Why Redington Checks Need Care
A portfolio can appear to satisfy Redington immunization conditions while failing a basic consistency check. A valuation date may have shifted, a rate may have been interpreted using the wrong compounding convention, or convexity may have been calculated for assets without comparing it with the liabilities. Even when all three conditions are calculated correctly, they describe local behavior around a valuation rate; they do not automatically guarantee a surplus under every possible rate change.
The audit starts by specifying the valuation date, the effective annual rate used at that date, and the cash flows included on each side. The formulas below assume dated cash flows and a flat effective annual valuation rate. For the standard Redington conditions, asset and liability cash flows are nonnegative, and the rate perturbation is a parallel change in that effective annual rate.
For assets and liabilities, write their present values as \(A(i)\) and \(L(i)\), and their durations and convexities as \(D_A,D_L\) and \(K_A,K_L\). Redington matching at a base rate \(i_0\) requires equal present values, equal durations, and strictly greater asset convexity:
The convexity definition here uses \(t(t+1)\). Its relation to the second derivative is \(V''(i)=V(i)K(i)/(1+i)^2\). Other presentations may report a convexity measure with a different normalization; the ranking must be compared using the same definition on both sides.
Derivation: Why the Three Conditions Matter
Let the surplus be \(S(i)=A(i)-L(i)\). For a cash flow at time \(t\), differentiating \(C_t(1+i)^{-t}\) gives a first derivative of \(-tC_t(1+i)^{-t-1}\) and a second derivative of \(t(t+1)C_t(1+i)^{-t-2}\). Summing over cash flows gives
When \(A(i_0)=L(i_0)\) and \(D_A(i_0)=D_L(i_0)\), the first derivatives of the two values are equal: both are \(-A(i_0)D_A(i_0)/(1+i_0)\). Therefore \(S(i_0)=0\) and \(S'(i_0)=0\). If asset convexity exceeds liability convexity, then
A Taylor expansion around \(i_0\) now gives \(S(i_0+h)=\tfrac12 S''(i_0)h^2+o(h^2)\). Thus, for sufficiently small nonzero changes \(h\), the surplus is positive. The positive second-order term explains the local protection; the remainder term also explains why this argument alone says nothing conclusive about every large change.
Worked Example: Unequal Valuation Dates and False Matching
Worked Example: Unequal Valuation Dates and False Matching
At a 5% effective annual valuation rate, an asset pays $105 at time 1 and $115.7625 at time 2. A liability pays $220.50 at time 2. Calculate present values and durations from time 0, then explain what goes wrong if the comparison is assumed to be matched.
Method. Discount every payment to the stated common valuation date, time 0, and use those present values as the duration weights.
Asset value. The time-1 payment has present value \(105/1.05=100\). The time-2 payment has present value \(115.7625/(1.05)^2=115.7625/1.1025=105\). Therefore the asset present value is \(100+105=205\).
Asset duration. Weight the payment times by their present values: \(D_A=(1\cdot100+2\cdot105)/205=310/205=62/41\approx1.5122\) years.
Liability value and duration. Its time-0 present value is \(220.50/1.1025=200\). Since the only liability payment is at time 2, \(D_L=2\) years.
The present values are $205 and $200, and the durations are approximately 1.5122 and 2 years. Neither value nor duration matches, so the Redington conditions are not satisfied. Common mistake avoided: treating the asset’s present value as $200, or reporting its duration as 1.5 years, without checking the discounted cash flows and denominator.
The valuation date matters too. If the same comparison is instead made at time 1, the asset has value \(105+115.7625/1.05=105+110.25=215.25\), while the liability has value \(220.50/1.05=210\). The time-1 durations, measured from time 1, are \(110.25/215.25\approx0.5122\) years for the asset and 1 year for the liability. Changing the date changes the values and the time coordinates; it does not make the two sides equal.
Use Present-Value Weights, and Include Both Sides
Duration and convexity are weighted summaries of discounted cash flows. The weight for a payment at time \(t\) is its present value divided by the total present value, not its undiscounted payment amount divided by the total payments. It is also not enough for an asset portfolio to have positive convexity: liabilities have convexity too, and the Redington test compares the two.
Worked Example: Convexity Must Be Compared With Liabilities
At a 10% effective annual valuation rate, an asset pays $121 at time 2. A liability pays $55 at time 1 and $66.55 at time 3. Check the value, duration, and convexity conditions.
Method. Discount each cash flow at 10%, then calculate duration and \(K\) using the discounted values as weights.
The asset payment has present value \(121/(1.1)^2=121/1.21=100\). Thus \(A=100\), \(D_A=2\), and \(K_A=2(2+1)=6\).
The liability payments have present values \(55/1.1=50\) and \(66.55/(1.1)^3=66.55/1.331=50\). Thus \(L=50+50=100\), and \(D_L=(1\cdot50+3\cdot50)/100=2\). Its convexity measure is \(K_L=(1\cdot2\cdot50+3\cdot4\cdot50)/100=(100+600)/100=7\).
Values and durations match, but \(K_A=6<7=K_L\). The strict convexity condition fails. Common mistake avoided: checking only that asset convexity is positive, or calculating it without including liability convexity.
For example, weighting liability payments by their undiscounted amounts would give \((55\cdot2+66.55\cdot12)/(55+66.55)=908.6/121.55\approx7.474\), not the correct value 7. The correct weights are the discounted amounts, 50 and 50. The incorrect calculation happens to preserve the conclusion in this example, but it is not the stated convexity measure and should not be used.
Keep Rate Conventions Consistent
The rate in the discount factors must be an effective rate for the length of the time period represented by \(t\). If \(t\) is measured in years, \(i\) in \((1+i)^{-t}\) is the effective annual rate. A nominal rate convertible more than once per year must first be converted to an effective annual rate, or used with a matching period-by-period formula. Mixing conventions can make values, durations, and convexities internally inconsistent.
Worked Example: Effective Annual Versus Nominal Rate
A $121 payment is due in two years. Compare its value using a 10% effective annual rate with the incorrect calculation that treats a nominal annual rate of 10% convertible semiannually as if it were an effective annual rate of 10%.
Correct calculation at 10% effective annual interest. There are two annual periods, so the present value is \(121/(1.10)^2=121/1.21=\$100.00\).
Convert the nominal rate. A nominal annual rate of 10% convertible semiannually has a half-year rate of \(0.10/2=0.05\), and an effective annual rate of \((1.05)^2-1=0.1025\), or 10.25%.
There are four half-year periods in two years. Using the nominal rate consistently gives \(121/(1.05)^4=121/1.21550625\approx\$99.55\), rounded to the nearest cent. This differs from $100.00 because the effective annual rates differ.
Common mistake avoided: entering a nominal rate directly as \(i\) in a formula defined for effective annual interest. When comparing asset and liability measures, use the same valuation rate and compounding convention throughout.
Local Stability Is Not a Global Guarantee
The Taylor argument establishes a positive surplus for sufficiently small parallel changes around the base rate, under the stated conditions. It does not establish that a portfolio remains adequate for every large shock or every change in the shape of the yield curve. Exact repricing is needed to evaluate a specified large shift. Redington matching also does not, by itself, protect against cash-flow changes, a different valuation basis, or other departures from the assumptions.
Worked Example: A Large Shock Can Reverse the Surplus
At a 10% effective annual base rate, an asset pays $1 at time 0 and $3.025 at time 2. Liabilities pay $3.025 at time 1 and $0.99825 at time 3. Check the Redington conditions, then revalue at an effective annual rate of \(-45\%\). Treat this as a hypothetical stress calculation.
Method. First express each cash flow’s present value at the base rate as its weight. Then compare value, duration, and convexity. For the stress, discount the actual payments at the new rate.
At the 10% base rate, the asset present values are \(1\) at time 0 and \(3.025/(1.1)^2=2.5\) at time 2. Its total value is \(3.5\), its duration is \((0\cdot1+2\cdot2.5)/3.5=5/3.5\), and its convexity is \(2\cdot3\cdot2.5/3.5=15/3.5\).
The liability present values are \(3.025/1.1=2.75\) at time 1 and \(0.99825/(1.1)^3=0.75\) at time 3. Its total value is also \(3.5\). Its duration is \((1\cdot2.75+3\cdot0.75)/3.5=5/3.5\), and its convexity is \((2\cdot2.75+12\cdot0.75)/3.5=14.5/3.5\). Thus values and durations match, and \(15/3.5>14.5/3.5\): the local convexity condition holds.
At \(-45\%\), the annual accumulation factor is \(1-0.45=0.55\). The asset value is \(1+3.025/(0.55)^2=1+10=11\). The liability value is \(3.025/0.55+0.99825/(0.55)^3=5.5+6=11.5\). The surplus is \(11-11.5=-0.5\), so the assets are short by $0.50 under this large shock.
This example satisfies the base-rate conditions but has a negative surplus at the specified stress rate. Common mistake avoided: interpreting a positive local second-order term as proof that the portfolio will have a nonnegative surplus for every rate change.
A Practical Error-Checking Sequence
Before accepting a Redington calculation, audit its inputs and comparisons in a fixed order. Each step addresses an error that can otherwise be hidden by a plausible-looking final duration or convexity.
Measure every payment time from one common date. If the date changes, re-express both the cash flows’ times and their present values.
Identify the effective rate per time unit and convert nominal rates when needed. Use the same basis for assets and liabilities.
Discount the dated cash flows individually and verify \(A(i_0)=L(i_0)\) before calculating normalized measures.
Calculate both durations and both convexities from present-value weights. Verify equal duration and the strict inequality \(K_A>K_L\).
Redington conditions support local stability under small parallel changes at the base rate. Reprice exactly for a specified large shift or assess separately any change outside that model.
These checks are useful because errors can cancel numerically. For example, using the same wrong rate on both sides may leave a reported duration comparison looking reasonable while distorting the values and convexities. A sound conclusion requires consistent inputs and a complete asset-versus-liability comparison, not merely a plausible final number. This method of diagnosing conditions and their limits is aligned with the SOA Exam FM (Financial Mathematics) syllabus.
Check Your Understanding
For each problem, check the valuation basis and use discounted cash flows as the duration and convexity weights.
- At 4% effective annual interest, an asset pays $104 at time 1 and $121.68 at time 2. A liability pays $216.32 at time 2. Find both present values and durations from time 0. (Answer: Asset value $212.50 and duration \(310/212.5\approx1.4588\) years; liability value $200 and duration 2 years; neither value nor duration matches.)
- At 5% effective annual interest, a $100 payment is due in three years. A candidate instead treats a nominal annual rate of 5% convertible semiannually as an effective annual rate. Give the correct effective annual rate and the correct present value under the stated nominal rate, to the nearest cent. (Answer: Effective annual rate \(1.025^2-1=5.0625\%\); present value \(100/(1.025)^6\approx\$74.42\).)
- At a 10% effective annual rate, an asset has present-value weights $40 at time 1 and $60 at time 3. A liability has present-value weights $100 at time 2. Compare their values, durations, and convexities. (Answer: Both values are $100 and both durations are 2 years; asset \(K=(40\cdot2+60\cdot12)/100=8\), liability \(K=6\), so the three Redington conditions hold.)
- At a 6% effective annual base rate, assets have present-value weights $50 at time 1 and $50 at time 3. Liabilities have present-value weights $100 at time 2. Find the durations and convexities and determine whether the convexity condition holds. (Answer: Both durations are 2 years; asset \(K=7\), liability \(K=6\), so the convexity condition holds.)
- Suppose equal present values and durations have been verified and asset convexity is strictly greater than liability convexity at the base rate. What does this establish, and what additional calculation is needed to assess a specified large parallel rate shock? (Answer: It establishes local positive-surplus behavior for sufficiently small nonzero parallel changes under the model assumptions; revalue all asset and liability cash flows exactly at the shocked rate and subtract liability value from asset value.)