Duration Misses $4.58 on a One-Point Move, and Convexity Hands Back $4.84

A 20-year 7 percent bond on a flat 10 percent curve prices at 744.5931, and a one-point rise costs exactly 63.1262. The tangent alone says 67.7028, and adding the second-order term of 4.8416 lands at 62.8612, inside 27 cents of the truth. Note that the correction and the error it corrects are two different numbers, which is why the estimate ends up on the wrong side of the answer.

A bond with twenty years to run pays a seven percent annual coupon, and the whole curve sits flat at ten percent. It is worth 744.5931. Rates rise by one point. How much do you lose?

The straight-line answer is 67.70 and it is arithmetically flawless, which is what makes it worth examining. The true loss is 63.1262. One extra term of a Taylor series recovers 94 percent of that gap and lands inside 27 cents, and the term has a name and a meaning of its own.

The bond, priced from its cash flows

Everything here is built from the payments themselves rather than from a quoted price, so no step is a black box. On a face of 1000 the bond pays 70 a year for twenty years and then 1000 back, and a flat annual yield of ten percent discounts all twenty-one cash flows:

P(y) \;=\; \sum_{k=1}^{20}\frac{70}{(1+y)^k} \;+\; \frac{1000}{(1+y)^{20}}, \qquad P(0.10) = 744.5931

(1)

Repricing the identical cash flows at eleven percent gives 681.4669, so the exact change is $\;-63.1262$ . That number is the target every approximation below is aiming at, and it required no calculus at all.

Duration is an average that happens to be a derivative

Duration and convexity

The Macaulay duration is the cash-flow-weighted average time to payment, $D = \sum_k t_k w_k$ with $w_k = \mathrm{PV}(\text{CF}_k)/P$ . The modified duration is $D_{\mathrm{mod}} = D/(1+y)$ , and it is exactly $-P'(y)/P$ . Convexity is $C = P''(y)/P$ .

The two readings of duration are worth holding at once. As an average it says when the money arrives; as a derivative it says how the price moves. For this bond the average lands at

D = 10.0018 \text{ years} \quad\Longrightarrow\quad D_{\mathrm{mod}} = \frac{10.0018}{1.10} = 9.0926

(2)

A twenty-year bond that moves like a ten-year zero-coupon bond, because four fifths of its present value arrives as coupons before maturity. The 1.8 thousandths past 10 are worth keeping. A rounded duration of exactly 10.00 gives a tangent of 67.6903 rather than 67.7028, and the 1.25 cent difference is the sort of thing that gets blamed on a convention later.

Duration also behaves as the average reading predicts. A zero-coupon bond of this maturity has a duration of exactly 20 years, and raising the coupon pulls the average forward: 11.3039 years at a four percent coupon, 10.0018 at seven, 9.3649 at ten, and 8.8105 at fifteen.

What the straight line predicts

The first-order term of the Taylor expansion of $P$ around ten percent is a straight line through the current price with slope $P'(y)$ :

\Delta P \;\approx\; -D_{\mathrm{mod}}\,P\,\Delta y \;=\; -9.0926 \times 744.5931 \times 0.01 \;=\; -67.7028

(3)

Against a truth of $-63.1262$ , the line overstates the loss by 4.5766. It has to overstate it, because the price curve is convex and a tangent to a convex curve lies below it everywhere. The error is not a mistake in the arithmetic. It is the price paid for replacing a curve by a line.

Fig. 1 — The tangent is below the curve at every yield except the point of contact. Over one point the gap is under two pixels at this scale, which is why it feels ignorable and why it is worth 4.58.

The second term, and why it overshoots

Differentiating (1) twice and dividing by the price gives convexity as another weighted sum over the same cash flows:

C \;=\; \frac{1}{P}\sum_k \frac{t_k(t_k+1)\,\text{CF}_k}{(1+y)^{t_k+2}} \;=\; 130.0468

(4)

Adding the second-order term to (3) gives the two-term estimate:

\Delta P \;\approx\; -D_{\mathrm{mod}}P\,\Delta y + \tfrac{1}{2}\,C\,P\,(\Delta y)^2 \;=\; -67.7028 + 4.8416 \;=\; -62.8612

(5)

Against the true $-63.1262$ that is 26.50 cents out, where the tangent alone was 4.5766 out. One extra term removed 94.21 percent of the error.

Here is the detail that gets garbled most often, and it is worth being pedantic about. The correction is 4.8416 and the error it was correcting was 4.5766. Those are two different numbers, differing by the 26.50 cents that remains. A sentence like “duration alone misprices the move by 4.84” states the size of the fix as the size of the problem. The correction slightly overshoots, which is exactly what a truncated alternating expansion does: the third-order term is negative here, and it is what the 26.50 cents is.

Fig. 2 — The correction is larger than the error it corrects, which is why the result sits on the wrong side of the truth by 26.50 cents.

The result is not a property of this one bond. Across twenty combinations of coupon and maturity the second-order term removes at least 90 percent of the first-order error every time, with a worst case of 90.09 percent and a best of 97.96.

The bend pays on both sides

Run the same move downwards. At nine percent the bond is worth 817.4290, a gain of 72.8360. The tangent predicts the same 67.7028 it predicted for the loss, so this time it understates by 5.1332, and the same positive correction of 4.8416 lands within 29.16 cents.

Look at the two exact numbers together. A one-point fall gains 72.8360 and a one-point rise loses 63.1262, a difference of 9.7098 in the holder's favour on moves of identical size. That asymmetry is convexity. It is the same quantity, read as a payoff rather than as a correction term, and it is why convexity is something a buyer will pay for rather than a numerical nuisance.

The correction degrades with the size of the move, since it is one term of a series. At three points the exact loss is 166.0782, the tangent says 203.1083, and the two-term estimate lands at 159.5340, still 6.54 out. That is 82.33 percent of the error removed rather than 94.21. Each further point of yield puts more weight on the terms that were dropped.

What duration matching actually protects

Matching durations between assets and liabilities is usually described as immunisation, and the word oversells what has been bought. Three things limit it, and none is a technicality.

The protection is first-order only. Equation (5) is the reason: two portfolios with the same duration and different convexities respond differently to the same shift, and the difference grows as the square of the move. Matching duration and leaving convexity unmatched leaves the second term unhedged.

The protection is also instantaneous. Duration changes as time passes and as yields move, so a match struck today is a match only today. Nothing in equation (2) is constant, and a portfolio left alone drifts out of alignment without anything happening in the market at all.

And the whole framework assumes the curve moves in parallel. A single yield $y$ stands in for the entire term structure here, which is what allows one derivative to summarise everything. When the curve steepens instead of shifting, two portfolios with identical duration and identical convexity can move in opposite directions, and the extra convexity someone paid for can cost them money. Convexity is worth having for what it does in equation (5), and it is worth being clear that this is the only thing it promises.

Sources and further reading

Wikipedia: Bond duration and Bond convexity, for the definitions and the standard conventions.
Wikipedia: Taylor series, since equation (5) is nothing more than the first two terms of one.
Frederick Macaulay, Some Theoretical Problems Suggested by the Movements of Interest Rates, Bond Yields and Stock Prices in the United States since 1856 (1938), where the weighted average in (2) was introduced.

Every price in this article was built cash flow by cash flow in exact rational arithmetic, so nothing quoted depends on a floating-point rounding. Duration and convexity were each computed twice by unrelated routes, once from the cash-flow weights and once as numerical first and second derivatives of the price with no duration formula anywhere in that code path, agreeing to six and four decimal places respectively.