Deriving the Local Surplus Minimum

Why a Local Surplus Minimum Matters

Redington immunization is concerned with how the difference between asset value and liability value responds to a change in yield. Matching present values removes a surplus or deficit at the valuation rate. Matching durations also removes the first-order effect of a small parallel yield shift. The remaining local question is whether the surplus curves upward or downward at that rate.

This tutorial derives the local minimum condition: if asset and liability present values and durations match, and asset convexity is strictly greater, then the surplus is positive for all sufficiently small, nonzero parallel shifts in either direction. “Sufficiently small” is essential. This is a local result, not a guarantee for every possible yield change.

Definition: Let $A(i)$ and $L(i)$ be the present values of assets and liabilities as functions of the annual effective yield $i$. Define the surplus by $S(i)=A(i)-L(i)$. For any differentiable present-value function $P(i)>0$, define its duration and convexity, with respect to $i$, by $D_P(i)=-P'(i)/P(i)$ and $C_P(i)=P''(i)/P(i)$, respectively. A parallel shift $h$ changes the yield from $i$ to $i+h$, using the same annual effective yield convention for both portfolios.

These definitions fix the rate convention: duration here is not the present-value-weighted mean payment time. For deterministic payments at times $t$, that mean is divided by $1+i$ to obtain $D_P(i)$. The derivative-based convexity in the definition likewise refers to derivatives with respect to the annual effective yield, rather than a differently scaled convexity measure.

Assume that the present values are twice continuously differentiable in an open interval around the valuation yield, that the values being compared are positive, and that the shifted yield remains in the domain of the valuation model. Finite deterministic cash-flow streams discounted at annual effective yield $i>-1$ satisfy the smoothness condition. The result below does not require random cash flows or a mortality model.

Deriving the Local Minimum Condition

Write $i_0$ for the valuation yield and evaluate the matching conditions there. Equal present values imply $A(i_0)=L(i_0)=V$. Equal durations then imply that the first derivatives agree as well, because $P'=-DP$. Thus the surplus starts at zero and has zero slope at the valuation yield.

S(i_0)=0,\qquad S'(i_0)=A'(i_0)-L'(i_0) =-D_A(i_0)A(i_0)+D_L(i_0)L(i_0)=0.

Convexity determines the second derivative. From its definition, $P''=C_P P$. At the matching point, therefore, the surplus second derivative is the common present value multiplied by the difference in convexities.

S''(i_0)=A''(i_0)-L''(i_0) =C_A(i_0)A(i_0)-C_L(i_0)L(i_0) =V\bigl(C_A(i_0)-C_L(i_0)\bigr).

Theorem: Suppose $A(i_0)=L(i_0)=V>0$, $D_A(i_0)=D_L(i_0)$, and $C_A(i_0)>C_L(i_0)$. If $S''$ is continuous near $i_0$, then there is an $\varepsilon>0$ such that $S(i_0+h)>0$ whenever $0<|h|<\varepsilon$. Thus the surplus has a strict local minimum of zero at the valuation yield.

The conclusion follows from continuity and the sign of the second derivative, not from assuming that a second-order approximation is exact. Since $V>0$ and $C_A(i_0)-C_L(i_0)>0$, we have $S''(i_0)>0$. Continuity ensures that $S''(i_0+t)>0$ throughout some interval $|t|<\varepsilon$. Within that interval, $S'$ is strictly increasing. Because $S'(i_0)=0$, it follows that $S'(i_0+t)>0$ for $0<t<\varepsilon$, and $S'(i_0+t)<0$ for $-\varepsilon<t<0$. Integrating the slope from the valuation yield in either direction gives $S(i_0+h)>S(i_0)=0$ for every nonzero $h$ in that interval.

This argument explains why the convexity condition is a strict inequality. If the convexities are equal, the second derivative of surplus is zero at the matching point, and this proof gives no sign for the surplus nearby. Higher-order behavior may matter; equal convexities alone do not establish a strict local minimum.

Worked Applications

Worked Example: Verifying a Local Minimum at a 4% Yield

A deterministic asset portfolio pays $52 at time 1 and $56.2432 at time 3. A deterministic liability pays $108.16 at time 2. Use an annual effective valuation yield of $i_0=0.04$. There is no mortality or randomness. Verify the matching conditions, calculate the derivative-based convexities, and determine the local surplus behavior.

Method. Discount each payment at the stated annual effective yield. For a finite deterministic stream with payments $c_t$ at times $t$, its present value is $\sum_t c_t(1+i)^{-t}$. Differentiating a term twice gives the convexity contribution $t(t+1)c_t(1+i)^{-t-2}$. Equivalently, convexity is the present-value-weighted average of $t(t+1)/(1+i)^2$.

Present values. The assets have present value $52/1.04=50$ at time 1 and $56.2432/1.04^3=56.2432/1.124864=50$ at time 3, for $A(i_0)=100$. The liability present value is $108.16/1.04^2=108.16/1.0816=100$. Hence the present values match at $V=100$.

Durations. The asset present-value weights are 50 and 50, so their weighted mean payment time is $(50(1)+50(3))/100=2$. The liability’s weighted mean payment time is 2. Under the stated derivative-based definition, divide each mean time by $1+i_0=1.04$: $D_A(i_0)=D_L(i_0)=2/1.04=1.9230769$. The durations match.

Convexities. For the assets, the weighted average of $t(t+1)$ is $\{50(1(2))+50(3(4))\}/100=(100+600)/100=7$. For the liability it is $2(3)=6$. Divide each by $1.04^2=1.0816$, giving $C_A(i_0)=7/1.0816=6.4719$ and $C_L(i_0)=6/1.0816=5.5473$, to four decimal places. Thus $S''(i_0)=100(7-6)/1.0816=92.4556$, to four decimal places.

Conclusion and check. The values and durations match, while asset convexity exceeds liability convexity. The theorem therefore gives positive surplus for sufficiently small nonzero parallel shifts in either direction. At the valuation yield, both values are exactly $100, so $S(i_0)=0$, as required.

Common mistake avoided: The weighted mean time is 2, but the derivative-based duration is not 2. It is $2/(1.04)$, because the yield derivative includes the factor $1/(1+i_0)$.

Worked Example: Equal Values and Durations Do Not Always Give a Minimum

At the annual effective valuation yield $i_0=0$, a deterministic asset pays $100 at time 2. Deterministic liabilities pay $50 at time 1 and $50 at time 3. There is no mortality or randomness. Check values and durations, compare convexities, and determine the surplus after a parallel shift $h$.

Values and durations. At $i_0=0$, all discount factors are 1, so $A(0)=100$ and $L(0)=50+50=100$. The asset’s weighted mean payment time is 2. The liabilities’ weighted mean time is $\{50(1)+50(3)\}/100=2$. Since $1+i_0=1$, both derivative-based durations are 2.

Convexities. The asset’s weighted $t(t+1)$ is $2(3)=6$. The liabilities’ weighted value is $\{50(1(2))+50(3(4))\}/100=(100+600)/100=7$. The denominator $(1+i_0)^2$ is 1, so $C_A(0)=6$ and $C_L(0)=7$. Asset convexity is lower, not greater.

Direct check after the shift. For $h>-1$, the repriced surplus is $S(h)=100(1+h)^{-2}-50(1+h)^{-1}-50(1+h)^{-3}$. Putting the terms over a common denominator gives $S(h)=50\{2(1+h)-(1+h)^2-1\}/(1+h)^3=-50h^2/(1+h)^3$. For every nonzero $h>-1$, this is negative. In particular, the surplus is not positive for small shifts; the valuation point is a local maximum, not a local minimum.

Check and common mistake avoided. At $h=0$, the exact formula gives $S(0)=0$, consistent with the matched values. Equal value and duration alone do not imply protection: the second-derivative test has the wrong sign here. Do not infer a local minimum merely because the first-order effect has been eliminated.

Worked Example: Computing the Surplus Curvature at a 5% Yield

At an annual effective valuation yield of $i_0=0.05$, deterministic assets pay $105 at time 1 and $115.7625 at time 3. A deterministic liability pays $220.50 at time 2. There is no mortality or randomness. Verify the conditions for a local minimum and compute $S''(i_0)$.

Present values. Since $1.05^3=1.157625$, the first asset payment has present value $105/1.05=100$, and the second has present value $115.7625/1.157625=100$. Thus $A(i_0)=200$. The liability present value is $220.50/1.05^2=220.50/1.1025=200$. The common value is $V=200$.

Durations. Asset present-value weights are 100 and 100, so the weighted mean payment time is $\{100(1)+100(3)\}/200=2$. The liability’s weighted mean time is 2. Divide by $1.05$: $D_A(i_0)=D_L(i_0)=2/1.05=1.9047619$. This verifies duration matching using the specified derivative convention.

Convexities and surplus curvature. The asset weighted average of $t(t+1)$ is $\{100(2)+100(12)\}/200=7$; the liability value is $2(3)=6$. Dividing by $1.05^2=1.1025$ gives $C_A(i_0)=7/1.1025=6.3492$ and $C_L(i_0)=6/1.1025=5.4422$, to four decimal places. Therefore $S''(i_0)=200(7-6)/1.1025=181.4059$, to four decimal places.

Conclusion and check. The portfolios have equal present values, equal durations, and greater asset convexity. The local surplus minimum theorem applies, so surplus is positive for sufficiently small, nonzero parallel shifts. The present-value check gives $S(i_0)=200-200=0$. The conclusion does not specify how large a shift is safe; the neighborhood depends on the cash flows and yield.

Common mistake avoided: The weighted mean time of 2 is not the derivative-based duration at a 5% yield. Both durations are $2/1.05$, not 2. The convexity comparison and local surplus conclusion are unchanged when duration is computed correctly.

Scope and Exam Traps

The result is aligned with the SOA Exam FM (Financial Mathematics) syllabus. Its practical value is that it separates three distinct checks: the value check matches the level of the portfolios at the valuation yield; the duration check matches their first-order response; and the strict convexity inequality gives a positive second derivative of surplus. Each condition has a different role in the proof.

Do not confuse weighted mean time with duration. Under the derivative convention used here, divide the present-value-weighted mean time by $1+i_0$.
Do not reverse the convexity inequality. For surplus defined as assets minus liabilities, $S''(i_0)=V(C_A-C_L)$. Greater asset convexity gives positive curvature.
Do not treat local as global. Continuity proves that some neighborhood exists; it does not identify its size or protect against a large shock.
Do not conclude from zero second derivative. If convexities match, this second-derivative argument is inconclusive about the sign of surplus nearby.

For exam calculations, write down the surplus sign convention first, then check the common present value, the derivative-based durations, and finally the convexity difference. This order makes it harder to overlook a rate-convention error or to claim more protection than the local result supports.

Check Your Understanding

For each problem, use the annual effective yield and derivative-based duration and convexity conventions defined above. Cash flows are deterministic; no mortality or stochastic assumptions are involved.

At $i_0=0$, assets pay $40 at time 1 and $60 at time 3. A liability pays $100 at time 2. Find the common present value, both durations, both convexities, and $S''(i_0)$. (Answer: $V=100$; $D_A=D_L=2$; $C_A=8$, $C_L=6$; $S''(i_0)=200$, so surplus has a strict local minimum.)
At $i_0=0.10$, assets pay $66 at time 1 and $79.86 at time 3, and a liability pays $145.20 at time 2. Verify matching values and durations, then find the convexity difference and surplus second derivative. (Answer: $V=120$; $D_A=D_L=2/1.10=1.8182$; $C_A-C_L=1/1.21=0.8264$; $S''(i_0)=120/1.21=99.1736>0$, so surplus has a strict local minimum.)
At $i_0=0$, assets pay $100 at time 2 and liabilities pay $50 at times 1 and 3. Are values and durations matched? Does the local minimum condition hold? (Answer: Values are both 100 and durations are both 2, but $C_A=6<C_L=7$, so $S''(i_0)=-100$; the local minimum condition does not hold.)
At $i_0=0.05$, a single deterministic payment has present value $300 and occurs at time 3. What is its derivative-based duration? (Answer: $3/1.05=2.8571$, not 3.)
Suppose asset and liability present values and durations match at $i_0$, and $C_A(i_0)=C_L(i_0)$. What can the second-derivative test conclude about a strict local surplus minimum? (Answer: $S(i_0)=0$, $S'(i_0)=0$, and $S''(i_0)=0$; this test is inconclusive and does not establish a strict local minimum.)

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