Unit 12, lesson 96

Liability hedging and immunization

How immunization protects an asset-liability surplus by matching value and duration, what normalized convexity adds, why a nonparallel curve shift can break the hedge, and how a key rate hedge matches several exposures.

  • Video 96
  • Growth, dynamic allocation and liabilities

A pension fund must pay 100 currency units in five years and holds zero-coupon bonds to meet the promise. When interest rates move, does the bond portfolio keep pace with the value of the promise, and against which rate moves can that be guaranteed? This lesson measures the gap by the asset-liability surplus, matches value and duration with three-year and seven-year zero-coupon bonds, and proves that the match protects the surplus against every parallel shift of a flat continuously compounded yield. It then shows how a nonparallel move breaks that protection and what a key rate hedge adds.

Key terms

Asset-liability surplus

Definition Consider deterministic cash flows without default risk, valued at time 00 by discounting with continuously compounded zero rates z(t)z(t), so that a payment cc at time tt in years is worth c e−z(t) tc\,e^{-z(t)\,t} in currency units. Most of this lesson uses a flat curve z(t)=yz(t) = y, under which a position paying cjc_j at times tjt_j is worth

V(y)=∑jcj e−ytj.V(y) = \sum_j c_j\, e^{-y t_j} .

Let AA be the value of the assets and LL the value of the liability, both computed with the same curve. The asset-liability surplus is

S=A−L,\mathcal{S} = A - L ,

and for L>0L > 0 the funding ratio is A/LA/L. A liability hedge aims to control the changes of S\mathcal{S}, not those of AA alone. Valuing the liability with the same curve as the assets is a modeling choice. A different discount rule for the liability changes both LL and its rate sensitivities.

In plain words What matters is whether the assets still cover what is owed, so the target of the hedge is the difference between the two values. A portfolio whose own value never moves can still leave the surplus exposed, because the liability moves.

Example In the constructed example of this lesson the liability pays 100100 at T=5T = 5 years and the flat yield is y=0.03y = 0.03, so

L=100 e−0.15≈86.070798.\begin{aligned} L &= 100\, e^{-0.15} \\[0.4em] &\approx 86.070798 . \end{aligned}

Holding 86.07079886.070798 in cash gives zero surplus today. If the yield falls to 0.020.02, the liability rises to 100 e−0.10≈90.483742100\, e^{-0.10} \approx 90.483742 while the cash keeps its value, and the surplus becomes about −4.412944-4.412944.

Duration matching

Definition For a position with value V(y)>0V(y) > 0 under the flat continuous yield, the duration is D=−V′(y)/V(y)D = -V'(y)/V(y) in years. For nonnegative cash flows it is the present-value-weighted mean payment time,

D=∑jwj tj,wj=cj e−ytjV(y),\begin{gathered} D = \sum_j w_j\, t_j , \\[0.4em] w_j = \frac{c_j\, e^{-y t_j}}{V(y)} , \end{gathered}

and under continuous compounding Macaulay and modified duration coincide. Duration matching chooses the assets so that

A(y)=L(y),A DA=L DL,A(y) = L(y), \qquad A\, D_A = L\, D_L ,

which gives S(y)=0\mathcal{S}(y) = 0 and S′(y)=−ADA+LDL=0\mathcal{S}'(y) = -A D_A + L D_L = 0. The surplus is then zero and has no first-order exposure to a parallel shift of the yield. When A≠LA \ne L, matching the dollar durations ADA=LDLA D_A = L D_L removes the first-order exposure of the surplus in currency units, while matching DA=DLD_A = D_L removes that of the funding ratio, since

ddyln⁡AL=−DA+DL.\frac{d}{dy} \ln \frac{A}{L} = -D_A + D_L .

With mm compounding periods per year, derivatives with respect to that yield carry the factor 1/(1+y/m)1/(1 + y/m), so the assets, the liability and the shift must all use one convention. Duration matching controls only the first-order effect of a parallel shift, and it does not match payment dates.

In plain words Give the assets the same value and the same average payment time as the liability. Then a small parallel move in rates changes both values by about the same amount.

Example Put half of L=86.070798L = 86.070798 into a three-year zero-coupon bond and half into a seven-year one. Each has present value 43.03539943.035399, so their faces are

43.035399 e0.09≈47.088227,43.035399 e0.21≈53.091827.\begin{aligned} 43.035399\, e^{0.09} &\approx 47.088227, \\[0.4em] 43.035399\, e^{0.21} &\approx 53.091827 . \end{aligned}

Equal present values need different faces because the discount horizons differ. The asset duration is 12⋅3+12⋅7=5\tfrac12 \cdot 3 + \tfrac12 \cdot 7 = 5, equal to the liability’s, so both values have the derivative −5⋅86.070798≈−430.354-5 \cdot 86.070798 \approx -430.354 per unit of yield, and a rise of one basis point lowers each of them by about 0.0430350.043035 to first order. Neither bond pays at year 5.

Normalized convexity

Definition The normalized convexity of a position with value V(y)≠0V(y) \ne 0 is

C=V′′(y)V(y),C = \frac{V''(y)}{V(y)} ,

in years squared, as distinct from the dollar convexity V′′(y)V''(y). Under the flat continuous yield and for nonnegative cash flows,

C=∑jwj tj2=D2+∑jwj (tj−D)2,\begin{aligned} C &= \sum_j w_j\, t_j^2 \\[0.4em] &= D^2 + \sum_j w_j\, (t_j - D)^2 , \end{aligned}

so C≥D2C \ge D^2, with equality exactly when all the value sits at one date. Under value and duration matching the surplus has

S′′(y)=L (CA−CL),\mathcal{S}''(y) = L\,(C_A - C_L) ,

and Taylor’s theorem gives, for a small parallel shift δ\delta,

S(y+δ)≈12 L (CA−CL) δ2,\begin{aligned} &\mathcal{S}(y + \delta) \\[0.4em] &\quad \approx \tfrac12\, L\,(C_A - C_L)\,\delta^2 , \end{aligned}

a local approximation with an error of order δ3\delta^3. Under periodic compounding the convexity with respect to that yield has extra factors, as in Duration convexity and key rate sensitivities.

In plain words Convexity measures how widely the payment dates are spread around the duration. A single payment has the smallest convexity possible for its duration, so assets spread around the liability date gain relative to the liability, to second order, whichever way rates move.

Example The three-year and seven-year bonds have CA=12⋅9+12⋅49=29C_A = \tfrac12 \cdot 9 + \tfrac12 \cdot 49 = 29, which is 525^2 plus the dispersion 44 of the dates around 55, while the single liability payment has CL=25C_L = 25. Hence S′′(0.03)=4L≈344.283\mathcal{S}''(0.03) = 4L \approx 344.283, and a parallel rise of δ=0.01\delta = 0.01 gives the estimate

12⋅4⋅86.070798⋅0.012≈0.017214.\begin{aligned} &\tfrac12 \cdot 4 \cdot 86.070798 \cdot 0.01^2 \\[0.4em] &\quad \approx 0.017214 . \end{aligned}

Exact repricing gives a surplus of about 0.0163750.016375. The third-order term 16S′′′(y) δ3=−10Lδ3≈−0.000861\tfrac16 \mathcal{S}'''(y)\,\delta^3 = -10 L \delta^3 \approx -0.000861 accounts for most of the difference.

Immunization

Definition A position is immunized against a stated family of curve changes if its surplus immediately after any change in the family is at least its surplus before. The following result covers one liability payment and parallel shifts. Let the liability be a single payment F>0F > 0 at TT, let the assets be deterministic cash flows cj≥0c_j \ge 0 at times tjt_j, not all zero, valued with a flat continuous yield yy, and suppose A(y)=L(y)A(y) = L(y) and DA=TD_A = T. Then for every real δ\delta

A(y+δ)L(y+δ)=∑jwj e−δ(tj−T)≥e−δ∑jwj(tj−T)=1.\begin{aligned} &\frac{A(y + \delta)}{L(y + \delta)} \\[0.4em] &\quad = \sum_j w_j\, e^{-\delta (t_j - T)} \\[0.4em] &\quad \ge e^{-\delta \sum_j w_j (t_j - T)} = 1 . \end{aligned}

The equality holds because A(y+δ)=A(y)∑jwje−δtjA(y + \delta) = A(y) \sum_j w_j e^{-\delta t_j} and L(y+δ)=L(y) e−δTL(y + \delta) = L(y)\, e^{-\delta T}. The inequality is Jensen’s inequality for the convex exponential, since the weights are nonnegative and add to 11, and the final exponent is 00 because ∑jwjtj=DA=T\sum_j w_j t_j = D_A = T. So S(y+δ)≥0\mathcal{S}(y + \delta) \ge 0 for every parallel shift, and for δ≠0\delta \ne 0 equality holds only when all the asset value sits at TT. The same proof works for a parallel shift of any continuously compounded zero curve, with the weights computed from that curve. For several liability payments, matching value and duration with CA>CLC_A > C_L makes δ=0\delta = 0 a strict local minimum of the surplus, so the surplus stays nonnegative for small parallel shifts, and in general the guarantee is only local. These are statements about values immediately after the shift. Funding the actual payment at TT also involves reinvesting early payments and selling later ones at rates that are not yet known. The result also shows that a one-period model in which the only possible change is an immediate parallel shift of a flat yield, nonzero with positive probability, cannot be free of arbitrage if five-year zero-coupon bonds trade at model prices. Buying the two bonds and selling a five-year bond of the same value costs nothing, never loses and gains after every nonzero shift.

In plain words When the asset value is spread around the liability date with the same average date, a parallel move helps one part of the assets relative to the liability and hurts the other, and the curvature of the exponential makes the gain larger than the loss in both directions. The argument says nothing about moves that are not parallel.

Example For the three-year and seven-year bonds the weights are 12\tfrac12 at tj−T=−2t_j - T = -2 and tj−T=2t_j - T = 2, so

A(y+δ)L(y+δ)=12e2δ+12e−2δ=cosh⁡(2δ)≥1.\begin{aligned} \frac{A(y + \delta)}{L(y + \delta)} &= \tfrac12 e^{2\delta} + \tfrac12 e^{-2\delta} \\[0.4em] &= \cosh(2\delta) \ge 1 . \end{aligned}

After a rise of δ=0.01\delta = 0.01 the liability is worth 100 e−0.20≈81.873075100\, e^{-0.20} \approx 81.873075 and the assets about 81.88945081.889450, a surplus of 81.873075 (cosh⁡0.02−1)≈0.01637581.873075 \,(\cosh 0.02 - 1) \approx 0.016375. After a fall of δ=−0.01\delta = -0.01 the surplus is about 0.0180970.018097, larger because it is the same factor applied to the larger liability value 90.48374290.483742. If the yield jumps once to y+δy + \delta and then stays at that flat level, reinvesting the three-year payment for two years and selling the seven-year bond at year 5 produce exactly 100cosh⁡(2δ)≥100100 \cosh(2\delta) \ge 100, so the payment is covered in that scenario.

Nonparallel curve shift

Definition A curve shift changes the continuously compounded zero rates by Δz(t)\Delta z(t). It is parallel when Δz(t)=δ\Delta z(t) = \delta for every tt and nonparallel otherwise, as in a steepening, a twist or a move concentrated at some maturities. For deterministic cash flows the exact change in value comes from repricing at the new rates. To first order, with asset present values pj=cje−ytjp_j = c_j e^{-y t_j} and one liability payment FF at TT,

ΔS≈−∑jtj pj Δz(tj)+TFe−yT Δz(T).\begin{aligned} \Delta \mathcal{S} &\approx -\sum_j t_j\, p_j\, \Delta z(t_j) \\[0.4em] &\quad + T F e^{-y T}\, \Delta z(T) . \end{aligned}

Duration matching makes this first-order change vanish when all the Δz\Delta z values are equal, and in general not otherwise. A duration is a derivative along the single parallel direction, and the curve has many other directions.

In plain words When rates at different maturities move by different amounts, positions with equal durations can move very differently. Convexity is a second-order correction along the same parallel direction, so it cannot repair a first-order exposure to another direction.

Example Raise the three-year and seven-year rates by one percentage point and leave the five-year rate at 0.030.03. The liability keeps its value 86.07079886.070798, the assets fall to about 81.88945081.889450, and the surplus is about −4.181347-4.181347. The first-order estimate is

−(3+7)⋅43.035399⋅0.01≈−4.303540.\begin{aligned} &-(3 + 7) \cdot 43.035399 \cdot 0.01 \\[0.4em] &\quad \approx -4.303540 . \end{aligned}

This does not contradict the cosh identity of the immunization card, which assumed the same shift at all three dates.

Key rate hedge

Definition Represent the curve by continuously compounded zero rates z1,…,zmz_1, \dots, z_m at chosen node maturities, and let a fixed interpolation rule give the zero rate at every other date. The key rate exposure of a position with value VV to node ii is −∂V/∂zi-\partial V/\partial z_i, in currency units per unit of rate, or 10−410^{-4} times that per basis point. Collect the exposures of one unit of each of KK hedge instruments as the columns of an m×Km \times K matrix HH, and those of the liability in a vector ℓ\ell. For holdings θ∈RK\theta \in \mathbb{R}^K and a move Δz\Delta z of the nodes, the first-order change of the surplus is

ΔS≈−(Hθ−ℓ)⊤Δz.\Delta \mathcal{S} \approx -(H\theta - \ell)^{\top} \Delta z .

A key rate hedge solves Hθ=ℓH\theta = \ell, usually together with the value condition v⊤θ=Lv^{\top}\theta = L, where vv holds the instrument prices,

[Hv⊤]θ=[ℓL].\begin{bmatrix} H \\ v^{\top} \end{bmatrix} \theta = \begin{bmatrix} \ell \\ L \end{bmatrix} .

This system has a solution exactly when its right side lies in the column space of the matrix on the left, and the solution is unique when those columns are linearly independent. Otherwise one minimizes a weighted residual (Hθ−ℓ)⊤Ω (Hθ−ℓ)(H\theta - \ell)^{\top} \Omega\, (H\theta - \ell) with Ω\Omega positive definite, and when HH has full column rank the minimizer is

θ=(H⊤ΩH)−1H⊤Ω ℓ,\theta = (H^{\top} \Omega H)^{-1} H^{\top} \Omega\, \ell ,

possibly with funding and position constraints added. When the interpolation weights ∂z(t)/∂zi\partial z(t)/\partial z_i add to 11 at every date, as with linear interpolation, the key rate exposures add to the exposure to a parallel shift, so matching every key rate implies dollar duration matching but not conversely. The nodes and the interpolation rule determine HH and ℓ\ell, so two systems that both report a five-year key rate can give different exposures. The framework is linear. It removes first-order exposure to the chosen node moves, leaves second-order effects, and does not address moves the nodes cannot express or uncertain liability cash flows.

In plain words Instead of one number for the whole curve, measure the exposure to each of several chosen maturities and match them all. Duration matching matches only their sum.

Example Take nodes at 33 and 77 years with linear interpolation, so z(5)=12z3+12z7z(5) = \tfrac12 z_3 + \tfrac12 z_7, both nodes at 0.030.03, and measure positions in present value. The liability has ∂L/∂z3=∂L/∂z7=−5⋅12⋅L\partial L/\partial z_3 = \partial L/\partial z_7 = -5 \cdot \tfrac12 \cdot L, so ℓ≈(215.1770,215.1770)\ell \approx (215.1770, 215.1770). The duration-matched bonds of the earlier cards have exposures (3⋅43.035399,7⋅43.035399)≈(129.1062,301.2478)(3 \cdot 43.035399, 7 \cdot 43.035399) \approx (129.1062, 301.2478), the same total 430.3540430.3540 split differently. A key rate hedge with present value aa in the three-year bond, bb in the seven-year bond and cc in cash, which has no rate exposure, solves 3a=2.5L3a = 2.5L, 7b=2.5L7b = 2.5L and a+b+c=La + b + c = L, so

a=56L≈71.7257,b=514L≈30.7396,c=−421L≈−16.3944,\begin{aligned} a &= \tfrac56 L \approx 71.7257, \\[0.4em] b &= \tfrac{5}{14} L \approx 30.7396, \\[0.4em] c &= -\tfrac{4}{21} L \approx -16.3944 , \end{aligned}

which means borrowing about 16.394416.3944 in cash. Its dollar convexity 9a+49b=25L9a + 49b = 25L equals the liability’s, yet the negative cash weight voids the hypothesis cj≥0c_j \ge 0 of the immunization result, and a parallel rise of one percentage point leaves a surplus of about −0.000276-0.000276. In this two-node model the move of the previous card is impossible, because the five-year rate must move by the average of the three-year and seven-year moves.

Check yourself

  1. A liability pays 200200 in 44 years and the flat continuous yield is 0.040.04. Find the present values and faces of a one-year and a six-year zero-coupon bond that match its value and duration, compare the convexities, and find the surplus after a parallel rise of two percentage points.
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The liability is worth L=200 e−0.16≈170.428758L = 200\, e^{-0.16} \approx 170.428758. A weight ww on the one-year bond must satisfy w⋅1+(1−w)⋅6=4w \cdot 1 + (1 - w) \cdot 6 = 4, so w=0.4w = 0.4. The present values are 68.17150368.171503 and 102.257255102.257255, and the faces are

68.171503 e0.04≈70.953635,102.257255 e0.24≈129.994448.\begin{aligned} 68.171503\, e^{0.04} &\approx 70.953635, \\[0.4em] 102.257255\, e^{0.24} &\approx 129.994448 . \end{aligned}

The asset convexity is 0.4⋅1+0.6⋅36=220.4 \cdot 1 + 0.6 \cdot 36 = 22, which exceeds the liability’s 1616 by the dispersion 0.4⋅9+0.6⋅4=60.4 \cdot 9 + 0.6 \cdot 4 = 6. After the rise to 0.060.06 the liability is worth 200 e−0.24≈157.325572200\, e^{-0.24} \approx 157.325572, and the funding ratio is

0.4 e0.06+0.6 e−0.04≈1.001208,\begin{aligned} &0.4\, e^{0.06} + 0.6\, e^{-0.04} \\[0.4em] &\quad \approx 1.001208 , \end{aligned}

at least 11 by Jensen’s inequality, so the surplus is about 0.1900940.190094. The second-order estimate 12⋅170.428758⋅6⋅0.022≈0.204515\tfrac12 \cdot 170.428758 \cdot 6 \cdot 0.02^2 \approx 0.204515 is close but too high.

  1. A colleague owes two payments whose present values are 43.03539943.035399 each at 33 and 77 years, a total of 86.07079886.070798 at the flat yield 0.030.03, and funds them with a single five-year zero-coupon bond of face 100100. Value and duration match, so the colleague claims the position is immunized against parallel shifts. Is it?
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No. The roles of the earlier example are reversed, so now

L(y+δ)A(y+δ)=12e2δ+12e−2δ=cosh⁡(2δ)≥1,\begin{aligned} \frac{L(y + \delta)}{A(y + \delta)} &= \tfrac12 e^{2\delta} + \tfrac12 e^{-2\delta} \\[0.4em] &= \cosh(2\delta) \ge 1 , \end{aligned}

and the surplus is negative after every nonzero parallel shift, about −0.016375-0.016375 after a rise of one percentage point and −0.018097-0.018097 after a fall of one percentage point. The immunization result needs a single liability payment, and matching value and duration is not enough by itself. Here the convexity condition fails, since the asset convexity 2525 is below the liability convexity 2929, so the surplus has a local maximum, not a minimum, at the current yield.

  1. Take the two-node curve of the key rate hedge card, with z(5)=12z3+12z7z(5) = \tfrac12 z_3 + \tfrac12 z_7 and both nodes at 0.030.03. The three-year rate falls by one percentage point and the seven-year rate rises by one percentage point. Find the surplus of the duration-matched bonds with present values 43.03539943.035399 each and of the key rate hedge, and compare the first case with its first-order estimate.
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The five-year rate moves by the average of the two node moves, which is zero, so the liability keeps its value 86.07079886.070798. The duration-matched bonds are worth 43.035399 (e0.03+e−0.07)≈84.47196243.035399\,(e^{0.03} + e^{-0.07}) \approx 84.471962, a surplus of about −1.598836-1.598836. Their exposure mismatch Hθ−ℓH\theta - \ell is about (−86.0708,86.0708)(-86.0708, 86.0708), so the first-order estimate is

−(−86.0708⋅(−0.01)+86.0708⋅0.01)≈−1.721416.\begin{aligned} &-\bigl(-86.0708 \cdot (-0.01) \\[0.4em] &\qquad + 86.0708 \cdot 0.01\bigr) \approx -1.721416 . \end{aligned}

The key rate hedge is worth 71.7257 e0.03+30.7396 e−0.07−16.3944≈86.17698471.7257\, e^{0.03} + 30.7396\, e^{-0.07} - 16.3944 \approx 86.176984, a surplus of about 0.1061870.106187. Its first-order exposure is zero, so only the second-order and higher terms remain. Equal durations hid the difference between the two portfolios, while the key rate exposures reveal it.

How this lesson connects