Local Stability Is Not a Global Guarantee
Redington immunization matches asset and liability values and first-order interest-rate sensitivity at a valuation yield, with asset curvature exceeding liability curvature there. These conditions make the asset-minus-liability surplus have a strict local minimum: small yield changes leave the surplus nonnegative. They do not promise the same result after an arbitrarily large change. To test a finite shock, reprice each dated cash flow at the new yield.
Throughout, assume deterministic cash flows paid at the ends of integer years, a single parallel change in the effective annual yield, and yields greater than \(-1\). There are no intervening changes to the cash flows. These assumptions let us compare local Redington conditions with exact repricing under one common yield.
The first condition matches present values. The second matches their local response to a yield change, equivalently matching dollar duration when present values are equal. The third requires the asset present value to have greater curvature than the liability present value at the valuation yield. A positive second derivative is a local condition: it describes the surplus near \(i_0\), not its sign everywhere in the yield domain.
Why the Redington Conditions Give Local Protection
The reason for the local result is the second-order Taylor expansion of the surplus around the valuation yield. If \(h\) is a small yield change and the present-value functions are twice continuously differentiable near \(i_0\), then
Under the first two Redington conditions, the first two terms are zero. If \(S''(i_0)>0\), the leading term is positive for nonzero \(h\), whether the small change is upward or downward. The remainder \(o(h^2)\), divided by \(h^2\), tends to zero as \(h\) tends to zero. Therefore, sufficiently close to \(i_0\), the remainder cannot outweigh the positive quadratic term. This proves local protection; it does not bound the remainder for a large \(h\).
Worked Example: Exact Repricing After a Large Upward Shock
Worked Example: Exact Repricing After a Large Upward Shock
A hypothetical portfolio is valued at an effective annual yield of \(i_0=0\). Its assets pay 2.8 at the end of year 2 and 1 at the end of year 4. Its liabilities pay 0.9 at the end of year 1 and 2.9 at the end of year 3. Check the Redington conditions, then find the surplus after a parallel increase in yield to 20%.
Method. Check the value, first time moment, and second-derivative condition at the valuation yield. Then reprice each dated cash flow at the new yield. This avoids treating a local condition as a guarantee for a large shock.
At \(i_0=0\), each cash flow has discount factor 1. The asset and liability present values are
Their first time moments are also equal:
For a cash flow \(C_t\), the second derivative of its present value \(C_t(1+i)^{-t}\) is \(t(t+1)C_t(1+i)^{-t-2}\). At \(i=0\), the asset second derivative is \(2(3)(2.8)+4(5)(1)=36.8\), while the liability second derivative is \(1(2)(0.9)+3(4)(2.9)=36.6\). Thus \(S(0)=0\), \(S'(0)=0\), and \(S''(0)=36.8-36.6=0.2>0\): the Redington conditions hold.
At the new yield \(i=0.20\), the discount factor per year is \(x=1/(1.20)=5/6\). Exact repricing gives
Therefore,
The portfolio satisfies the local Redington tests but has a negative surplus after this finite shock. As a check, the exact asset value is less than the exact liability value by \(2/1296\). The common mistake this calculation avoids is inferring the sign after a large shock from the sign of \(S''(0)\) alone.
Factoring the Surplus Reveals the Sign Change
A useful way to see how local stability can fail is to express the exact surplus as a function of the discount factor \(x=1/(1+i)\). For the preceding cash flows, assets minus liabilities are
The factor \((x-1)^2\) reflects the matched value and first-order sensitivity at \(i=0\), where \(x=1\). Near \(x=1\), the remaining factors \(x\) and \(x-0.9\) are positive, so the surplus is positive except at the valuation yield itself. But when \(x<0.9\), the surplus is negative. The boundary \(x=0.9\) corresponds to \(1/(1+i)=0.9\), or \(i=1/9\). An upward yield change beyond \(1/9\) therefore produces a negative surplus in this example.
Use the same effective annual yield to discount every asset and liability cash flow, and specify the finite yield change being tested.
Verify equal present values, equal first derivatives, and a strictly positive second derivative of the surplus.
Discount every dated payment at the new yield, subtract the liability value from the asset value, and assess the resulting surplus.
Worked Example: A Different Surplus Boundary
Worked Example: A Different Surplus Boundary
A hypothetical portfolio has assets paying 2.6 at time 2 and 1 at time 4; its liabilities pay 0.8 at time 1 and 2.8 at time 3. The valuation yield is 0%. Find the exact surplus at a new yield of 30%, and identify the yield at which the surplus is zero apart from the matched valuation yield.
Method. First verify the matching conditions, then factor the exact surplus in powers of the discount factor. The factorization identifies the nonlocal zero and shows on which side of it the surplus is negative.
At the valuation yield, both present values equal 3.6. The first time moments are \(2(2.6)+4(1)=9.2\) for assets and \(1(0.8)+3(2.8)=9.2\) for liabilities. The second derivatives at 0 are \(2(3)(2.6)+4(5)(1)=35.6\) for assets and \(1(2)(0.8)+3(4)(2.8)=35.2\) for liabilities. The second-derivative difference is \(0.4>0\), so all three Redington conditions hold.
Let \(x=1/(1+i)\). The surplus as a function of \(x\) is
At a 30% yield, \(x=1/1.3=10/13\). Substitution into the factored form gives
The additional zero occurs at \(x=0.8\), which means \(1/(1+i)=0.8\), so \(i=0.25\). For yields above 25%, \(x<0.8\), and the surplus is negative. The exact calculation confirms the sign at 30%. The trap is to assume that matching and positive curvature at 0% prevent a sign change beyond the local neighborhood.
Worked Example: The Local Approximation Can Miss the Sign
Worked Example: The Local Approximation Can Miss the Sign
For the first portfolio in this tutorial, use the second-order Taylor approximation at \(i_0=0\) to estimate the surplus after a 20% yield increase. Compare that estimate with exact repricing and explain why they differ in sign.
Method. Use the matched value and first derivative and the calculated second derivative to form the local approximation. Then compare it with the exact surplus, which is necessary for a finite change.
The first portfolio has \(S(0)=0\), \(S'(0)=0\), and \(S''(0)=0.2\). For \(h=0.20\), its second-order approximation is
This approximation is positive. Exact repricing, however, gives \(S(0.20)=-1/648\approx-0.001543\). One can see why by writing the exact surplus for that portfolio in terms of the yield:
At \(i=0.20\), the factor \(0.1-0.9i\) equals \(-0.08\), so the exact surplus is negative. Near zero, this factor is positive and the leading term of the surplus is \(0.1i^2\), consistent with the Taylor approximation. At a larger change, higher-order terms matter and can reverse the sign. The second-order approximation describes local behavior; it is not a substitute for exact repricing when the shock is large.
Practical Interpretation and Common Traps
“Large” is not a fixed number of percentage points. The relevant issue is whether a yield change remains in the neighborhood where the local expansion is informative and the surplus retains its sign. The examples show that the boundary depends on the cash-flow pattern: one portfolio changes sign above \(1/9\), another above \(25\%\). A positive surplus curvature at the base yield does not specify the location of such a boundary.
- Confusing a local minimum with a global minimum. Redington conditions describe derivatives at one yield. They do not establish the surplus sign at every yield.
- Using the Taylor estimate as an exact value. A second-order approximation omits higher-order terms, which can be decisive for a finite shock.
- Changing compounding conventions mid-calculation. If yields are effective annual rates and payments are annual, use \(1/(1+i)^t\) consistently for every cash flow.
- Checking only upward shocks. Local convexity protects in both directions sufficiently near the valuation yield, but exact repricing is needed to test any specified finite upward or downward change.
For an exam-style calculation, state the valuation yield and shock, reprice each asset and liability cash flow using the same convention, and compute \(A(i)-L(i)\). If a factored surplus is available, use it to explain the sign and locate any break-even yield. These steps distinguish the local Redington result from the exact outcome under a large shock and are aligned with the SOA Exam FM (Financial Mathematics) syllabus.
Check Your Understanding
For each problem, assume deterministic year-end cash flows and a parallel change in the effective annual yield. Use exact discounting for finite shocks.
- A portfolio has assets 2.7 at time 2 and 1 at time 4, and liabilities 0.7 at time 1 and 2.7 at time 3. It is valued at \(i_0=0\). Write its surplus in terms of \(x=1/(1+i)\), then find the yield at which its nonvaluation surplus zero occurs. (Answer: \(S=x(x-1)^2(x-0.7)\); the additional zero is at \(i=3/7\), approximately 42.86%.)
- For assets 2.5 at time 2 and 1 at time 4, and liabilities 0.75 at time 1 and 2.75 at time 3, verify the equal present values and first time moments at \(i_0=0\), then identify the second-derivative surplus. (Answer: both present values are 3.5, both first time moments are 9, and \(S''(0)=0.5>0\).)
- A portfolio has exact surplus \(S=x(x-1)^2(x-0.85)\), where \(x=1/(1+i)\). Find the surplus at \(i=20\%\). (Answer: \(-1/2592\approx-0.000386\).)
- For \(S=x(x-1)^2(x-0.9)\), find the exact surplus at \(i=5\%\). (Answer: \(22/194481\approx0.0001131\).)
- A portfolio has exact surplus \(S=x(x-1)^2(x-0.75)\). Find the interval of upward yields, starting from the valuation yield \(i_0=0\), for which the surplus is positive. (Answer: \(0<i<1/3\); it is zero at \(i=0\) and \(i=1/3\), and negative for \(i>1/3\).)