Short-Rate Discounting and Zero-Coupon Bond Prices

Short Rates, Accumulation, and Discounting

A short rate describes the instantaneous rate of return on a money-market account. A zero-coupon bond, by contrast, promises one payment at a specified future date and makes no interim payments. To value it, one discounts that future payment at the short rates encountered along the way, then takes a conditional expectation under the pricing measure. When rates are deterministic, this procedure reduces to an ordinary exponential formula; when rates are random, the conditional expectation matters.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$. For pricing on a horizon $[0,T]$, suppose there is an equivalent risk-neutral measure $\mathbb{Q}$ on $\mathcal{F}_T$. The physical measure $\mathbb{P}$ describes the real-world probability model; $\mathbb{Q}$ is the measure used for risk-neutral valuation. The formulas below assume that the money-market account is the numeraire and that the relevant discounted traded prices are martingales under $\mathbb{Q}$, as in risk-neutral pricing.

No Ito stochastic integral is needed here. A stochastic integral, if used, is the Ito integral $\int_0^t H_s\,dB_s$ for a progressively measurable integrand $H_s$ in $\mathcal{L}^2$, meaning $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. The time integrals of rates below are ordinary pathwise Lebesgue integrals.

Definition: The short rate $r_t$ is the instantaneous rate of return on the money-market account. Given an adapted rate process with $\int_0^T |r_s|\,ds<\infty$ almost surely, the account with initial value $A_0=1$ is $$ A_t=\exp\left(\int_0^t r_s\,ds\right). $$ It is positive and satisfies $dA_t=r_tA_t\,dt$ for almost every $t$, by ordinary calculus for absolutely continuous functions. The discount factor from $t$ to $T$ is $D(t,T)=A_t/A_T=\exp(-\int_t^T r_s\,ds)$.

The account is an accumulation process: a unit invested at time $t$ becomes $A_T/A_t$ units at time $T$. Thus $D(t,T)$ converts one unit payable at $T$ into its discounted amount at $t$. This definition allows negative rates; in that case the account can decrease over time and the discount factor can exceed one.

Zero-Coupon Bond Price

Definition: A zero-coupon bond maturing at $T$ pays one unit of currency at $T$ and nothing before then. Provided $D(t,T)$ is integrable under $\mathbb{Q}$, its risk-neutral price at $t\le T$ is $$ P(t,T)=\mathbb{E}^{\mathbb{Q}}\left[\left.\exp\left(-\int_t^T r_s\,ds\right)\right|\mathcal{F}_t\right]. $$ In particular, $P(T,T)=1$. If the bond pays a fixed face amount $N$ rather than one unit, its price is $N P(t,T)$.

The conditional expectation is essential: at time $t$, the future path of a random short rate need not be known. The formula discounts the payoff along each possible path and averages those discounted payoffs using the conditional probabilities under $\mathbb{Q}$. It is not, in general, the exponential of the conditional expected rate.

Worked Example: A Constant Short Rate

Suppose $r_s=\tfrac{1}{20}$ for all $s$, and find the price at $t=1$ of a unit zero-coupon bond maturing at $T=3$.

The rate is constant, so the accumulated rate from 1 to 3 is $$ \int_1^3 r_s\,ds=\int_1^3 \frac{1}{20}\,ds=\frac{1}{20}(3-1)=\frac{1}{10}. $$ The discount factor is deterministic, so its conditional expectation is itself: $$ P(1,3)=\mathbb{E}^{\mathbb{Q}}\left[\left.e^{-1/10}\right|\mathcal{F}_1\right]=e^{-1/10}. $$

Check: the account grows from $A_1=e^{1/20}$ to $A_3=e^{3/20}$. Their ratio is $A_1/A_3=e^{1/20-3/20}=e^{-1/10}$, exactly the bond price. For a positive short rate, the unit bond price is below one.

Theorem: If $r_s=r(s)$ is a deterministic, integrable function on $[0,T]$, then for every $0\le t\le T$, $$ P(t,T)=\exp\left(-\int_t^T r(s)\,ds\right). $$ This formula holds under any probability measure because its right-hand side is deterministic.

Proof. Since $r(s)$ is deterministic and integrable, the random variable $D(t,T)=\exp(-\int_t^T r(s)\,ds)$ is a finite deterministic number. The conditional expectation of a deterministic integrable random variable given $\mathcal{F}_t$ equals that same number: it is $\mathcal{F}_t$-measurable and its integral over every event in $\mathcal{F}_t$ equals the integral of the original variable. Substituting $D(t,T)$ into the bond-pricing definition gives the stated formula, including $t=T$, when the integral is zero and the price is one. $\square$

Computing Deterministic-Rate Prices

For a deterministic rate curve, the task is to integrate the rate over the remaining life of the bond. The rate before $t$ does not enter $P(t,T)$; it has already affected the account’s value, but the price is expressed in currency units at time $t$. A piecewise rate must be integrated separately over each interval it covers.

Worked Example: A Piecewise Constant Rate

Let $r(s)=\tfrac{1}{20}$ for $0\le s<2$ and $r(s)=\tfrac{1}{10}$ for $2\le s\le4$. Find $P(1,4)$.

The interval from 1 to 4 has length 1 at the first rate and length 2 at the second rate. Hence $$ \int_1^4 r(s)\,ds =\int_1^2 \frac{1}{20}\,ds+\int_2^4 \frac{1}{10}\,ds =\frac{1}{20}(2-1)+\frac{1}{10}(4-2) =\frac{1}{20}+\frac{2}{10} =\frac{1}{4}. $$ Therefore $P(1,4)=e^{-1/4}$.

Check: the corresponding account ratio is $A_1/A_4$. The account has accumulated $\tfrac{1}{20}$ from 0 to 1, then $\tfrac{1}{20}$ from 1 to 2, and $\tfrac{2}{10}$ from 2 to 4. Thus $A_4/A_1=e^{1/20+2/10}=e^{1/4}$, so $A_1/A_4=e^{-1/4}$, as obtained from the bond formula.

For $t<T$, the continuously compounded yield $y(t,T)$ of a positive unit bond is defined by $P(t,T)=e^{-y(t,T)(T-t)}$, or equivalently $y(t,T)=-\log P(t,T)/(T-t)$. Under deterministic rates, the theorem gives $$ y(t,T)=\frac{1}{T-t}\int_t^T r(s)\,ds. $$ Thus the yield is the average short rate over the remaining interval, not necessarily the rate at time $t$. The definition excludes $t=T$, where the denominator is zero.

Worked Example: A Linearly Changing Short Rate

Suppose $r(s)=\tfrac{1}{10}+\tfrac{s}{100}$ for $0\le s\le3$. Find the price at $t=1$ of a unit bond maturing at $T=3$, and its yield.

Integrating the constant and linear terms separately gives $$ \int_1^3 r(s)\,ds =\frac{1}{10}(3-1)+\frac{1}{100}\left[\frac{s^2}{2}\right]_1^3 =\frac{2}{10}+\frac{1}{200}(9-1) =\frac{1}{5}+\frac{1}{25} =\frac{6}{25}. $$ The bond price is $P(1,3)=e^{-6/25}$. Since the remaining life is $3-1=2$, the yield is $$ y(1,3)=-\frac{\log(e^{-6/25})}{2}=\frac{3}{25}. $$

Check: the average rate over the interval is $(6/25)/2=3/25$, matching the yield formula. The endpoint rates are $r(1)=\tfrac{11}{100}$ and $r(3)=\tfrac{13}{100}$; their average is $\tfrac{12}{100}=\tfrac{3}{25}$, also consistent with a linear rate curve.

Discounted Bond Prices and the Martingale Property

Proposition: Suppose $D(0,T)=1/A_T$ is integrable under $\mathbb{Q}$, and define $P(t,T)$ by risk-neutral conditional discounting for $0\le t\le T$. Then the discounted bond price $P(t,T)/A_t$ is a $\mathbb{Q}$-martingale. This statement concerns the price divided by the money-market account; the undiscounted bond price need not be a martingale.

Proof. Since $A_t$ is positive and $\mathcal{F}_t$-measurable, and $D(t,T)=A_t/A_T$, the defining conditional expectation gives $$ \frac{P(t,T)}{A_t} =\frac{1}{A_t}\mathbb{E}^{\mathbb{Q}}\left[\left.\frac{A_t}{A_T}\right|\mathcal{F}_t\right] =\mathbb{E}^{\mathbb{Q}}\left[\left.\frac{1}{A_T}\right|\mathcal{F}_t\right]. $$ Write $X=1/A_T$. The assumption $\mathbb{E}^{\mathbb{Q}}[X]<\infty$ makes each conditional expectation integrable. For $0\le s\le t\le T$, the tower property gives $$ \mathbb{E}^{\mathbb{Q}}\left[\left.\frac{P(t,T)}{A_t}\right|\mathcal{F}_s\right] =\mathbb{E}^{\mathbb{Q}}\left[\left.\mathbb{E}^{\mathbb{Q}}[X\mid\mathcal{F}_t]\right|\mathcal{F}_s\right] =\mathbb{E}^{\mathbb{Q}}[X\mid\mathcal{F}_s] =\frac{P(s,T)}{A_s}. $$ This is precisely the martingale property, including the case $s=t$. $\square$

In the deterministic-rate case, $P(t,T)/A_t=1/A_T$, a constant in time. With stochastic rates, it is instead the conditional expectation of the future discounted payoff, which can change as information arrives. The martingale assertion does not say that $P(t,T)$ itself is a martingale: the account $A_t$ is the numeraire used for discounting.

Why Averaging Rates First Can Fail

Worked Example: A Random Future Rate

Suppose a one-year bond is valued at time 0, and the short rate throughout the year is either $0$ or $\log 2$, each with $\mathbb{Q}$-probability $\tfrac12$. Assume the outcome is not known at time 0. Find the bond price and compare it with discounting at the expected rate.

In the first outcome the discount factor is $e^0=1$; in the second it is $e^{-\log 2}=\tfrac12$. Therefore $$ P(0,1)=\frac12\cdot1+\frac12\cdot\frac12=\frac34. $$ The expected rate is $\tfrac12\log 2$. Substituting that rate into a deterministic discount formula would give $e^{-(\log 2)/2}=1/\sqrt{2}$, which differs from $\tfrac34$. Indeed, $1/\sqrt{2}\ne\tfrac34$, since their squares are $\tfrac12$ and $\tfrac{9}{16}$, respectively.

Check: the conditional-expectation formula averages the two pathwise discount factors, $1$ and $\tfrac12$, and so gives $\tfrac34$. The alternative discounts once at the average rate and does not perform that average. The two procedures are not interchangeable.

Common mistake. A tempting but generally incorrect calculation is $$ P(t,T)\stackrel{\text{wrong}}{=}\exp\left(-\int_t^T \mathbb{E}^{\mathbb{Q}}[r_s\mid\mathcal{F}_t]\,ds\right). $$ The bond price is the conditional expectation of an exponential, not the exponential of an expected integral. The random-rate example gives the concrete discrepancy $\tfrac34\ne1/\sqrt{2}$. The deterministic formula is valid because the integrated rate is already known, not because expectation and exponentiation can generally be interchanged.

These definitions provide the starting point for pricing bonds when rates evolve randomly. In that setting, calculating the conditional expectation may require a model for the rate process. The essential distinction remains: the drift and diffusion of a short-rate model describe the rate’s dynamics, while the bond price is the risk-neutral conditional value of its discounted payoff.

Check Your Understanding

Use the definitions and results above; in each question, assume the stated conditional expectations are well-defined.

If $r_s=c$ is a constant, derive $P(t,T)$ and identify the conditions under which it is below, equal to, or above one.
For a deterministic rate $r(s)=a+bs$, integrate over $[t,T]$ and express the bond yield as an average rate.
Explain why the discounted price $P(t,T)/A_t$, rather than necessarily $P(t,T)$, is a martingale under $\mathbb{Q}$.
For a deterministic rate that is negative throughout $[t,T]$, determine whether the unit bond price is above or below one and justify your answer from the discounting formula.
In the random-rate example, compute the variance of the discount factor and explain how it confirms that replacing the random rate by its mean loses information.