implies a lower mean for \(B(\tau)r(T)\), which implies a lower mean for \(P(T, T_S) = e^{A + Br(T)}\).

This means the mistaken \(\mathbb{Q}\)-calculation would:

  • Underestimate the expected bond price \(\mathbb{E}[P(T, T_S)]\)
  • Underestimate the probability that the call option finishes in the money
  • Produce a downward-biased call option price (the call would be underpriced)

For a put option on \(P(T, T_S)\), the bias would be reversed: the put would be overpriced.

The error arises because the formula \(V = P(t_0, T)\,\mathbb{E}^T[\cdot]\) is derived specifically under the \(T\)-forward measure. Using the \(\mathbb{Q}\)-distribution in this formula is mathematically inconsistent -- it conflates the discount factor (which is correct for \(\mathbb{Q}\)) with the payoff expectation (which must be evaluated under \(\mathbb{Q}^T\)). The correct \(\mathbb{Q}\)-based formula would be \(V = \mathbb{E}^{\mathbb{Q}}[e^{-\int r\,ds}\max(P - K, 0)]\), which accounts for the correlation between the discount factor and the payoff, but this is harder to evaluate.