T-Forward Measures¶

Beyond the risk-neutral measure, it is often convenient to price derivatives under a forward measure, associated with a specific maturity $T$.

Definition of the T-forward measure¶

Let $P(t,T)$ be the zero-coupon bond maturing at $T$. The T-forward measure $\mathbb{Q}^T$ is defined by choosing $P(t,T)$ as numéraire.

Under $\mathbb{Q}^T$,

\[ \frac{S_t}{P(t,T)} \text{ is a martingale} \]

for any tradable asset $S_t$ that pays off at or before $T$.

Pricing under the forward measure¶

For a payoff $V_T$ at time $T$,

\[ V_t = P(t,T)\,\mathbb{E}^{\mathbb{Q}^T}[V_T \mid \mathcal{F}_t]. \]

Discounting disappears because the numéraire already matures at $T$.

Dynamics under the forward measure¶

Changing from $\mathbb{Q}$ to $\mathbb{Q}^T$: - alters drift terms, - leaves volatilities unchanged, - simplifies pricing of forwards, FRAs, and caps.

Many rates become martingales under their natural forward measures.

Practical importance¶

Forward measures are especially useful for: - caplets and floorlets, - forward-starting contracts, - simplifying drift terms in HJM and LMM.

Key takeaways¶

Forward measures use zero-coupon bonds as numeraires.
Pricing simplifies to expectation without discounting.
Measure choice is a powerful modeling tool.

QuantPie Derivation: Change of Numeraire¶

Instantaneous Forward Rate Dynamics under Different Measures¶

Risk Neutral Measure:

\[\begin{array}{lllll} \text{Risk Neutral}&& \displaystyle df(t,T) &=&\displaystyle \left(\sigma(t,T)\int_t^T\sigma(t,T')dT'\right)dt+\sigma(t,T)dW^\mathbb{Q}(t)\\ \end{array}\]

T Forward Measure:

\[\begin{array}{lllll} \text{$T$ Forward}&& \displaystyle df(t,T) &=&\displaystyle \sigma(t,T)dW^T(t)\\ \end{array}\]

$T_f$ Forward Measure:

\[\begin{array}{lllll} \text{$T_f$ Forward}&& \displaystyle df(t,T) &=&\displaystyle -\left(\sigma(t,T)\int_T^{T_f}\sigma(t,T')dT'\right)dt +\sigma(t,T)dW^{T_f}(t)\\ \end{array}\]

Forward Rate as a Markov Process¶

$f(t,T)$, as a function of $t$, is a Markov process.

\[\begin{array}{lllll} \displaystyle f(t,T) = f(0,T)+ \int_0^t\left(\sigma(t',T)\int_{t'}^T\sigma(t',T')dT'\right)dt'+\int_0^t\sigma(t',T)dW^\mathbb{Q}(t') \end{array}\]

\[\begin{array}{lllll} \displaystyle f(t+\Delta,T) = f(0,T)+ \int_0^{t+\Delta}\left(\sigma(t',T)\int_{t'}^T\sigma(t',T')dT'\right)dt'+\int_0^{t+\Delta}\sigma(t',T)dW^\mathbb{Q}(t') \end{array}\]

\[\begin{array}{lllll} \displaystyle f(t+\Delta,T)-f(t,T) = \int_t^{t+\Delta}\left(\sigma(t',T)\int_{t'}^T\sigma(t',T')dT'\right)dt'+\int_t^{t+\Delta}\sigma(t',T)dW^\mathbb{Q}(t') \end{array}\]

T-Forward Measure: Direct Computation¶

Instantaneous Forward Rate Dynamics

\[\begin{array}{lllll} \displaystyle df(t,T) = \mu^\mathbb{Q}(t,T)dt+\sigma(t,T)dW^{\mathbb{Q}}(t) \end{array}\]

ZCB Dynamics

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

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

\[\begin{array}{lllll} \displaystyle \log P(t,T)-\log P(0,T) = \int_0^t\left(r(t')-\frac{1}{2}\sigma_P^2(t',T)\right)dt'+\int_0^t\sigma_P(t',T)dW^{\mathbb{Q}}(t')\\ \end{array}\]

\[\begin{array}{lllll} \displaystyle \frac{P(t,T)}{P(0,T)} = \text{exp}\left(\int_0^t\left(r(t')-\frac{1}{2}\sigma_P^2(t',T)\right)dt'+\int_0^t\sigma_P(t',T)dW^{\mathbb{Q}}(t')\right) \end{array}\]

Radon-Nikodym Derivative

\[\begin{array}{lllll} \displaystyle \lambda_\mathbb{Q}^T(t) &=&\displaystyle \displaystyle \frac{d\mathbb{Q}^T}{d\mathbb{Q}}\Big{|}_{{\cal F}(t)}\\ &=&\displaystyle \frac{P(t,T)/P(0,T)}{M(t)/M(0)}\\ &=&\displaystyle \text{exp}\left( -\int_0^tr(t')dt' +\int_0^t\left(r(t')-\frac{1}{2}\sigma_P^2(t',T)\right)dt'+\int_0^t\sigma_P(t',T)dW^{\mathbb{Q}}(t')\right)\\ &=&\displaystyle \text{exp}\left( -\frac{1}{2}\int_0^t\sigma_P^2(t',T)dt'+\int_0^t\sigma_P(t',T)dW^\mathbb{Q}(t') \right)\\ \end{array}\]

Girsanov Theorem

\[\begin{array}{lllll} \displaystyle dW^T(t)=dW^{\mathbb{Q}}(t)-\sigma_P(t,T)dt \end{array}\]

where

\[\begin{array}{lllll} \displaystyle \sigma_P(t,T) = -\int_t^T\sigma(t',T)dt' \end{array}\]

Forward Rate Dynamics under $\mathbb{Q}^T$

\[\begin{array}{lllll} \displaystyle df(t,T) &=&\displaystyle \left(\sigma(t,T)\int_t^T\sigma(t,T')dT'\right)dt +\sigma(t,T)dW^{\mathbb{Q}}(t)\\ &=&\displaystyle \left(\sigma(t,T)\int_t^T\sigma(t,T')dT'\right)dt +\sigma(t,T)\left( dW^T(t)-\left(\int_t^T\sigma(t,T')dT'\right)dt \right)\\ &=&\displaystyle \sigma(t,T)dW^T(t) \end{array}\]

Exercises¶

Exercise 1. Let $P(0, 1) = 0.96$ and $P(0, 3) = 0.88$. A forward rate agreement (FRA) pays $L(1, 3) - K$ at time $T = 3$, where $L(1, 3)$ is the simply-compounded rate for the period $[1, 3]$. Using the $T$-forward measure with $T = 3$, show that the fair FRA rate equals the forward rate $F(0; 1, 3) = \frac{1}{2}\left(\frac{P(0,1)}{P(0,3)} - 1\right)$. Compute its numerical value.

Solution to Exercise 1

Setting up the FRA pricing. The FRA pays $L(1, 3) - K$ at time $T = 3$, where the simply-compounded rate is

\[ L(1, 3) = \frac{1}{\delta}\left(\frac{P(1, 1)}{P(1, 3)} - 1\right) = \frac{1}{2}\left(\frac{1}{P(1, 3)} - 1\right) \]

with $\delta = T_2 - T_1 = 3 - 1 = 2$.

Pricing under the $T$-forward measure with $T = 3$. Using numéraire $P(t, 3)$:

\[ V_0 = P(0, 3)\,\mathbb{E}^{\mathbb{Q}^3}[L(1, 3) - K] \]

The fair FRA rate $K^*$ is the value of $K$ that makes $V_0 = 0$:

\[ K^* = \mathbb{E}^{\mathbb{Q}^3}[L(1, 3)] \]

Now, $L(1, 3) = \frac{1}{2}\left(\frac{1}{P(1, 3)} - 1\right)$. Under the $\mathbb{Q}^3$-measure, the forward LIBOR rate $L(t; 1, 3)$ is a martingale (since it can be written as a ratio involving $P(t, 3)$ as numéraire). Therefore:

\[ \mathbb{E}^{\mathbb{Q}^3}[L(1, 3)] = L(0; 1, 3) = \frac{1}{2}\left(\frac{P(0, 1)}{P(0, 3)} - 1\right) \]

This gives the forward rate:

\[ F(0; 1, 3) = \frac{1}{2}\left(\frac{P(0, 1)}{P(0, 3)} - 1\right) \]

Numerical computation. Substituting $P(0, 1) = 0.96$ and $P(0, 3) = 0.88$:

\[ F(0; 1, 3) = \frac{1}{2}\left(\frac{0.96}{0.88} - 1\right) = \frac{1}{2}(1.09091 - 1) = \frac{1}{2} \times 0.09091 = 0.04545 \]

The fair FRA rate is approximately 4.545%.

Exercise 2. Under the risk-neutral measure $\mathbb{Q}$, the instantaneous forward rate satisfies

\[ df(t, T) = \sigma(t, T)\int_t^T \sigma(t, T')\,dT'\,dt + \sigma(t, T)\,dW^{\mathbb{Q}}(t) \]

Show that under the $T$-forward measure $\mathbb{Q}^T$, the drift vanishes and $f(t, T)$ satisfies $df(t, T) = \sigma(t, T)\,dW^T(t)$. Identify the Girsanov kernel used in the change of measure.

Solution to Exercise 2

Starting point. Under $\mathbb{Q}$, the HJM drift condition gives

\[ df(t, T) = \sigma(t, T)\int_t^T \sigma(t, T')\,dT'\,dt + \sigma(t, T)\,dW^{\mathbb{Q}}(t) \]

Girsanov kernel. The $T$-forward measure $\mathbb{Q}^T$ is defined by the numéraire $P(t, T)$. Under $\mathbb{Q}$, the bond price dynamics are

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

where $\sigma_P(t, T) = -\int_t^T \sigma(t, u)\,du$. The Girsanov kernel (the volatility of the numéraire) is $\sigma_P(t, T)$, and the relationship between Brownian motions is

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

Deriving the drift-free dynamics. Substituting $dW^{\mathbb{Q}}(t) = dW^T(t) - \int_t^T \sigma(t, u)\,du\,dt$ into the $\mathbb{Q}$-dynamics:

\[ df(t, T) = \sigma(t, T)\int_t^T \sigma(t, T')\,dT'\,dt + \sigma(t, T)\left(dW^T(t) - \int_t^T \sigma(t, u)\,du\,dt\right) \]

\[ = \sigma(t, T)\int_t^T \sigma(t, T')\,dT'\,dt - \sigma(t, T)\int_t^T \sigma(t, u)\,du\,dt + \sigma(t, T)\,dW^T(t) \]

The two drift terms cancel exactly:

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

This confirms that $f(t, T)$ is a martingale under its own $T$-forward measure, a fundamental result of the HJM framework. The instantaneous forward rate for maturity $T$ is driftless when viewed under the measure whose numéraire matures at $T$.

Exercise 3. In the Hull--White model, $dr_t = (theta(t) - ar_t)\,dt + \sigma\,dW_t^{\mathbb{Q}}$, and the bond price volatility is $\Sigma(t, T) = -\frac{\sigma}{a}(1 - e^{-a(T-t)})$. Write down the Radon--Nikodym derivative $d\mathbb{Q}^T/d\mathbb{Q}|_{\mathcal{F}_t}$ in terms of $\Sigma(t, T)$ and the risk-neutral Brownian motion. Then derive the dynamics of $r_t$ under $\mathbb{Q}^T$ and identify the new drift.

Solution to Exercise 3

Bond price volatility in the Hull--White model. We are given

\[ \Sigma(t, T) = -\frac{\sigma}{a}(1 - e^{-a(T-t)}) \]

(Note: this is $\sigma_P(t, T)$ in the notation of the ZCB dynamics $dP/P = r\,dt + \Sigma(t,T)\,dW^{\mathbb{Q}}$.)

Radon--Nikodym derivative. The change from $\mathbb{Q}$ to $\mathbb{Q}^T$ is

\[ \frac{d\mathbb{Q}^T}{d\mathbb{Q}}\bigg|_{\mathcal{F}_t} = \exp\!\left(-\frac{1}{2}\int_0^t \Sigma(s, T)^2\,ds + \int_0^t \Sigma(s, T)\,dW_s^{\mathbb{Q}}\right) \]

Substituting the Hull--White expression:

\[ \frac{d\mathbb{Q}^T}{d\mathbb{Q}}\bigg|_{\mathcal{F}_t} = \exp\!\left(-\frac{1}{2}\int_0^t \frac{\sigma^2}{a^2}(1 - e^{-a(T-s)})^2\,ds - \int_0^t \frac{\sigma}{a}(1 - e^{-a(T-s)})\,dW_s^{\mathbb{Q}}\right) \]

This is a valid density process (positive $\mathbb{Q}$-martingale with initial value 1) by the Novikov condition.

Dynamics of $r_t$ under $\mathbb{Q}^T$. The Girsanov change gives

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

Substituting $dW_t^{\mathbb{Q}} = dW_t^T + \Sigma(t, T)\,dt$ into the $\mathbb{Q}$-dynamics:

\[ dr_t = (\theta(t) - ar_t)\,dt + \sigma\,dW_t^{\mathbb{Q}} \]

\[ = (\theta(t) - ar_t)\,dt + \sigma\left(dW_t^T + \Sigma(t, T)\,dt\right) \]

\[ = \left(\theta(t) - ar_t + \sigma\Sigma(t, T)\right)dt + \sigma\,dW_t^T \]

Substituting $\Sigma(t, T) = -\frac{\sigma}{a}(1 - e^{-a(T-t)})$:

\[ dr_t = \left(\theta(t) - ar_t - \frac{\sigma^2}{a}(1 - e^{-a(T-t)})\right)dt + \sigma\,dW_t^T \]

The new drift under $\mathbb{Q}^T$ is

\[ \mu^T(t, r_t) = \theta(t) - ar_t - \frac{\sigma^2}{a}\left(1 - e^{-a(T-t)}\right) \]

The additional drift term $-\frac{\sigma^2}{a}(1 - e^{-a(T-t)})$ represents the convexity adjustment from changing numéraire. It is always negative, reflecting the negative correlation between bond prices and interest rates: higher rates mean lower bond prices (the numéraire), which biases the measure change toward lower rate paths.

Exercise 4. Explain why the $T$-forward measure is particularly well-suited for pricing a European option with payoff $g(r_T)$ at time $T$, where $g$ is an arbitrary function of the short rate. Specifically, show that the option price simplifies to

\[ V_0 = P(0, T)\,\mathbb{E}^{\mathbb{Q}^T}[g(r_T)] \]

and that no stochastic discounting appears. Why would the same computation under $\mathbb{Q}$ require knowledge of the joint distribution of $\int_0^T r_s\,ds$ and $r_T$?

Solution to Exercise 4

Under $\mathbb{Q}^T$. Choose numéraire $N_t = P(t, T)$. The pricing formula gives

\[ V_0 = P(0, T)\,\mathbb{E}^{\mathbb{Q}^T}\!\left[\frac{g(r_T)}{P(T, T)}\;\middle|\;\mathcal{F}_0\right] \]

Since $P(T, T) = 1$ (a zero-coupon bond at its own maturity equals par):

\[ V_0 = P(0, T)\,\mathbb{E}^{\mathbb{Q}^T}[g(r_T)] \]

No stochastic discounting appears. The price is the bond price times a simple expectation of the payoff function.

Why this is advantageous. To evaluate $\mathbb{E}^{\mathbb{Q}^T}[g(r_T)]$, we only need the marginal distribution of $r_T$ under $\mathbb{Q}^T$. In many models (Vasicek, Hull--White, CIR), $r_T$ has a known distribution under $\mathbb{Q}^T$ (Gaussian for Vasicek/Hull--White, non-central chi-squared for CIR), making the expectation computable analytically or by simple numerical integration.

Why the $\mathbb{Q}$-computation is harder. Under $\mathbb{Q}$:

\[ V_0 = \mathbb{E}^{\mathbb{Q}}\!\left[e^{-\int_0^T r_s\,ds}\,g(r_T)\right] \]

The discount factor $D = e^{-\int_0^T r_s\,ds}$ depends on the entire path of $r_s$ over $[0, T]$, while $g(r_T)$ depends on the terminal value. Since $D$ and $r_T$ are not independent (they are functionally related through the path of $r$), one cannot separate the expectation:

\[ \mathbb{E}^{\mathbb{Q}}[D \cdot g(r_T)] \neq \mathbb{E}^{\mathbb{Q}}[D] \cdot \mathbb{E}^{\mathbb{Q}}[g(r_T)] \]

Computing $\mathbb{E}^{\mathbb{Q}}[D \cdot g(r_T)]$ requires the joint distribution of $\left(\int_0^T r_s\,ds,\; r_T\right)$. In Gaussian models, these are jointly normal (since both are linear functionals of the Gaussian process $r$), so the joint distribution is available. But even then, the computation involves a bivariate normal integration, which is more complex than the univariate expectation under $\mathbb{Q}^T$.

For non-Gaussian models, the joint distribution may not be tractable at all, making the forward measure approach essential.

Exercise 5. Consider two forward measures $\mathbb{Q}^{T_1}$ and $\mathbb{Q}^{T_2}$ with $T_1 < T_2$. Write down the Radon--Nikodym derivative for changing from $\mathbb{Q}^{T_2}$ to $\mathbb{Q}^{T_1}$ and the corresponding Girsanov drift adjustment. A forward LIBOR rate $L(t; T_1, T_2)$ is a martingale under $\mathbb{Q}^{T_2}$. What drift does it acquire under $\mathbb{Q}^{T_1}$?

Solution to Exercise 5

Radon--Nikodym derivative between forward measures. The Radon--Nikodym derivative for changing from $\mathbb{Q}^{T_2}$ to $\mathbb{Q}^{T_1}$ is

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

Under $\mathbb{Q}^{T_2}$, using the bond dynamics, the log of this ratio involves:

\[ \ln\frac{P(t, T_1)}{P(t, T_2)} - \ln\frac{P(0, T_1)}{P(0, T_2)} = \int_0^t \left[\Sigma(s, T_1) - \Sigma(s, T_2)\right]\,dW_s^{T_2} + \text{drift terms} \]

where $\Sigma(t, U) = -\int_t^U \sigma(t, u)\,du$. By Girsanov's theorem:

\[ dW_t^{T_1} = dW_t^{T_2} + \left[\Sigma(t, T_2) - \Sigma(t, T_1)\right]dt \]

The drift adjustment is $\gamma(t) = \Sigma(t, T_2) - \Sigma(t, T_1) = \int_{T_1}^{T_2} \sigma(t, u)\,du$.

Since $T_1 < T_2$ and $\sigma(t, u) > 0$, this drift adjustment is positive.

Drift of $L(t; T_1, T_2)$ under $\mathbb{Q}^{T_1}$. Under $\mathbb{Q}^{T_2}$, the forward LIBOR rate $L(t; T_1, T_2)$ is a martingale with dynamics

\[ dL(t; T_1, T_2) = \sigma_L(t) L(t; T_1, T_2)\,dW_t^{T_2} \]

Substituting $dW_t^{T_2} = dW_t^{T_1} - \gamma(t)\,dt$:

\[ dL(t; T_1, T_2) = \sigma_L(t) L(t; T_1, T_2)\left[dW_t^{T_1} - \gamma(t)\,dt\right] \]

\[ = -\sigma_L(t)\,\gamma(t)\,L(t; T_1, T_2)\,dt + \sigma_L(t)\,L(t; T_1, T_2)\,dW_t^{T_1} \]

The drift under $\mathbb{Q}^{T_1}$ is

\[ \mu_L^{T_1}(t) = -\sigma_L(t)\,\gamma(t)\,L(t; T_1, T_2) = -\sigma_L(t)\left(\int_{T_1}^{T_2}\sigma(t, u)\,du\right)L(t; T_1, T_2) \]

This negative drift reflects the fact that $L(t; T_1, T_2)$ is naturally a martingale under $\mathbb{Q}^{T_2}$ (the measure corresponding to the payment date), and acquires a drift under any other measure.

Exercise 6. Under the $T_f$-forward measure (where $T_f \neq T$), the instantaneous forward rate has dynamics

\[ df(t, T) = -\sigma(t, T)\int_T^{T_f}\sigma(t, T')\,dT'\,dt + \sigma(t, T)\,dW^{T_f}(t) \]

Verify that when $T_f = T$, the drift vanishes (recovering the result under the own $T$-forward measure). When $T_f > T$, determine the sign of the drift and provide a financial interpretation.

Solution to Exercise 6

Case $T_f = T$. When $T_f = T$, the drift term is

\[ -\sigma(t, T)\int_T^{T_f}\sigma(t, T')\,dT' = -\sigma(t, T)\int_T^{T}\sigma(t, T')\,dT' = 0 \]

since the integral over an empty interval (from $T$ to $T$) vanishes. This gives

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

recovering the driftless dynamics under the own $T$-forward measure, consistent with Exercise 2.

Case $T_f > T$: sign of the drift. When $T_f > T$, the drift coefficient is

\[ \alpha(t, T) = -\sigma(t, T)\int_T^{T_f}\sigma(t, T')\,dT' \]

Since $\sigma(t, T) > 0$ and $\sigma(t, T') > 0$ for all $T' \in [T, T_f]$, the integral $\int_T^{T_f}\sigma(t, T')\,dT' > 0$. Therefore

\[ \alpha(t, T) < 0 \]

The drift is negative.

Financial interpretation. Under the $T_f$-forward measure (with $T_f > T$), the forward rate $f(t, T)$ has a negative drift. This can be understood as follows:

The numéraire is $P(t, T_f)$, a longer-dated bond. Bond prices and interest rates move in opposite directions.
When forward rates increase, bond prices decrease. A longer-dated bond ($T_f > T$) is more sensitive to rate changes than a shorter-dated one.
Under the $T_f$-forward measure, the probability weighting favors scenarios where $P(t, T_f)$ is large, i.e., where rates are low.
This tilting toward low-rate scenarios manifests as a negative drift for $f(t, T)$.

Equivalently, this is a convexity adjustment: the measure associated with the longer bond $P(t, T_f)$ assigns more probability mass to low-rate states, pulling the expected forward rate downward relative to the $T$-forward measure.

Case $T_f < T$: By the same argument, the integral $\int_T^{T_f}$ has reversed limits with $T_f < T$, giving a negative integral. Thus $\alpha(t, T) > 0$ and the drift is positive. Under a shorter-dated numéraire, the forward rate is biased upward.