The Pricing PDE for Short-Rate Bonds

From Discounted Payoffs to a Backward Equation

A zero-coupon bond with maturity $T$ pays one unit at $T$ and nothing before then. Its price at an earlier time depends on the future short rate: the payoff is discounted by the accumulated rate between the pricing time and maturity. When that rate is itself a diffusion, the conditional-expectation formula connects bond pricing to a backward partial differential equation (PDE). The drift term in that equation is the drift of the short-rate process under the risk-neutral measure; the discounting term is separate.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{Q})$ carrying a Brownian motion $B_t$. Here $\mathbb{Q}$ is the risk-neutral measure for pricing, not necessarily the physical measure $\mathbb{P}$. Suppose the short rate $R_t$ satisfies, under $\mathbb{Q}$,

dR_t=\mu(t,R_t)\,dt+\sigma(t,R_t)\,dB_t.

Thus $\mu$ is the drift and $\sigma$ is the diffusion coefficient under $\mathbb{Q}$. Assume $\mu$ and $\sigma$ are continuous in time and state, globally Lipschitz in the state uniformly over time, and satisfy a uniform linear-growth bound: for some constant $C$, $|\mu(t,r)|+|\sigma(t,r)|\le C(1+|r|)$. These standard SDE conditions give a unique strong solution for each initial time and rate; a strong solution is built on the given Brownian motion, whereas a weak solution may use a different probability space. The conditions also give pathwise uniqueness, not merely uniqueness in law. We use the standard existence theorem for Lipschitz SDEs here.

All stochastic integrals below are Itô integrals. Their integrands are progressively measurable, and whenever an Itô isometry or a true-martingale conclusion is used, the integrand is in $\mathcal{L}^2$: $\mathbb{E}^{\mathbb{Q}}\int_0^T H_s^2\,ds<\infty$. We also assume the needed discounted quantities are integrable; for example, it is sufficient for the pricing representation that $\mathbb{E}^{\mathbb{Q}}\exp(\int_0^T|R_s|\,ds)<\infty$.

Definition: The time-$t$ price of a unit zero-coupon payment at $T$, given $R_t=r$, is $$ P(t,r;T)=\mathbb{E}^{\mathbb{Q}}_{t,r}\!\left[\exp\!\left(-\int_t^T R_s\,ds\right)\right]. $$ Here $\mathbb{E}^{\mathbb{Q}}_{t,r}$ means expectation for the short-rate diffusion started at time $t$ from rate $r$. The terminal condition is $P(T,r;T)=1$ for every rate $r$, because the bond pays one at maturity.

The Markov property of the strong solution and independent Brownian increments identify this function with the conditional price: $P(t,R_t;T)=\mathbb{E}^{\mathbb{Q}}[\exp(-\int_t^T R_s\,ds)\mid\mathcal{F}_t]$. We use this standard Markov property of Lipschitz SDEs.

The Risk-Neutral Pricing PDE

Theorem: Suppose the preceding assumptions hold, the discounted payoff is integrable, and the pricing function $P$ is $C^{1,2}$ on $[0,T)\times\mathbb{R}$ and continuous up to $T$. Then, for $t<T$ and every $r\in\mathbb{R}$, $$ P_t(t,r;T)+\mu(t,r)P_r(t,r;T)+\frac12\sigma^2(t,r)P_{rr}(t,r;T)-rP(t,r;T)=0, \qquad P(T,r;T)=1. $$ The continuity of the coefficients matters: it supports the pointwise equation. With merely measurable time dependence, the same argument need not give the PDE at every time.

Proof. Write $D_t=\exp(-\int_0^t R_s\,ds)$. The conditional pricing formula gives $D_tP(t,R_t;T)=\mathbb{E}^{\mathbb{Q}}[D_T\mid\mathcal{F}_t]$. The right side is a true martingale: $D_T$ is integrable by assumption, and conditional expectations of an integrable random variable form a martingale. Apply Itô’s formula to $P(t,R_t;T)$. Its partial derivatives are $P_t,P_r,P_{rr}$, and the quadratic-variation correction is $\tfrac12 P_{rr}\sigma^2dt$, since $dB_t\,dB_t=dt$, $dB_t\,dt=0$, and $dt\,dt=0$. The product rule with $dD_t=-R_tD_t\,dt$ then gives $$ d\big(D_tP(t,R_t;T)\big) =D_t\sigma(t,R_t)P_r(t,R_t;T)\,dB_t +D_t A(t,R_t)\,dt, $$ where $A(t,r)=P_t+\mu P_r+\tfrac12\sigma^2P_{rr}-rP$, with all terms evaluated at $(t,r;T)$.

The first term is a local martingale after localization; the full left side is a true martingale. Uniqueness of the continuous local-martingale and finite-variation decomposition therefore implies $\int_0^sD_vA(v,R_v)\,dv=0$ for every $s$, almost surely. To obtain the equation at each point, start the diffusion from an arbitrary pair $(t,r)$. By continuity of the coefficients, $P$ and its displayed derivatives, and the sample paths, the integrand $D_vA(v,R_v)$ is continuous at the starting time and equals $A(t,r)$ there. Its integral is identically zero, so its right derivative at the starting time is zero. Since $D_t=1$ for this restarted discount factor, that derivative is $A(t,r)$. Hence $A(t,r)=0$ for every $t<T$ and every $r$. Finally, as $t$ increases to $T$, the time interval in the pricing expectation shrinks to zero, so the terminal payoff is one; this agrees with the stated terminal condition. $\square$

The PDE is backward because the terminal value is specified at $T$ and the equation determines prices at earlier times. Its second-order term comes from Brownian quadratic variation, while $-rP$ comes from discounting. These are distinct contributions.

Worked Checks of the Equation

Worked Example: A Constant Short Rate

Let $R_s=r$ for all $s$, so $\mu=0$ and $\sigma=0$. For $h=T-t$, the accumulated short rate is $rh$, and therefore $P(t,r;T)=e^{-rh}$. The derivatives are $P_t=re^{-rh}=rP$, $P_r=-he^{-rh}=-hP$, and $P_{rr}=h^2e^{-rh}=h^2P$. The PDE reduces to $P_t-rP=rP-rP=0$; the terminal value is $e^0=1$.

Check: with $r=\tfrac{1}{10}$, $T=2$, and $t=0$, the pricing formula gives $P(0,\tfrac{1}{10};2)=e^{-(1/10)(2)}=e^{-1/5}$. The PDE check at these parameters is still $rP-rP=0$, and at $t=T$ the formula gives $e^0=1$.

Worked Example: A Deterministic Rate with Drift

Let $dR_s=a\,ds$, with $\sigma=0$, so a rate starting at $r$ at time $t$ is $R_s=r+a(s-t)$. Put $h=T-t$. Direct integration gives $\int_t^T R_s\,ds=rh+\tfrac12ah^2$, hence $P(t,r;T)=\exp(-rh-\tfrac12ah^2)$. Its derivatives are $P_t=(r+ah)P$, $P_r=-hP$, and $P_{rr}=h^2P$. The PDE left side is $$ (r+ah)P+a(-hP)-rP=0. $$ The terminal value follows by setting $h=0$.

Check: take $r=\tfrac12$, $a=\tfrac14$, and $h=2$. The integrated rate is $(\tfrac12)(2)+\tfrac12(\tfrac14)(2^2)=1+\tfrac12=\tfrac32$, so the price is $e^{-3/2}$. Substituting these parameters into the PDE terms gives $(\tfrac12+\tfrac14\cdot2)e^{-3/2}-(\tfrac14)(2)e^{-3/2}-(\tfrac12)e^{-3/2}=0$.

Worked Example: A Brownian Short Rate

Let $dR_s=dB_s$, so $\mu=0$ and $\sigma=1$, and start at rate $r$ at time $t$. For $h=T-t$, the random part of $\int_t^T R_s\,ds$ is $Y=\int_t^T(B_s-B_t)\,ds$. This is centered Gaussian. Its variance, using the Brownian covariance $\mathbb{E}^{\mathbb{Q}}[(B_{t+u}-B_t)(B_{t+v}-B_t)]=\min(u,v)$, is $$ \mathbb{E}^{\mathbb{Q}}[Y^2] =\int_0^h\int_0^h\min(u,v)\,du\,dv =2\int_0^h\int_0^v u\,du\,dv =2\int_0^h\frac{v^2}{2}\,dv =\frac{h^3}{3}. $$ The normal moment-generating formula therefore gives $P(t,r;T)=\mathbb{E}^{\mathbb{Q}}[e^{-rh-Y}]=e^{-rh+h^3/6}$. For $P=e^{-rh+h^3/6}$, the derivatives are $P_t=(r-\tfrac12h^2)P$, $P_r=-hP$, and $P_{rr}=h^2P$. Thus $P_t+\tfrac12P_{rr}-rP=(r-\tfrac12h^2)P+\tfrac12h^2P-rP=0$, and $P(T,r;T)=1$.

Check: at $r=0$ and $h=1$, the expectation gives $e^{1/6}$. The PDE terms are $P_t=-\tfrac12e^{1/6}$, $\tfrac12P_{rr}=\tfrac12e^{1/6}$, and $-rP=0$, whose sum is zero. This check also shows why a stochastic rate can produce a bond price above one, even though the payoff is one: some paths have negative accumulated rates.

Verification and the Role of the Assumptions

Verification theorem: Let $u\in C^{1,2}([0,T)\times\mathbb{R})$ be continuous up to $T$, solve $u_t+\mu u_r+\tfrac12\sigma^2u_{rr}-ru=0$, and satisfy $u(T,r)=1$. If, for each starting pair $(t,r)$, the process $D_{t,s}u(s,R_s)$, where $D_{t,s}=\exp(-\int_t^sR_v\,dv)$, is a true martingale through $T$, then $u(t,r)=P(t,r;T)$. A sufficient integrability condition is that its Itô integrand $D_{t,s}\sigma(s,R_s)u_r(s,R_s)$ belongs to $\mathcal{L}^2$ and the terminal value is integrable.

Proof. Itô’s formula, with the derivatives $u_t,u_r,u_{rr}$ and correction term $\tfrac12\sigma^2u_{rr}ds$, followed by the product rule with $dD_{t,s}=-R_sD_{t,s}ds$, gives $$ d\big(D_{t,s}u(s,R_s)\big) =D_{t,s}\sigma(s,R_s)u_r(s,R_s)\,dB_s +D_{t,s}\big(u_t+\mu u_r+\tfrac12\sigma^2u_{rr}-R_su\big)(s,R_s)\,ds. $$ The PDE makes the $ds$ term zero. By the assumed true-martingale condition, conditional expectation at time $t$ gives $u(t,r)=\mathbb{E}^{\mathbb{Q}}_{t,r}[D_{t,T}u(T,R_T)]$. The terminal condition makes $u(T,R_T)=1$, so this conditional expectation is exactly $P(t,r;T)$. A local martingale alone would not justify taking expectations: without integrability or uniform integrability it need not be a true martingale. $\square$

The continuity assumptions address a precise issue. If the time dependence of a coefficient is merely measurable, an integral drift identity can imply that the PDE holds only for almost every time. It cannot in general establish the equation at a prescribed exceptional time. For example, changing a drift coefficient at one time does not change the time integral defining the SDE, but it can change the value of the drift term in a pointwise PDE at that time. Continuity of the coefficients, together with the stated classical regularity of the price, prevents that gap in the derivation.

Common Pitfalls

Omitting discounting: the incorrect equation $P_t+\mu P_r+\tfrac12\sigma^2P_{rr}=0$ leaves out the effect of accumulating the short rate. For the constant-rate example it gives $P_t=rP\ne0$ whenever $rP\ne0$. The correction is the term $-rP$, which makes $P_t-rP=0$ in that example.

Using the wrong drift: the PDE uses $\mu$ under $\mathbb{Q}$. A drift specified under the physical measure $\mathbb{P}$ cannot be inserted unchanged unless it is also the risk-neutral drift. A change of measure can alter the Brownian drift and hence the short-rate drift; pricing here is an expectation under $\mathbb{Q}$.

Claiming too much regularity: Lipschitz and linear-growth conditions give a well-behaved SDE, but by themselves they do not assert that its price function is $C^{1,2}$. The pointwise PDE theorem above explicitly assumes that classical regularity. If it is unavailable, a weaker PDE interpretation requires additional theory and is not being asserted here.

Check Your Understanding

Use the PDE and conditional-expectation definition to organize each response.

For a deterministic constant rate $r$, calculate $P_t$ and verify the PDE and terminal condition.
Identify separately the terms caused by the short-rate drift, Brownian quadratic variation, and discounting.
Explain why continuity in time helps turn an integral drift identity into a pointwise PDE.
For $dR_s=dB_s$, derive the variance of $\int_t^T(B_s-B_t)\,ds$ and use it to recover the bond-price formula.
State why a local-martingale conclusion alone is insufficient for the conditional-expectation step in the verification theorem.