Convexity Matching and the Second Derivative

Why the Second Derivative Matters

Redington immunization examines how the surplus between assets and liabilities responds to a change in the valuation yield. Matching present values makes the surplus zero at the valuation yield, and matching first derivatives makes its initial response to a small yield change zero. The next question is about curvature: if the yield moves slightly in either direction, does the surplus curve bend upward or downward near the matching point?

For fixed cash flows valued using one annual effective yield, curvature is measured by the second derivative of present value with respect to that yield. The difference between the asset and liability second derivatives is the second derivative of surplus. When present values and first derivatives match, a positive surplus second derivative is the standard strict convexity condition used in Redington immunization.

Definition: Let $A(i)$ and $L(i)$ be the present values of fixed asset and liability cash flows at annual effective yield $i$. Define surplus by $S(i)=A(i)-L(i)$. At a valuation yield $i_0$, present-value matching means $A(i_0)=L(i_0)$, and first-derivative matching means $A'(i_0)=L'(i_0)$. The surplus second derivative is $S''(i_0)=A''(i_0)-L''(i_0)$.

Second Derivatives of Fixed Cash Flows

Assume each cash flow is fixed in amount and occurs at a fixed time $t$, measured in years. All cash flows are discounted using the same annual effective yield $i$, with $i>-1$. For an asset cash flow $C_t$, its present value is $C_t(1+i)^{-t}$. The same form applies to a liability payment. The formulas below also allow noninteger times $t$, provided the cash flows are discounted under this yield model.

Differentiating one discounted payment once gives $-tC_t(1+i)^{-t-1}$. Differentiating again gives $t(t+1)C_t(1+i)^{-t-2}$. Thus, for finite cash-flow streams, the second derivative is obtained by adding the second derivatives of their discounted payments.

\frac{d}{di}\left[C_t(1+i)^{-t}\right] =-tC_t(1+i)^{-t-1}, \qquad \frac{d^2}{di^2}\left[C_t(1+i)^{-t}\right] =t(t+1)C_t(1+i)^{-t-2}.

The factor $t(t+1)$ is specific to differentiating with respect to the annual effective yield under annual compounding. It is not just $t^2$. For a stream of payments $C_t$, define its present value $P(i)$ by summing the discounted payments. Then

P''(i)=\sum_t t(t+1)C_t(1+i)^{-t-2}.

A convenient normalized convexity at $i_0$ is the present-value second derivative divided by the present value. This convention means that $P''(i_0)=P(i_0)$ times the normalized convexity. Some settings use other scaling conventions, so the yield and definition must be kept consistent when comparing convexities.

Definition: For a fixed cash-flow stream with positive present value $P(i_0)$, its normalized annual-compounding convexity at $i_0$ is $C(i_0)=P''(i_0)/P(i_0)$. Equivalently, $$ C(i_0)=\frac{\sum_t t(t+1)C_t(1+i_0)^{-t-2}}{P(i_0)}. $$ Under this definition, $P''(i_0)=P(i_0)C(i_0)$.

To compare asset and liability convexities after present-value matching, it is often easiest to use the present-value contributions at the valuation yield. If $w_t=C_t(1+i_0)^{-t}$ is the present value of the payment at time $t$, then $P''(i_0)=\sum_t t(t+1)w_t/(1+i_0)^2$. The common denominator makes the comparison direct.

Deriving the Surplus Convexity Inequality

Apply the fixed-cash-flow formula separately to assets and liabilities. With $S(i)=A(i)-L(i)$, differentiation gives $S''(i_0)=A''(i_0)-L''(i_0)$. If $A(i_0)=L(i_0)=V>0$, and $C_A$ and $C_L$ denote their normalized convexities, then $A''(i_0)=VC_A$ and $L''(i_0)=VC_L$. Therefore the sign of the surplus curvature is the sign of the difference between those convexities.

$$ S''(i_0) =A''(i_0)-L''(i_0) =V C_A-V C_L =V(C_A-C_L). $$

The same result follows directly from cash-flow present-value contributions. Write $w_t^A$ and $w_t^L$ for the asset and liability present values at time $t$, respectively. Then

S''(i_0) =\frac{\sum_t t(t+1)w_t^A-\sum_t t(t+1)w_t^L}{(1+i_0)^2}.

Since the denominator is positive for $i_0>-1$, surplus has positive curvature precisely when the asset $t(t+1)$-weighted present-value total exceeds the liability total. When the present values are equal, this is equivalent to $C_A>C_L$. First-derivative matching is also part of the Redington setting: it ensures the surplus has no initial slope at the valuation yield. It does not change the algebra for $S''$, but it is what makes the curvature comparison relevant to the local surplus response.

If $S'(i_0)=0$, the second-derivative test says that $S''(i_0)>0$ gives a strict local minimum of surplus, assuming the cash-flow present values are twice continuously differentiable near $i_0$. If $S''(i_0)<0$, it gives a local maximum instead. A zero second derivative does not settle the question. Thus the standard strict convexity inequality for this Redington check is $C_A>C_L$, or equivalently $A''(i_0)>L''(i_0)$. A positive second derivative is a sufficient local-minimum test; it should not be confused with a claim that every possible strict local minimum must have a positive second derivative.

Worked Example: Matched Value and First Derivative, Greater Asset Convexity

A hypothetical fund has asset payments of $52.50 thousand at time 1 and $57.88125 thousand at time 3. Its liability is a payment of $110.25 thousand at time 2. Use annual effective yield $i_0=5\%$. Verify present-value and first-derivative matching, then find the surplus second derivative and compare normalized convexities. All amounts are in thousands of dollars.

Method. Discount each payment at $5\%$, then compare the present-value contributions and their time-weighted totals. The common mistake to avoid is comparing the undiscounted payments as though they were present values.

The asset present values at times 1 and 3 are $52.50/1.05=50$ and $57.88125/(1.05)^3=50$. The liability present value at time 2 is $110.25/(1.05)^2=100$. Hence both total present values equal 100.

The asset first-derivative numerator is $1(50)+3(50)=200$; the liability numerator is $2(100)=200$. Since a first derivative has the common factor $-1/(1+i_0)$, both derivatives equal $-200/1.05=-190.4762$, to four decimal places. First-derivative matching holds.

For the second derivatives, the asset weighted total is $1(2)(50)+3(4)(50)=100+600=700$. The liability weighted total is $2(3)(100)=600$. Therefore $A''(i_0)=700/(1.05)^2=700/1.1025=634.9206$ and $L''(i_0)=600/1.1025=544.2177$, to four decimal places. Subtracting gives $S''(i_0)=100/1.1025=90.7029$, to four decimal places.

The normalized convexities are $C_A=634.9206/100=6.3492$ and $C_L=544.2177/100=5.4422$, to four decimal places. Their difference is positive, consistent with $S''(i_0)=100(C_A-C_L)>0$, subject to rounding. The final check is that the asset convexity exceeds the liability convexity and the directly calculated surplus second derivative is positive.

How to Check the Condition in Practice

The second derivative comparison is compact, but a reliable calculation keeps three checks separate. First verify present-value matching. Second verify that the first-derivative totals match; for annual effective yield these totals are the present-value contributions multiplied by time. Third compare the second-derivative totals, which use $t(t+1)$, and include the common factor $(1+i_0)^{-2}$ when calculating the derivative itself.

Value at the valuation yield.
Discount every fixed payment using the stated annual effective yield and confirm $A(i_0)=L(i_0)$ if value matching is required.

Check first-order matching.
Compare $\sum_t t w_t^A$ with $\sum_t t w_t^L$. Their negatives divided by $1+i_0$ are the respective first derivatives.

Compare curvature.
Compare $\sum_t t(t+1)w_t^A$ with $\sum_t t(t+1)w_t^L$, then divide their difference by $(1+i_0)^2$ to obtain $S''(i_0)$.

Worked Example: Matching First Order but Failing the Convexity Test

At annual effective yield $i_0=5\%$, a hypothetical asset has a single payment at time 2 with present value $100 thousand. The liabilities have present values of $50 thousand at time 1 and $50 thousand at time 3. Verify the first-order match and determine whether the strict convexity condition holds. All amounts are in thousands of dollars.

Method. Work with the stated present-value contributions; no further discounting is needed to verify their totals. The common mistake to avoid is assuming that matched present value and duration automatically imply matched or adequate convexity.

Both sides have present value $100$. The asset time-weighted total is $2(100)=200$. The liability total is $1(50)+3(50)=200$, so both first derivatives equal $-200/1.05=-190.4762$, to four decimal places.

The asset second-derivative weighted total is $2(3)(100)=600$. The liability total is $1(2)(50)+3(4)(50)=100+600=700$. Thus $S''(i_0)=(600-700)/(1.05)^2=-100/1.1025=-90.7029$, to four decimal places.

The asset normalized convexity is $600/(1.1025\cdot100)=5.4422$; the liability normalized convexity is $700/(1.1025\cdot100)=6.3492$, to four decimal places. Assets have less convexity, so the strict inequality fails. The negative surplus second derivative is consistent with the arithmetic and indicates downward curvature at the matching point, not the positive curvature sought by the strict Redington condition.

Worked Example: Comparing Two Matched Portfolios at a Four-Percent Yield

A hypothetical asset portfolio has present-value contributions of $150 thousand at times 1 and 4. A liability portfolio has present-value contributions of $150 thousand at times 2 and 3. The valuation yield is $i_0=4\%$ effective annually. Verify value and first-derivative matching, calculate both second derivatives, and state which side has greater normalized convexity. All amounts are in thousands of dollars.

Method. Use the present-value contributions to form the time and $t(t+1)$ totals. The common mistake to avoid is omitting the factor $(1+i_0)^{-2}$ from the second derivatives.

Each portfolio has present value $150+150=300$. The asset time-weighted total is $1(150)+4(150)=750$. The liability total is $2(150)+3(150)=750$. Consequently the first derivatives match, both equal to $-750/1.04=-721.1538$, to four decimal places.

The asset second-derivative weighted total is $1(2)(150)+4(5)(150)=300+3000=3300$. The liability total is $2(3)(150)+3(4)(150)=900+1800=2700$. Since $(1.04)^2=1.0816$, $A''(i_0)=3300/1.0816=3051.0355$ and $L''(i_0)=2700/1.0816=2496.3018$, to four decimal places. Thus $S''(i_0)=600/1.0816=554.7337$, to four decimal places.

Dividing the second derivatives by the common present value 300 gives normalized convexities $C_A=10.1701$ and $C_L=8.3210$, to four decimal places. Assets have greater convexity, and $300(10.1701-8.3210)$ is approximately $554.7$; the small difference from the reported second derivative is due to rounding the convexities. The direct calculation and normalized comparison agree.

Common Mistakes and Scope of the Result

Using $t^2$ instead of $t(t+1)$. Under annual effective yield, the second derivative of $(1+i)^{-t}$ contains $t(t+1)$. Substituting $t^2$ gives the wrong derivative and the wrong convexity convention.
Forgetting the yield factor. If calculations use present-value contributions $w_t$, the second derivative is the weighted total divided by $(1+i_0)^2$, not just the weighted total.
Comparing unnormalized convexities when values differ. In general, the sign of $S''$ is the sign of $A(i_0)C_A-L(i_0)C_L$. It reduces to comparing $C_A$ and $C_L$ only when the present values are equal.
Ignoring a failed first-order check. The second derivative can be calculated even if first derivatives do not match, but the usual local Redington interpretation assumes first-order neutrality at the valuation yield.
Applying the result to changing or signed cash flows without checking assumptions. The derivation assumes fixed cash-flow amounts and dates under one common yield. For signed cash flows the derivative algebra still applies, but a positive present value alone does not guarantee positive convexity. Different yield conventions also require their own derivative formulas.

Key takeaway: Under annual effective yield and fixed cash flows, $S''(i_0)=A''(i_0)-L''(i_0)$. With matched present values $V>0$, this becomes $S''(i_0)=V(C_A-C_L)$. Once first derivatives also match, the strict convexity condition for the standard local surplus test is $C_A>C_L$.

Check Your Understanding

For each problem, use annual effective yield and fixed cash flows. Treat stated amounts as present-value contributions at the stated valuation yield unless the problem says otherwise.

At $i_0=3\%$, assets have present-value contributions of 40 at time 1 and 60 at time 3; liabilities have a contribution of 100 at time 2. Find $S''(i_0)$, and state whether the first derivatives match. (Answer: $S''(i_0)=200/(1.03)^2=188.5192$; first derivatives do not match.)
At $i_0=4\%$, assets have a present-value contribution of 120 at time 2; liabilities have contributions of 40 at time 1 and 80 at time 3. Find the surplus second derivative and state whether assets have greater convexity. (Answer: $S''(i_0)=-320/(1.04)^2=-295.8580$; no, asset convexity is lower.)
At $i_0=5\%$, assets have contributions of 100 at time 1 and 150 at time 3. Liabilities have contributions of 50 at time 1, 100 at time 2, and 100 at time 3. Verify value and first-derivative matching, then find the surplus second derivative. (Answer: Both present values are 250, both time-weighted totals are 550, and $S''(i_0)=100/(1.05)^2=90.7029$.)
At a valuation yield, both asset and liability present values are 500. Their normalized convexities are 9.2 and 8.7, respectively. Find $S''$ at that yield and interpret its sign, assuming first-derivative matching. (Answer: $S''=500(9.2-8.7)=250$, so the surplus has positive curvature and passes the strict convexity comparison.)

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