Case study: bond prices under a deterministic short rate

A bank account that follows the short rate

The short rate $r(t)$ is the interest rate a bank pays at time $t$ on money left with it for the next instant. It is a rate per year, compounded continuously (case study 1.4.6). This case study takes it as known in advance, a deterministic short rate: $r(t) = 0.03 + 0.01t$, with $t$ in years.

Let $B(t)$ be the balance of an account that starts with $B(0) = 1$. Over a short time $\Delta t$ it earns about $r(t)\,B(t)\,\Delta t$. Dividing by $\Delta t$ and letting it shrink gives a differential equation.

The letter $s$ inside the integral is the time of each instant, since $t$ is already the upper limit. The integral is the area under the short rate (lesson 2.5.3), a pure number: a rate per year times years. For the rising rate,

After 10 years that area is 0.8, so the account holds $B(10) = e^{0.8} \approx 2.225541$.

The price of a zero-coupon bond

A zero-coupon bond pays 1 at its maturity $T$ and nothing before (case study 2.2.4). Write $P(0, T)$ for its price today. The rates are known, so the bond and the account are both riskless, and they must grow alike.

Put $P(0, T)$ in the account today: at $T$ it has grown to $P(0, T)\,B(T)$, which must equal the bond's 1. Any other price allows an arbitrage (case study 3.3.4). If the bond were cheaper, you could borrow from the account, buy bonds, and keep a profit with nothing to pay later.

At one year the two are close. The gap widens with maturity, because each extra year is discounted at a higher rate than the last.

The price solves an equation in the maturity

Seen as a function of its maturity, the price solves a differential equation of its own. The fundamental theorem (lesson 2.5.3) gives $\frac{d}{dT}\int_0^T r(s)\,ds = r(T)$, and the chain rule then gives

A bond maturing now is worth its payment. Each extra instant of maturity is discounted at the short rate of that instant.

The explorer below draws this equation, $P' = -(0.03 + 0.01T)\,P$, with $y$ for $P$ and $t$ for $T$; the exact curve ends at 0.44933. Euler's method with $h = 1$ lands on 0.45637, too high by 0.00705. Try $h = 0.5$ and $h = 0.1$: the error falls to 0.00331, then 0.00063.

Yields and forward rates

The continuous yield $y(T)$ is the one constant rate that gives the same price: $P(0, T) = e^{-y(T)\,T}$. It is case study 1.6.4's continuous zero rate. Taking logs gives it from the price.

For the rising rate, $y(T) = 0.03 + 0.005T$: 3.5% at one year, 8% at ten and 13% at twenty. The yield curve, the yield against maturity, rises, but half as steeply as the short rate.

The instantaneous forward rate

Case study 1.6.4 read one-year forward rates off a curve, $1 + f = \frac{D_{n-1}}{D_n}$: the growth the curve implies for one year alone. With continuous rates, the forward rate from $T_1$ to $T_2$ is $\frac{1}{T_2 - T_1}\ln\frac{P(0, T_1)}{P(0, T_2)}$. Shrink the gap to an instant and it becomes a derivative.

At $T = 10$ the forward rate is 13%, and a central difference of $-\ln P$ (lesson 2.8.1) agrees. The one-year forward rate from 9 to 10 is 12.5%, the short rate at mid-year, 9.5. The yield is the average of the forward rates up to $T$, so on a rising curve it trails them: 8% against 13% at ten years.

Checking with solve_ivp

The exact price needs $\int r$ in closed form; solve_ivp (lesson 6.1.2) needs only $r$ itself. On $\frac{dP}{dT} = -r(T)\,P$ from $P = 1$, with $\text{rtol} = 10^{-10}$ and $\text{atol} = 10^{-12}$, it matches the table at every maturity to about $10^{-10}$. With the default tolerances the 10-year price comes out 0.449235, off by about $10^{-4}$: too rough for a price.

Backwards from maturity

A bond also has a value at each date $t$ before maturity: $V(t) = \exp\big(-\int_t^T r(s)\,ds\big)$. It solves $\frac{dV}{dt} = r(t)\,V$, and what is known is its end: $V(T) = 1$. A condition at the final time is a terminal condition, and the equation is solved backwards from it.

Given a t_span that runs from 10 down to 0, solve_ivp steps backwards and lands on 0.449329. Option values are found the same way, backwards from their payoff (lesson 6.3.2 and case study 6.3.4).

A short rate from a table

A desk's short-rate curve comes from market quotes at a few dates, joined by straight lines with np.interp (lesson 1.6.1). Take rates of 3.0%, 4.2%, 5.5%, 6.0% and 6.2% at 0, 1, 3, 5 and 10 years: there is no formula to integrate.

The area under straight-line pieces is exactly the trapezoid rule on the knots (lesson 2.8.2), so $e^{-\text{area}}$ is the exact price: 0.575222 at 10 years, a yield of 5.53%. solve_ivp agrees to within $10^{-8}$; the corners in the rate cost it some of the accuracy that rtol asks for.

A stock that pays dividends

The same equations price a forward contract on a stock: an agreement today to buy the stock at time $T$ for a price $F$, fixed now and paid at $T$. Nothing changes hands today, so the fair $F$ is the one that makes the deal worth nothing now.

Copy the deal with the bank: borrow to buy the stock today, hold it to $T$, and repay the loan then. Whatever you owe at $T$ is the fair forward price, because at $T$ you hold one share either way. Dividends paid along the way change both sides.

A dividend yield

Say the stock pays dividends continuously at a dividend yield $q$: per year, $q$ times its price, reinvested at once in more shares. The number of shares $N(t)$ then grows by $\frac{dN}{dt} = qN$, a separable equation with $N(T) = N(0)\,e^{qT}$.

To hold one share at $T$, buy $e^{-qT}$ shares today: 0.818731 of a share for $q = 2\%$ and $T = 10$. Borrow their cost, $S_0 e^{-qT}$, from the account; by the account equation the debt grows to $S_0e^{-qT}B(T)$ at $T$.

With $S_0 =$ $100 and $T = 10$, the forward price is $222.5541 without dividends and $182.2119 with $q = 2\%$. That is $100e^{0.6}$: the yield takes 0.2 off the area of 0.8 under the short rate.

Dividends in dollars

Now say the stock pays a fixed $2 a year in cash, continuously, instead of a yield. Buy one share today with a loan of $S_0 =$ $100, and use the dividends to pay the loan down. The loan $L(t)$ grows with interest and shrinks with the dividends:

This is linear (lesson 6.1.2), with $p = -r(t)$. Its integrating factor $e^{-\int_0^t r(s)\,ds}$ is the discount factor $P(0, t)$ itself. So $\big(P(0, t)\,L\big)' = -c\,P(0, t)$, and integrating from 0 to $T$ gives the forward price, the loan left at $T$:

Completing the square in $-(0.03t + 0.005t^2)$ turns $P(0, t)$ into a bell curve, which has no antiderivative formula (lesson 2.5.2), so quad finds the area: $\int_0^{10} P(0, t)\,dt \approx 7.480262$. The dividends pay $20 in all, worth $14.9605 today, and $F \approx$ $189.2588. Solving the loan equation with solve_ivp gives the same to about $10^{-9}$.

import math

import numpy as np


def rising(t):
    return 0.03 + 0.01 * t


def test_rising_rate_price():
    """Under the short rate 0.03 + 0.01t, 1 paid in 10 years costs 0.449329 today"""
    expect_close(zero_price(rising, 10), math.exp(-0.8), what="zero_price(rising, 10)", rel=1e-8)


def test_flat_rate_price():
    """A flat 3% for 5 years gives the price e^(-0.15)"""
    expect_close(zero_price(lambda t: 0.03, 5), math.exp(-0.15), what="zero_price(flat 3%, 5)", rel=1e-8)


def test_maturity_now():
    """A bond maturing now is worth 1"""
    expect_close(zero_price(rising, 0), 1.0, what="zero_price(rising, 0)")


def test_negative_maturity():
    """A negative maturity raises ValueError"""
    try:
        zero_price(rising, -1)
    except ValueError:
        return
    raise AssertionError("expected ValueError for T = -1")


def test_negative_rate():
    """A flat -0.5% for 5 years makes the bond cost more than 1: e^0.025"""
    expect_close(zero_price(lambda t: -0.005, 5), math.exp(0.025), what="zero_price(flat -0.5%, 5)", rel=1e-8)


def test_yield_is_average_rate():
    """The 10-year yield under the rising rate is its average, 8%"""
    expect_close(zero_yield(rising, 10), 0.08, what="zero_yield(rising, 10)", rel=1e-7)


def test_yield_needs_positive_maturity():
    """A yield at maturity 0 raises ValueError"""
    try:
        zero_yield(rising, 0)
    except ValueError:
        return
    raise AssertionError("expected ValueError for T = 0: the yield divides by T")


def test_forward_rate():
    """The forward rate from 9 to 10 years is 12.5%, the short rate at 9.5"""
    expect_close(forward_rate(rising, 9, 10), 0.125, what="forward_rate(rising, 9, 10)", rel=1e-6)


def test_forward_rate_order():
    """A forward rate from 5 back to 2 raises ValueError"""
    try:
        forward_rate(rising, 5, 2)
    except ValueError:
        return
    raise AssertionError("expected ValueError when T2 is not after T1")


def test_rate_from_a_table():
    """A rate joined by straight lines between quotes gives e^(-area) by the trapezoid rule: 0.575222"""
    knots, rates = [0, 1, 3, 5, 10], [0.030, 0.042, 0.055, 0.060, 0.062]
    price = zero_price(lambda t: np.interp(t, knots, rates), 10)
    expect_close(price, math.exp(-np.trapezoid(rates, knots)), what="zero_price(table rate, 10)", rel=1e-7)