Unit 3, lesson 19

Duration convexity and key rate sensitivities

How Macaulay duration, modified duration and convexity follow from yield derivatives of a bond's price, how the second-order price approximation compares with repricing, and what key rate durations add to a parallel yield shift.

  • Video 19
  • Bonds, discount curves and rate sensitivities

If a bond’s yield rises by one basis point, how much does its price fall, and how far can that answer be trusted for a move of a whole percentage point? For fixed payments the price is a smooth function of the yield, so the answer comes from its first two derivatives, duration and convexity, each tied to a stated compounding convention. This lesson derives both, compares the resulting approximations with exact repricing, and shows why one parallel duration cannot describe how a bond responds when different maturities of the curve move by different amounts.

Key terms

Macaulay duration

Definition Consider fixed payments C1,…,CnC_1, \dots, C_n at times 0<t1<⋯<tn0 < t_1 < \dots < t_n in years, with no default and no dependence of the payments on rates, and the price-yield function under annual compounding

V(y)=∑k=1nCk (1+y)−tkV(y) = \sum_{k=1}^{n} C_k \,(1+y)^{-t_k}

for y>−1y > -1, as in Bond cash flows and yield mathematics. At a yield yy with V(y)>0V(y) > 0, the present-value weights and the Macaulay duration are

wk=Ck (1+y)−tkV(y),DMac=∑k=1ntk wk,\begin{gathered} w_k = \frac{C_k \,(1+y)^{-t_k}}{V(y)}, \\[0.4em] D_{\text{Mac}} = \sum_{k=1}^{n} t_k \, w_k , \end{gathered}

measured in years. The weights always add up to 11. If every Ck≥0C_k \ge 0 and at least one is positive, the weights are nonnegative, so DMacD_{\text{Mac}} is a weighted average of payment times and lies between the earliest and the latest date with a positive payment. It equals TT for a zero-coupon bond maturing at TT. If some payments are negative, some weights are negative, and DMacD_{\text{Mac}} can lie outside the range of payment dates.

In plain words Macaulay duration is the average time at which the bond pays, with each date weighted by its share of today’s price. It is a time, not yet a sensitivity, and the next card links the two.

Example For the constructed bond paying 55 at year 1 and 105105 at year 2 at y=0.06y = 0.06, the present values are about 4.7169814.716981 and 93.44962693.449626, the price is about 98.16660798.166607, and the time-weighted present values add up to 1⋅4.716981+2⋅93.449626≈191.6162341 \cdot 4.716981 + 2 \cdot 93.449626 \approx 191.616234. Hence

DMac≈191.61623498.166607≈1.951949\begin{aligned} D_{\text{Mac}} &\approx \frac{191.616234}{98.166607} \\[0.4em] &\approx 1.951949 \end{aligned}

years. About 95.2 percent of the value sits at year 2, so the duration is close to 2 and below it. For a contrast, a contract paying 230230 at year 1 and requiring the holder to pay 132132 at year 2 is worth 100100 at y=0.10y = 0.10, with present values of about 209.0909209.0909 and −109.0909-109.0909. Its Macaulay duration is (209.0909−218.1818)/100≈−0.0909(209.0909 - 218.1818)/100 \approx -0.0909 years, outside the dates 11 and 22.

Modified duration

Definition The modified duration is the negative logarithmic derivative of the price in the yield,

Dmod=−V′(y)V(y)=DMac1+y.D_{\text{mod}} = -\frac{V'(y)}{V(y)} = \frac{D_{\text{Mac}}}{1+y} .

The second equality follows from V′(y)=−(1+y)−1∑ktkCk(1+y)−tkV'(y) = -(1+y)^{-1} \sum_k t_k C_k (1+y)^{-t_k} under annual compounding. With mm compounding periods per year, V(y)=∑kCk(1+y/m)−mtkV(y) = \sum_k C_k (1 + y/m)^{-m t_k} and Dmod=DMac/(1+y/m)D_{\text{mod}} = D_{\text{Mac}}/(1 + y/m), with the weights of that convention. Under continuous compounding, V(y)=∑kCke−ytkV(y) = \sum_k C_k e^{-y t_k} and the two durations coincide. Modified duration is the relative price change per unit of decimal yield, to first order, and the DV01 of a long position is

DV01⁡=−V′(y)⋅10−4=V Dmod⋅10−4,\begin{aligned} \operatorname{DV01} &= -V'(y) \cdot 10^{-4} \\[0.4em] &= V \, D_{\text{mod}} \cdot 10^{-4}, \end{aligned}

in currency units, positive for positive payments. After a rise of 11 bp the price changes by approximately −DV01⁡-\operatorname{DV01}. These formulas assume that the payments do not change when the yield changes. A callable bond, whose issuer may repay early when rates fall, needs an effective duration computed from a model of that decision, which this lesson does not build.

In plain words Modified duration turns the average payment time into a price sensitivity. A modified duration of 1.841.84 means that a small rise Δy\Delta y in the yield lowers the price by about 1.84 Δy1.84 \, \Delta y times the price, about 0.01840.0184 percent of the price per basis point.

Example For the bond paying 55 and 105105 at y=0.06y = 0.06, Dmod=1.951949/1.06≈1.841462D_{\text{mod}} = 1.951949/1.06 \approx 1.841462, and

DV01⁡≈98.1666⋅1.8415⋅10−4≈0.018077\begin{aligned} &\operatorname{DV01} \\[0.4em] &\approx 98.1666 \cdot 1.8415 \cdot 10^{-4} \\[0.4em] &\approx 0.018077 \end{aligned}

per 100 of principal. Repricing at a yield of 6.01 percent changes the price by about −0.018074-0.018074, close to the estimate. Using the Macaulay duration in place of the modified duration would overstate this sensitivity by the factor 1.061.06.

Convexity

Definition Under annual compounding, the convexity of fixed payments is the normalized second derivative

C=V′′(y)V(y)=1V∑k=1ntk(tk+1) Ck(1+y)tk+2,\begin{aligned} \mathcal{C} &= \frac{V''(y)}{V(y)} \\[0.4em] &= \frac{1}{V} \sum_{k=1}^{n} \frac{t_k (t_k + 1)\, C_k}{(1+y)^{t_k + 2}}, \end{aligned}

measured in years squared. If every Ck≥0C_k \ge 0 and at least one is positive, then V′′>0V'' > 0, so the price-yield function is strictly convex on (−1,∞)(-1, \infty). Under continuous compounding the convexity is ∑ktk2wk\sum_k t_k^2 w_k with the weights of that convention. For a zero-coupon bond maturing at TT, annual compounding gives Dmod=T/(1+y)D_{\text{mod}} = T/(1+y) and C=T(T+1)/(1+y)2\mathcal{C} = T(T+1)/(1+y)^2, while continuous compounding gives duration TT and convexity T2T^2. Each pair is correct in its own rate coordinate, and the change of variables ez=1+ye^{z} = 1 + y connects them.

In plain words Convexity measures how the price-yield curve bends away from its tangent line. For positive payments it bends upward, so the tangent line lies below the true price on both sides of the current yield.

Example For the bond paying 55 and 105105 at y=0.06y = 0.06, the second derivative is

V′′(0.06)=2⋅51.063+6⋅1051.064≈8.396193+499.019008≈507.415201,\begin{aligned} &V''(0.06) \\[0.4em] &= \frac{2 \cdot 5}{1.06^{3}} + \frac{6 \cdot 105}{1.06^{4}} \\[0.4em] &\approx 8.396193 + 499.019008 \\[0.4em] &\approx 507.415201 , \end{aligned}

and dividing by the price gives C≈507.415201/98.166607≈5.168919\mathcal{C} \approx 507.415201/98.166607 \approx 5.168919 in years squared. Almost all of it comes from the payment at year 2, which carries most of the value and also the larger factor t(t+1)=6t(t+1) = 6.

Second-order price approximation

Definition For fixed payments the function VV is infinitely differentiable on (−1,∞)(-1, \infty). For a yield change Δy\Delta y with y+Δy>−1y + \Delta y > -1, Taylor’s theorem with the Lagrange remainder gives

V(y+Δy)=V(y)+V′(y) Δy+12V′′(y) (Δy)2+R3,\begin{aligned} &V(y + \Delta y) \\[0.4em] &\quad = V(y) + V'(y)\,\Delta y \\[0.4em] &\quad + \tfrac12 V''(y)\,(\Delta y)^2 + R_3 , \end{aligned}

with R3=16V′′′(ξ) (Δy)3R_3 = \tfrac16 V'''(\xi)\,(\Delta y)^3 for some ξ\xi between yy and y+Δyy + \Delta y. Dividing by V(y)V(y) gives the second-order price approximation

ΔVV≈−Dmod Δy+12 C (Δy)2,\begin{aligned} \frac{\Delta V}{V} &\approx -D_{\text{mod}}\,\Delta y \\[0.4em] &\quad + \tfrac12\, \mathcal{C}\,(\Delta y)^2 , \end{aligned}

whose error is R3/V(y)R_3/V(y). If every Ck≥0C_k \ge 0 and at least one is positive, then V′′′<0V''' < 0, so the remainder is negative after a rise in the yield and positive after a fall. The quadratic estimate then overstates the new price after a rise and understates it after a fall, while the first-order estimate understates the new price in both directions because VV is strictly convex. The approximation is local, and for a large move the exact repriced value is the scenario answer.

In plain words Duration gives the straight-line estimate of the price change, and convexity adds a correction for the curvature that is positive whichever way the yield moves. The result is still an estimate, and repricing the bond at the new yield is the exact answer.

Example Raise the yield of the bond paying 55 and 105105 from 6 to 7 percent. Exact repricing gives about 96.38396496.383964. The first-order estimate is 98.166607−1.807700≈96.35890798.166607 - 1.807700 \approx 96.358907, and the convexity term adds

12⋅98.166607⋅5.168919×0.012≈0.025371,\begin{aligned} &\tfrac12 \cdot 98.166607 \cdot 5.168919 \\[0.4em] &\quad \times 0.01^2 \approx 0.025371 , \end{aligned}

for a second-order estimate of about 96.38427896.384278. The second-order estimate is far closer, yet it lies above the exact price by about 0.0003140.000314, as the sign of the remainder predicts for a rise.

Parallel yield shift

Definition A parallel shift moves every rate of a chosen rate coordinate by the same amount Δ\Delta. For one bond described by its own yield it is the move from yy to y+Δy + \Delta, which is a parallel shift of the flat curve at level yy. For a zero curve with continuously compounded zero rates z(t)z(t) it is the move to z(t)+Δz(t) + \Delta at every maturity, which multiplies each discount factor P(0,t)P(0, t) by e−Δte^{-\Delta t}. Durations and convexities are derivatives along such a direction, so they are defined only once the coordinate is named. A shift that is parallel in one coordinate is generally not parallel in another. Since 1+ya(t)=ez(t)1 + y_a(t) = e^{z(t)} for the annually compounded zero rate ya(t)y_a(t), a shift of every z(t)z(t) by Δ\Delta moves ya(t)y_a(t) by (1+ya(t))(eΔ−1)(1 + y_a(t))(e^{\Delta} - 1), an amount that depends on tt unless the curve is flat.

In plain words The phrase “rates up one basis point” has no single meaning until it says which rates. A duration answers that question for the coordinate in which the shift is parallel, and only for that one.

Example The bond paying 55 and 105105, priced at a flat annual yield of 6 percent, has continuously compounded zero rates ln⁡1.06≈0.058269\ln 1.06 \approx 0.058269 at both dates. A rise of 11 bp in the annual yield lowers its price by about 0.0180740.018074. A rise of 11 bp in both continuous zero rates lowers it by about 0.0191600.019160, close to 1.061.06 times as much, because here a rise of 0.00010.0001 in the continuous rates is a rise of 1.06 (e0.0001−1)≈0.000106011.06 \, (e^{0.0001} - 1) \approx 0.00010601 in the annual yield, about 1.06011.0601 basis points.

Key rate duration

Definition Represent the curve by continuously compounded zero rates z1,…,zmz_1, \dots, z_m at chosen nodes, and let the zero rate z(t)z(t) at each payment date be a fixed differentiable function of the node rates, for example through linear interpolation between nodes. The price is V=∑kCke−z(tk) tkV = \sum_k C_k e^{-z(t_k)\, t_k}, and the key rate duration of node jj is

KRD⁡j=−1V ∂V∂zj,\operatorname{KRD}_j = -\frac{1}{V}\,\frac{\partial V}{\partial z_j},

the relative price sensitivity to a bump of node jj alone with every other node held fixed. By the chain rule,

KRD⁡j=1V∑k=1ntk Ck×e−z(tk) tk ∂z(tk)∂zj.\begin{aligned} &\operatorname{KRD}_j = \frac{1}{V} \sum_{k=1}^{n} t_k \, C_k \\[0.4em] &\quad \times e^{-z(t_k)\, t_k} \, \frac{\partial z(t_k)}{\partial z_j} . \end{aligned}

When the nodes are the payment dates themselves and each node rate moves alone, only one term is left and KRD⁡j=tjwj\operatorname{KRD}_j = t_j w_j, with the continuous-compounding weights wj=Cje−zjtj/Vw_j = C_j e^{-z_j t_j}/V. The key rate durations add up to the duration for a parallel shift of the continuous zero curve, ∑ktkwk\sum_k t_k w_k, whenever ∑j∂z(tk)/∂zj=1\sum_j \partial z(t_k)/\partial z_j = 1 at every payment date, as with linear interpolation between nodes. That sum need not equal a modified duration in annual yield. A key rate duration is incomplete without its nodes, its rate coordinate and its bump rule.

In plain words Duration asks what happens when the whole curve moves up together. Key rate durations ask what happens when one maturity moves while the others stay put. Two positions with the same duration can react very differently when short and long rates move in opposite directions.

Example For the bond paying 55 and 105105, with independent nodes at the two payment dates, both at ln⁡1.06\ln 1.06,

KRD⁡1≈1⋅4.71698198.166607≈0.048051,KRD⁡2≈2⋅93.44962698.166607≈1.903898.\begin{aligned} \operatorname{KRD}_1 &\approx \frac{1 \cdot 4.716981}{98.166607} \\[0.4em] &\approx 0.048051, \\[0.4em] \operatorname{KRD}_2 &\approx \frac{2 \cdot 93.449626}{98.166607} \\[0.4em] &\approx 1.903898 . \end{aligned}

Their sum, about 1.9519491.951949, is the duration for a parallel shift of the continuous zero rates, which here equals the Macaulay duration, and not the modified duration 1.8414621.841462 for a shift of the annual yield. A payment at 1.51.5 years under linear interpolation between nodes at 11 and 22 would have ∂z(1.5)/∂z1=∂z(1.5)/∂z2=1/2\partial z(1.5)/\partial z_1 = \partial z(1.5)/\partial z_2 = 1/2, so it would contribute to both key rates.

Check yourself

  1. A five-year zero-coupon bond pays 100100 and has an annual yield of 4 percent with annual compounding. Find its price, its Macaulay and modified durations, its convexity and its DV01.
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The price is 100/1.045≈82.192711100/1.04^5 \approx 82.192711. With a single payment, DMac=5D_{\text{Mac}} = 5 years, and

Dmod=51.04≈4.807692,C=5⋅61.042≈27.736686.\begin{gathered} D_{\text{mod}} = \frac{5}{1.04} \approx 4.807692, \\[0.4em] \mathcal{C} = \frac{5 \cdot 6}{1.04^2} \approx 27.736686 . \end{gathered}

The DV01 is 82.192711⋅4.807692⋅10−4≈0.03951682.192711 \cdot 4.807692 \cdot 10^{-4} \approx 0.039516, and repricing at 4.01 percent changes the price by about −0.039504-0.039504. With the equivalent continuously compounded yield ln⁡1.04≈0.039221\ln 1.04 \approx 0.039221, the duration is 55 and the convexity is 2525, different numbers for the same bond in a different coordinate.

  1. The yield of the bond paying 55 and 105105 falls from 6 to 5 percent. Using Dmod≈1.841462D_{\text{mod}} \approx 1.841462 and C≈5.168919\mathcal{C} \approx 5.168919, find the first-order and second-order estimates of the new price, compare them with the exact price, and explain the sign of each error.
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Here Δy=−0.01\Delta y = -0.01. The first-order term is −VDmodΔy≈+1.8077-V D_{\text{mod}} \Delta y \approx +1.8077, so the first-order estimate is about 98.1666+1.8077=99.974398.1666 + 1.8077 = 99.9743. The convexity term is 12VC(Δy)2≈0.0254\tfrac12 V \mathcal{C} (\Delta y)^2 \approx 0.0254. It is added, not subtracted, because (Δy)2>0(\Delta y)^2 > 0 whichever way the yield moves, so the second-order estimate is about 99.999799.9997. The exact price at 5 percent is 100100, since the coupon rate equals the yield at a coupon date. Both estimates are too low. The first-order error, about 0.02570.0257, is positive because the convex price curve lies above its tangent line. The second-order error, about 0.00030.0003, is positive because the remainder 16V′′′(ξ)(Δy)3\tfrac16 V'''(\xi)(\Delta y)^3 has V′′′<0V''' < 0 and (Δy)3<0(\Delta y)^3 < 0. After the rise to 7 percent the second-order error had the opposite sign.

  1. Position A is a two-year zero-coupon bond. Position B has the same value, split equally between a one-year and a three-year zero-coupon bond. Use independent continuously compounded zero-rate nodes at 11, 22 and 33 years. Show that A and B have the same parallel duration, find their key rate durations, and compare their relative price changes when the one-year rate falls by 1010 bp and the three-year rate rises by 1010 bp, with the two-year rate unchanged.
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Position A has all its value at year 2, so its only key rate duration is KRD⁡2=2\operatorname{KRD}_2 = 2 and its parallel duration is 22. Position B has weight 12\tfrac12 at years 1 and 3, so KRD⁡1=12\operatorname{KRD}_1 = \tfrac12, KRD⁡2=0\operatorname{KRD}_2 = 0 and KRD⁡3=32\operatorname{KRD}_3 = \tfrac32, and its parallel duration is 12+32=2\tfrac12 + \tfrac32 = 2 as well. In the twist A does not move at all, because its only rate is unchanged. For B the first-order estimate is

−12⋅(−0.001)−32⋅0.001=−0.001,\begin{aligned} &-\tfrac12 \cdot (-0.001) \\[0.4em] &\quad - \tfrac32 \cdot 0.001 = -0.001 , \end{aligned}

a fall of about 0.1 percent, and exact repricing, 12e0.001+12e−0.003−1≈−0.000998\tfrac12 e^{0.001} + \tfrac12 e^{-0.003} - 1 \approx -0.000998, agrees closely. Equal durations hid this difference. Under a parallel rise of 1010 bp both fall by about 0.2 percent, B slightly less because its convexity 12⋅1+12⋅9=5\tfrac12 \cdot 1 + \tfrac12 \cdot 9 = 5 exceeds the convexity 44 of A.

How this lesson connects