Forward Measures for Zero-Coupon Bonds

Why Use a Bond as Numeraire?

A zero-coupon bond maturing at time $T$ pays one unit at $T$. When a claim also pays at $T$, measuring its value in units of that bond makes its maturity payoff especially simple: the payoff, expressed in bond units, is just the claim itself. The probability measure associated with this numeraire is called the $T$-forward measure. Its central pricing identity replaces a bank-account-discounted expectation under the risk-neutral measure with an expectation under the forward measure multiplied by the current bond price.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying Brownian motion $B_t$. Let $\mathbb{Q}$ be the risk-neutral measure, equivalent to $\mathbb{P}$ on the time horizon considered. The bank account is $A_t=\exp(\int_0^t r_s\,ds)$, where $r_s$ is the short rate and the integral is finite. Write $P(t,T)$ for the time-$t$ price of a zero-coupon bond paying one at $T$, with $P(T,T)=1$. Assume $A_t>0$ and $P(t,T)>0$.

Definition: The $T$-forward measure $\mathbb{Q}^T$, on $\mathcal{F}_T$, is defined by $$ \frac{d\mathbb{Q}^T}{d\mathbb{Q}}\Big|_{\mathcal{F}_T} =\frac{1}{A_TP(0,T)}. $$ Here $A_0=1$, and the risk-neutral bond-pricing condition is $P(0,T)=\mathbb{E}^{\mathbb{Q}}[1/A_T]$. The associated density process for $0\le t\le T$ is $$ Z_t^T=\mathbb{E}^{\mathbb{Q}}\left[\frac{1}{A_TP(0,T)}\middle|\mathcal{F}_t\right] =\frac{P(t,T)}{A_tP(0,T)}. $$

The pricing condition makes the terminal density have expectation one. Positivity of the bond and bank account makes the density strictly positive, so $\mathbb{Q}^T$ and $\mathbb{Q}$ are equivalent on $\mathcal{F}_T$: they assign probability zero to the same events. The density process formula follows from risk-neutral pricing of the bond at time $t$, namely $P(t,T)=A_t\mathbb{E}^{\mathbb{Q}}[1/A_T\mid\mathcal{F}_t]$.

The Density and the Pricing Conversion

Theorem: Suppose the discounted bond $P(t,T)/A_t$ is a true $\mathbb{Q}$-martingale on $0\le t\le T$, with $P(0,T)>0$. Then $Z_t^T=P(t,T)/(A_tP(0,T))$ is a true $\mathbb{Q}$-martingale with $Z_0^T=1$, and its terminal value defines $\mathbb{Q}^T$. For any $\mathcal{F}_T$-measurable claim $X_T$ such that the expectations below are finite, its time-$t$ value is $$ V_t=A_t\mathbb{E}^{\mathbb{Q}}\left[\frac{X_T}{A_T}\middle|\mathcal{F}_t\right] =P(t,T)\mathbb{E}^{\mathbb{Q}^T}[X_T\mid\mathcal{F}_t]. $$

Proof. The martingale assumption gives $\mathbb{E}^{\mathbb{Q}}[Z_t^T]=Z_0^T=1$ and $\mathbb{E}^{\mathbb{Q}}[Z_t^T\mid\mathcal{F}_s]=Z_s^T$ for $s\le t$. Its terminal value is $Z_T^T=1/(A_TP(0,T))$, since $P(T,T)=1$. A positive, mean-one terminal density defines a probability measure equivalent to $\mathbb{Q}$. For the pricing identity, conditional Bayes’ formula gives $$ \mathbb{E}^{\mathbb{Q}^T}[X_T\mid\mathcal{F}_t] =\frac{\mathbb{E}^{\mathbb{Q}}[Z_T^T X_T\mid\mathcal{F}_t]}{Z_t^T} =\frac{\mathbb{E}^{\mathbb{Q}}[X_T/A_T\mid\mathcal{F}_t]}{P(t,T)/A_t}. $$ Multiplying by $P(t,T)$ gives the claimed value. The conditional formula is meaningful when its denominator $Z_t^T$ is positive, which holds here, and when the conditional expectations are finite. $\square$

The equality compares the same value in two forms. $V_t/A_t$ is the claim’s value in bank-account units, while $V_t/P(t,T)$ is its value in units of the $T$-maturity bond. Thus $$ \frac{V_t}{A_t}=\frac{P(t,T)}{A_t}\mathbb{E}^{\mathbb{Q}^T}[X_T\mid\mathcal{F}_t], \qquad \frac{V_t}{P(t,T)}=\mathbb{E}^{\mathbb{Q}^T}[X_T\mid\mathcal{F}_t]. $$

The theorem requires a true martingale, not merely a local martingale. If discounted bond prices are only known to be local martingales, their expectations need not be preserved, and the proposed density need not have expectation one. In applications, the true-martingale property may follow directly from the pricing model’s assumptions.

Worked Example: Deterministic Interest Rate

Let the short rate be constant $r=1/20$, let $T=2$, and take $A_0=1$. Find the bond price and the $T$-forward density process.

The bank account is $A_t=e^{t/20}$. The zero-coupon bond price is $P(t,2)=e^{-(2-t)/20}$, so $P(0,2)=e^{-2/20}=e^{-1/10}$. Therefore, for $0\le t\le2$,

Z_t^2=\frac{e^{-(2-t)/20}}{e^{t/20}e^{-2/20}}=1.

At maturity, the same result follows directly: $A_2=e^{1/10}$, hence $Z_2^2=1/(e^{1/10}e^{-1/10})=1$. Thus $\mathbb{Q}^2=\mathbb{Q}$ in this deterministic-rate model. For any integrable payoff $X_2$, its value is $P(t,2)\mathbb{E}^{\mathbb{Q}}[X_2\mid\mathcal{F}_t]$.

Check: for the unit payoff $X_2=1$, the forward-measure formula gives $P(t,2)\mathbb{E}^{\mathbb{Q}^2}[1\mid\mathcal{F}_t]=P(t,2)$, exactly the zero-coupon bond price.

Worked Example: A Two-State Measure Change and Claim Value

At time $T$, suppose there are two states, each with $\mathbb{Q}$-probability $1/2$. Let $A_T=1$ in the first state and $A_T=2$ in the second, and let $A_0=1$. Compute $P(0,T)$, the forward probabilities, and the value of a claim paying $0$ in the first state and $3$ in the second.

The bond-price condition gives $$ P(0,T)=\mathbb{E}^{\mathbb{Q}}[1/A_T] =\frac12(1)+\frac12\left(\frac12\right)=\frac34. $$ The density $Z_T^T=1/(A_TP(0,T))$ is $4/3$ in the first state and $2/3$ in the second. Its expectation is $\frac12(4/3)+\frac12(2/3)=1$. The forward probabilities are therefore $\frac12(4/3)=2/3$ and $\frac12(2/3)=1/3$.

The risk-neutral bank-account calculation gives $$ V_0=\mathbb{E}^{\mathbb{Q}}[X_T/A_T] =\frac12(0/1)+\frac12(3/2)=\frac34. $$ Under the forward measure, $\mathbb{E}^{\mathbb{Q}^T}[X_T]=(2/3)(0)+(1/3)(3)=1$, so $P(0,T)\mathbb{E}^{\mathbb{Q}^T}[X_T]=(3/4)(1)=3/4$, as required.

Check: the forward probabilities sum to $2/3+1/3=1$, and both pricing calculations give $3/4$. The state with the larger bank account receives less forward-measure probability because the density weights states by $1/A_T$.

Brownian Motion Under the Forward Measure

The density also describes how Brownian motion changes when the bond is used as numeraire. Suppose that, under $\mathbb{Q}$, the bond has dynamics $$ \frac{dP(t,T)}{P(t,T)}=r_t\,dt+\sigma_P(t)\,dB_t^{\mathbb{Q}}, $$ where $\sigma_P$ is progressively measurable. Assume the stochastic integral with integrand $\sigma_P$ is defined in the usual Itô sense: on each finite interval it is an Itô integral of a progressively measurable integrand in $\mathcal{L}^2$, the class satisfying $\mathbb{E}^{\mathbb{Q}}\int_0^T H_s^2\,ds<\infty$. In particular, assume this integrability for $\sigma_P$. Since $dA_t/A_t=r_t\,dt$, the quotient rule gives $dZ_t^T=Z_t^T\sigma_P(t)\,dB_t^{\mathbb{Q}}$.

Result: If Novikov’s condition $$ \mathbb{E}^{\mathbb{Q}}\left[\exp\left(\frac12\int_0^T\sigma_P(s)^2\,ds\right)\right]<\infty $$ holds, the stochastic exponential defining $Z_t^T$ is a true martingale. Girsanov’s theorem then says that $$ B_t^T=B_t^{\mathbb{Q}}-\int_0^t\sigma_P(s)\,ds $$ is Brownian motion under $\mathbb{Q}^T$, for $0\le t\le T$. Novikov’s condition is a sufficient condition, not a necessary one. This use of Novikov’s condition and Girsanov’s theorem is a standard result; its proof is beyond this tutorial.

Consequently, $dB_t^{\mathbb{Q}}=dB_t^T+\sigma_P(t)\,dt$. Replacing the Brownian increment in the bond dynamics shows that under $\mathbb{Q}^T$, $$ \frac{dP(t,T)}{P(t,T)} =\big(r_t+\sigma_P(t)^2\big)\,dt+\sigma_P(t)\,dB_t^T. $$ This is the bond’s drift under the new measure; it is not the drift under the bank-account risk-neutral measure. The volatility coefficient is unchanged, while the drift changes with the numeraire.

Worked Example: Constant Bond Volatility

Suppose $\sigma_P(t)=1/5$ on $[0,1]$. Compute the density process and the Brownian shift. Here the integrand is deterministic and belongs to $\mathcal{L}^2$, since $\int_0^1(1/5)^2\,dt=1/25$.

The stochastic exponential solution to $dZ_t^T=Z_t^T(1/5)dB_t^{\mathbb{Q}}$, $Z_0^T=1$, is $$ Z_t^T=\exp\left(\frac15B_t^{\mathbb{Q}}-\frac{t}{50}\right). $$ For an explicit Itô check, set $f(t,x)=\exp(x/5-t/50)$. Its derivatives are $f_t=-f/50$, $f_x=f/5$, and $f_{xx}=f/25$. Itô’s formula, including the second-derivative correction, gives $$ df(t,B_t^{\mathbb{Q}}) =\left(-\frac{f}{50}+\frac12\frac{f}{25}\right)dt+\frac{f}{5}\,dB_t^{\mathbb{Q}} =\frac{f}{5}\,dB_t^{\mathbb{Q}}. $$ Thus $dZ_t^T=Z_t^T(1/5)dB_t^{\mathbb{Q}}$.

Novikov’s expectation is $\mathbb{E}^{\mathbb{Q}}[e^{1/50}]=e^{1/50}<\infty$. The density is a true martingale; directly, the normal moment-generating function gives $\mathbb{E}^{\mathbb{Q}}[Z_t^T]=e^{-t/50}e^{t/50}=1$. Girsanov’s theorem yields $B_t^T=B_t^{\mathbb{Q}}-t/5$. Substitution in the bond equation changes its drift from $r_t$ to $r_t+1/25$.

Check: the drift addition is $\sigma_P^2=(1/5)^2=1/25$, and the exponent’s compensating term is $-\tfrac12(1/25)t=-t/50$, consistent with the density having mean one.

Common Mistakes and Practical Use

Do not confuse the two measures. $\mathbb{Q}$ is the bank-account risk-neutral measure; $\mathbb{Q}^T$ is the measure associated with the bond maturing at $T$. The conversion formula uses a conditional expectation under the measure matching the numeraire: $V_t=P(t,T)\mathbb{E}^{\mathbb{Q}^T}[X_T\mid\mathcal{F}_t]$.

A common wrong calculation is to use $P(t,T)\mathbb{E}^{\mathbb{Q}}[X_T\mid\mathcal{F}_t]$ without changing measure. In the two-state example, that would give $(3/4)(3/2)=9/8$, not the correct value $3/4$. The correction is to use the forward expectation, which is $1$, giving $(3/4)(1)=3/4$. Equivalently, keep $\mathbb{Q}$ and discount by the bank account: $\mathbb{E}^{\mathbb{Q}}[X_T/A_T]=3/4$.

The forward measure is useful when a payoff is naturally expressed at a bond maturity, and when its distribution is simpler under that measure. The change of measure does not change the payoff or the claim’s value; it changes the probability weights used to calculate the expectation and the numeraire used to express the value.

Check Your Understanding

Use the density definition and the pricing identity to answer each question.

Why must the discounted bond be a true martingale, rather than merely a local martingale, for its normalized price to define the density process?
For a payoff $X_T$, write its value in bank-account units and in $T$-bond units.
If $A_T$ takes positive values in finitely many states, how do the $T$-forward probabilities relate to the risk-neutral probabilities?
Under Novikov’s condition, what Brownian motion is obtained under $\mathbb{Q}^T$ when the bond volatility is $\sigma_P(t)$?
In the two-state example, verify the value again by calculating $\mathbb{E}^{\mathbb{Q}}[X_T/A_T]$ state by state.

Forward Measures for Zero-Coupon Bonds

What You'll Learn