Why Strict Convexity Matters
Matching an asset portfolio’s present value and duration to a liability stream makes the surplus locally insensitive to a small change in the valuation rate to first order. It does not, by itself, show whether the surplus stays nonnegative when the rate moves slightly up or down. The second derivative answers that local question: it determines whether the matched point is a strict minimum, a strict maximum, or a case requiring further analysis.
The comparison is between asset and liability curvature, not between the asset portfolio’s convexity and zero. Both may have positive convexity while the asset portfolio has less curvature than the liabilities. The relevant sign is the second derivative of the surplus.
Model and Second-Derivative Test
Assume assets and liabilities are represented by fixed, deterministic cash flows \(C_t\) paid at times \(t\), and that the same effective annual rate applies to each period under consideration. A parallel rate change means replacing the valuation rate \(i_0\) by \(i_0+h\), using that rate for all cash flows. The cash flows and their payment times do not change with the rate. For the standard interpretation of convexity, take the asset and liability cash flows to be nonnegative and the rates to satisfy \(i>-1\). This local test does not, by itself, cover uncertain cash flows, embedded options that change payments, or nonparallel yield-curve shifts.
For a cash flow \(C_t\) at time \(t\), its present value at \(i\) is \(C_t(1+i)^{-t}\). Differentiating with respect to \(i\) gives the following formulas for a stream of cash flows:
The first derivative records the sensitivity of present value to the rate. Under positive cash flows, it is nonpositive. The second derivative is nonnegative, and it measures the curvature of that present-value function with respect to the annual effective rate. For surplus, however, the relevant derivative is the difference between the asset and liability derivatives.
The result follows from the Taylor expansion about \(i_0\). Because the value and first derivative are zero, the leading term, when the second derivative is nonzero, is the quadratic term:
Since \(h^2\) is positive on either side of zero, a positive \(S''(i_0)\) makes the leading change positive for either direction of a sufficiently small rate shift. A negative value makes it negative on both sides. If the second derivative is zero, this quadratic term vanishes; the remaining terms must be examined before drawing a conclusion.
Comparing Asset and Liability Convexity
In this tutorial, dollar convexity means the second derivative of present value, \(PV''(i_0)\). Thus the surplus curvature is
If both present values match and are positive, the same comparison can be made using convexity normalized by present value: dividing each second derivative by its present value gives the same ordering because \(A(i_0)=L(i_0)\). Be consistent about the rate variable and convexity convention; the comparison here is with respect to the effective annual rate \(i\).
Duration matching is the first-derivative condition \(A'(i_0)=L'(i_0)\). It aligns the first-order responses, but says nothing by itself about which second derivative is larger. The three possible comparisons have different meanings:
| Curvature comparison at \(i_0\) | Surplus interpretation, given value and first-derivative matches |
|---|---|
| \(A''(i_0)>L''(i_0)\) | \(S''(i_0)>0\): strict local minimum |
| \(A''(i_0)=L''(i_0)\) | \(S''(i_0)=0\): second-derivative test is inconclusive |
| \(A''(i_0)<L''(i_0)\) | \(S''(i_0)<0\): strict local maximum, not a local surplus minimum |
Worked Example: Greater Asset Convexity Gives a Strict Local Minimum
At an annual effective valuation rate of \(i_0=0.10\), a liability payment of $242 is due at time 2. An asset portfolio pays $110 at time 1 and $133.10 at time 3. Verify the value and duration matches, then determine the surplus curvature. Treat all cash flows as certain and fixed.
Method and common mistake avoided. Compare present values and derivatives at the same valuation rate. Do not infer the surplus curvature merely from the fact that each stream has positive convexity.
The liability present value is \(242/(1.1)^2=200\). Its first derivative is \(-2(242)/(1.1)^3=-400/1.1=-363.63636\). Its second derivative is \(2(3)(242)/(1.1)^4=1200/(1.1)^2=991.73554\).
For the assets, the present values are \(110/1.1=100\) at time 1 and \(133.10/(1.1)^3=100\) at time 3. Their total present value is also \(200\). Their first derivative is \[ -\frac{1(100)+3(100)}{1.1}=-\frac{400}{1.1}=-363.63636, \] so the first derivatives match. Value matching and first-derivative matching are both verified.
The asset second derivative is \[ \frac{1(2)(100)+3(4)(100)}{(1.1)^2} =\frac{1400}{1.21} =1157.02479. \] Therefore, \[ S''(i_0)=1157.02479-991.73554=165.28925>0. \]
The surplus has a strict local minimum of zero at the valuation rate. For sufficiently small nonzero rate changes in either direction, the surplus is positive. The conclusion is local; this calculation does not establish the surplus for large rate changes.
Worked Example: Equal Convexity Is Inconclusive
At \(i_0=0.10\), liabilities have present-value contributions of $100 at time 1 and $100 at time 3. Assets have present-value contributions of \(100/3\) at time 0, $100 at time 2, and \(200/3\) at time 3. Compare their value, first derivative, and second derivative, and determine whether the second-derivative test establishes a local minimum.
Method and common mistake avoided. Compare the derivatives of the two complete streams. Do not treat equality of second derivatives as proof of a strict local minimum; it only removes the quadratic term.
Both streams have present value \(200\). For liabilities, the time-weighted sum of present-value contributions is \(1(100)+3(100)=400\). For assets, it is \(0(100/3)+2(100)+3(200/3)=400\). By the first-derivative formula, both first derivatives are \(-400/1.1=-363.63636\).
For liabilities, the second-derivative numerator, before division by \((1.1)^2\), is \(1(2)(100)+3(4)(100)=1400\). For assets it is \(0+2(3)(100)+3(4)(200/3)=1400\). Both second derivatives are therefore \(1400/1.21=1157.02479\), and \(S''(i_0)=0\). The second-derivative test is inconclusive.
In this particular example the next derivative distinguishes the streams. The third derivative of a present-value stream is \[ PV'''(i_0)=-\sum_t\frac{t(t+1)(t+2)c_t}{(1+i_0)^{t+3}}, \] where \(c_t\) denotes the present-value contribution at \(i_0\). The liability numerator is \(1(2)(3)(100)+3(4)(5)(100)=6600\); the asset numerator is \(2(3)(4)(100)+3(4)(5)(200/3)=6400\). Thus \[ S'''(i_0)=\frac{6600-6400}{(1.1)^3}=\frac{200}{1.331}>0. \] Because the first two surplus derivatives are zero, the leading nonzero term is cubic: \(S(i_0+h)\) is approximately \(S'''(i_0)h^3/6\). It is negative for sufficiently small negative \(h\) and positive for sufficiently small positive \(h\). So this equal-convexity case is not a local minimum, even though the second-derivative test alone could not determine that.
Worked Example: Lower Asset Convexity Produces a Local Maximum
At an annual effective valuation rate of \(i_0=0.10\), assets pay $242 at time 2. Liabilities pay $110 at time 1 and $133.10 at time 3. Check the value and first-derivative matches, then assess the surplus.
Method and common mistake avoided. Subtract liability curvature from asset curvature. A positive asset second derivative alone does not establish that the surplus has a minimum.
The assets have present value \(242/(1.1)^2=200\), first derivative \(-2(242)/(1.1)^3=-400/1.1=-363.63636\), and second derivative \(2(3)(242)/(1.1)^4=1200/1.21=991.73554\).
The liabilities have present values \(110/1.1=100\) and \(133.10/(1.1)^3=100\), for a total of \(200\). Their first derivative is \(-[1(100)+3(100)]/1.1=-363.63636\). Their second derivative is \([1(2)(100)+3(4)(100)]/1.21=1400/1.21=1157.02479\).
Thus \(S''(i_0)=991.73554-1157.02479=-165.28925<0\). The matched point is a strict local maximum of the surplus, not a local minimum. For sufficiently small nonzero rate changes, the surplus is negative. The error to avoid is declaring the portfolio locally immunized just because the assets’ own second derivative is positive.
Interpreting the Result Carefully
A strict surplus minimum is a local statement. It supports the conclusion that a small parallel rate shift in either direction creates positive surplus when value and first derivative are matched and \(S''(i_0)>0\). It does not guarantee protection against every possible shift, a large change in rates, or a change in the shape of the yield curve.
Equality of asset and liability convexity requires particular care. When \(S''(i_0)=0\), the second-order approximation predicts no curvature difference. Higher-order terms may produce a local minimum, a local maximum, a sign change across the valuation rate, or a still flatter surplus. The result cannot be decided from second derivatives alone.
A useful diagnostic is to write the conditions in order and test each one at the same rate:
Check \(A(i_0)=L(i_0)\), so \(S(i_0)=0\).
Check \(A'(i_0)=L'(i_0)\), the first-order neutrality condition.
Compute \(A''(i_0)-L''(i_0)\). A positive result gives a strict local minimum; zero is inconclusive; a negative result gives a local maximum.
A common exam trap is to read “strict convexity” as “the assets have positive convexity.” The relevant requirement is stricter: asset curvature must exceed liability curvature at the matched valuation rate. Another trap is to call the equality case a pass. Equality does not fail the second-derivative test in the same way as negative curvature, but it also does not prove a strict local minimum. These distinctions are aligned with the SOA Exam FM (Financial Mathematics) syllabus.
Check Your Understanding
For each problem, use the stated valuation rate and interpret the surplus curvature after checking the matching conditions.
- At \(i_0=0.08\), assets and liabilities each have present value $500 and equal first derivatives. Their second derivatives are $5,400 and $5,200, respectively. What does the second-derivative test conclude? (Answer: \(S''(i_0)=200>0\); the matched point is a strict local minimum.)
- At a matched valuation point, \(A''(i_0)=L''(i_0)=3,250\). What can be concluded from the second-derivative test alone? (Answer: It is inconclusive; equality does not establish a strict local minimum.)
- Assets have present-value contributions $150 at time 1 and $150 at time 3. Liabilities have a present-value contribution of $300 at time 2. At \(i_0=0.10\), compare present value, first derivative, and second derivative. (Answer: Both present values are $300; both first derivatives are \(-600/1.1\); \(S''(i_0)=(2,100-1,800)/1.21=300/1.21>0\), so there is a strict local minimum.)
- At \(i_0=0.05\), value and first derivative match. Asset dollar convexity is 1,080 and liability dollar convexity is 1,125. What is the local interpretation? (Answer: \(S''(i_0)=-45<0\); the matched point is a strict local maximum, not a local minimum.)
- Present values and first derivatives match, and \(S''(i_0)=0\), while \(S'''(i_0)=24\). What is the sign of the leading surplus change on each side of \(i_0\)? (Answer: The leading term is \(24h^3/6=4h^3\); it is negative for small \(h<0\) and positive for small \(h>0\), so there is no local minimum.)