Case study: bond price sensitivity and duration

Price as a function of yield

A bond pays a coupon $c$ at the end of each year and its face value $F$ after $T$ years. At the yield $y$ (a decimal: 5% is 0.05), its price is the present value of those payments (case study 1.5.5):

A higher yield divides every payment by more, so the price falls as the yield rises. Its derivative $P'(y)$, written $\frac{dP}{dy}$, is negative: the graph slopes down (lesson 2.2.2).

The tangent at 5% is the duration picture: its slope is $P'(0.05)$. The rest of this case study measures that slope, puts it in traders' units, and tests how far the tangent holds.

Bumping the yield

The formula is a sum of powers of $1 + y$. Differentiating it exactly needs the rules of module 2.3. Instead, do what lesson 2.2.2 did with a function you can only evaluate: move the yield a little each way and reprice.

How big a bump?

Lesson 2.2.2 showed that a finite difference fails if $h$ is too big or too small. The same holds here.

Units: per unit, per 1%, per basis point

The slope $-420.98$ is in price per unit of yield. One unit is a move of 1 in $y$: 100 percentage points. No market moves that far, so the slope is rescaled to a realistic move.

The last column has a name. The DV01 (dollar value of an 01, a basis point) is the price fall for a 1 bp rise in yield. It is quoted as a positive number.

A slope of $-420.98$, a change of $-4.21$ per 1% and a DV01 of 0.0421 describe the same bond. Always check which unit a number is in before comparing two of them.

Duration

A slope in dollars grows with the size of the position. Dividing by the price turns it into a percentage, which compares bonds of any size.

For the 5-year 4% bond, $D = 420.98 / 95.67 \approx 4.4003$. Other bonds at the same 5% yield:

Longer bonds are more sensitive. For the same maturity, a lower coupon means a higher duration, because more of the money arrives at the end. Duration also moves with the yield: the 5-year 4% bond has 4.5042 at 3% and 4.2997 at 7%.

The zero-coupon bond: the clean case

A zero-coupon bond pays only its face value, at time $T$: $P(y) = \frac{F}{(1 + y)^T}$. Its duration is exactly $\frac{T}{1 + y}$: for 5 years at 5%, $\frac{5}{1.05} = 4.7619$, as the bump found.

So for a zero, duration is its maturity divided by $1 + y$. A coupon bond is a bundle of payments at different times. Weight each payment's time by its share of the price and average: that is the Macaulay duration, $D_{\text{Mac}}$.

For the 5-year 4% bond, 85% of the value arrives at year 5 and the rest earlier, so the average time is 4.62 years. Dividing by $1 + y$ gives 4.40, the bumped duration. For yearly payments, $D = \frac{D_{\text{Mac}}}{1 + y}$; case study 2.3.7 shows why, with the differentiation rules.

How far does the tangent go?

The derivative gives the tangent line of lesson 2.2.1: $P(y + \Delta y) \approx P(y) + P'(y)\,\Delta y$, where $\Delta y$ (delta y) is the change in yield. Here it is against the true price, both ways.

For a 1 bp move the tangent is off by only 0.00001. For 1% it is off by about 0.12, and for 2% by about 0.45 to 0.49. Doubling the move roughly quadruples the error, like the $(x - 3)^2$ gap in lesson 2.2.1.

The true price is above the tangent in every row. So duration overstates the loss when yields rise, and understates the gain when they fall. That bend is convexity, the subject of case study 2.4.5.

The $-1\%$ row is a yield of 4%, equal to the coupon rate. There the price is exactly 100: a bond whose yield equals its coupon rate is worth its face value.

def test_slope():
    """The 5-year 4% bond at 5% has dP/dy of about −420.98"""
    expect_close(price_slope(4, 100, 5, 0.05), -420.97969897625376, what="price_slope(4, 100, 5, 0.05)", rel=1e-6)


def test_slope_is_negative():
    """Prices fall as yields rise, so the slope is negative"""
    s = price_slope(6, 100, 10, 0.03)
    assert s < 0, f"expected a negative slope, got {s}"


def test_dv01():
    """Its DV01 is about 0.0421 per 100 of face value, a positive number"""
    expect_close(dv01(4, 100, 5, 0.05), 0.042097969897625376, what="dv01(4, 100, 5, 0.05)", rel=1e-6)


def test_dv01_scales_with_face():
    """On 10 million of face value (coupon 400,000 a year) the DV01 is about 4,209.80"""
    expect_close(dv01(400_000, 10_000_000, 5, 0.05), 4209.796989762537, what="dv01(400000, 10000000, 5, 0.05)", rel=1e-6)


def test_duration():
    """Its modified duration is about 4.40"""
    expect_close(modified_duration(4, 100, 5, 0.05), 4.4003072662927565, what="modified_duration(4, 100, 5, 0.05)", rel=1e-6)


def test_zero_coupon():
    """A 5-year zero-coupon bond at 5% has duration 5 / 1.05"""
    expect_close(modified_duration(0, 100, 5, 0.05), 5 / 1.05, what="modified_duration(0, 100, 5, 0.05)", rel=1e-6)


def test_zero_yield():
    """At a yield of exactly 0 the 5-year 4% bond has duration about 4.6667 (no division by zero)"""
    expect_close(modified_duration(4, 100, 5, 0.0), 4.666666981666623, what="modified_duration(4, 100, 5, 0.0)", rel=1e-6)


def test_longer_is_riskier():
    """A 10-year bond has a higher duration than a 5-year one"""
    d5, d10 = modified_duration(4, 100, 5, 0.05), modified_duration(4, 100, 10, 0.05)
    assert d10 > d5, f"expected the 10-year duration ({d10:.3f}) above the 5-year ({d5:.3f})"