Taylor Expansion of the Immunized Surplus

Why Expand Immunized Surplus?

Redington immunization is assessed near a specified valuation yield. At that yield, matching the present values makes surplus zero; matching the first derivatives removes the surplus’s first-order response to a small yield change. A Taylor expansion makes the next part explicit: the second-order term describes the leading local effect when those two matches hold, while a remainder accounts for what the quadratic approximation leaves out.

The expansion is a local calculation, not an exact repricing rule for every yield shift. It is especially useful for distinguishing three quantities that are easy to confuse: the second derivative, the coefficient of the quadratic term, and the exact surplus after a finite change in yield.

Definition: At yield $i$, let $A(i)$ be the present value of an asset cash-flow stream and $L(i)$ the present value of a liability cash-flow stream. The surplus is $S(i)=A(i)-L(i)$. A base yield $i_0$ is the yield about which the expansion is made, and $h$ is the change from that yield, so the new yield is $i_0+h$.

Model and Taylor Expansion

Assume deterministic cash flows $C_t$ and $D_t$, respectively, for assets and liabilities, at times $t$ measured in years. Under an annual effective yield $i>-1$, their present values are $A(i)=\sum_t C_t(1+i)^{-t}$ and $L(i)=\sum_t D_t(1+i)^{-t}$. The sums are finite, and positive cash flows are inflows to the asset or liability stream as appropriate. Surplus is signed: a liability cash flow is subtracted through $L(i)$. The yield shift must keep $i_0+h>-1$.

Each present-value term is differentiable in this domain. Taylor’s theorem around $i_0$, through second order, gives

S(i_0+h)=S(i_0)+S'(i_0)h+\frac{1}{2}S''(i_0)h^2+R_3(h).

For these finite cash-flow streams, a third-order remainder can be written as

R_3(h)=\frac{S'''(i_0+\theta h)}{6}h^3 \quad\text{for some }\theta\in(0,1),

when $h\ne 0$ and the interval between $i_0$ and $i_0+h$ is in the domain. Thus the expansion is exact when the remainder is included. Dropping $R_3(h)$ gives a second-order approximation, not an identity for a finite yield change.

For one cash flow $C$ at time $t$, differentiating its present value gives

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

The same formulas apply to liability present value, using its cash flows. For surplus, liability derivatives are subtracted from asset derivatives. In particular, $S''(i_0)$ is the difference of the second derivatives at the base yield; the coefficient multiplying $h^2$ is $S''(i_0)/2$, not $S''(i_0)$.

What Matching Removes

The Redington value and first-derivative matches are $S(i_0)=0$ and $S'(i_0)=0$. Substituting them into Taylor’s theorem removes the constant and linear terms:

S(i_0+h)=\frac{1}{2}S''(i_0)h^2+R_3(h).

This is the key second-order expansion of immunized surplus. The matched first-order term is zero, so for a small shift the quadratic term is the leading term unless its coefficient is also zero. The remainder is generally not zero; its size and sign depend on how the derivatives behave between the base and shifted yields.

Takeaway: Under value and first-derivative matching, the quadratic approximation to surplus is $\tfrac12 S''(i_0)h^2$. A positive $S''(i_0)$ gives a positive quadratic contribution for either sign of $h$, but the approximation alone does not give the exact surplus for a finite shift.

Worked Expansions

Worked Example: Finding the Quadratic Coefficient After Matching

At base yield $i_0=0$, an asset pays \$50 at time 1 and \$50 at time 3. A liability pays \$100 at time 2. Find the Taylor expansion of surplus through second order in $h$, and identify the coefficient of $h^2$.

Method. Write each present value as a function of yield, then calculate the value and derivatives at the base yield. The common trap is to report $S''(0)$ as the quadratic coefficient without dividing by 2.

The present values and surplus are

A(i)=\frac{50}{1+i}+\frac{50}{(1+i)^3}, \qquad L(i)=\frac{100}{(1+i)^2}, \qquad S(i)=A(i)-L(i).

At $i=0$, $A(0)=50+50=100$ and $L(0)=100$, so $S(0)=0$. The first derivatives are $A'(0)=-50(1)-50(3)=-200$ and $L'(0)=-100(2)=-200$. Therefore, $S'(0)=-200-(-200)=0$.

For the second derivatives, a payment $C$ at time $t$ contributes $t(t+1)C$ at $i=0$. Thus $A''(0)=50(1)(2)+50(3)(4)=100+600=700$, while $L''(0)=100(2)(3)=600$. Hence $S''(0)=700-600=100$, and the quadratic coefficient is $100/2=50$.

The expansion is $S(h)=50h^2+R_3(h)$. Its constant and linear terms vanish because value and first derivative match. The coefficient is \$50 per squared unit of yield change when cash flows are measured in dollars; it is not \$100.

Worked Example: Detecting a Remaining Linear Term

At base yield $i_0=0$, an asset pays \$30 at time 1 and \$70 at time 3, while a liability pays \$100 at time 2. Find the expansion through second order. Does the first-order term vanish?

Method. Calculate each derivative separately before subtracting liability values. The trap is assuming that equal present values alone imply first-derivative matching.

At the base yield, both present values equal \$100, so $S(0)=0$. The first derivatives are $A'(0)=-30(1)-70(3)=-30-210=-240$ and $L'(0)=-100(2)=-200$. Therefore, $S'(0)=-240-(-200)=-40$, which does not vanish.

The second derivatives are $A''(0)=30(1)(2)+70(3)(4)=60+840=900$ and $L''(0)=100(2)(3)=600$. It follows that $S''(0)=900-600=300$, so the quadratic coefficient is $300/2=150$.

The expansion is $S(h)=-40h+150h^2+R_3(h)$. The nonzero linear term remains and can dominate the quadratic term for sufficiently small nonzero $h$. Equal present values do not by themselves make a portfolio first-order immunized.

Worked Example: Comparing the Approximation With Exact Repricing

Use the cash flows from the first worked example: \$50 at times 1 and 3 for the asset, and \$100 at time 2 for the liability. The base yield is $0$. Calculate the exact surplus at a new yield of $0.01$, compare it with the second-order approximation, and find the remainder.

Method. Reprice each stream at the new yield and then subtract. The common trap is to treat the quadratic approximation as the exact repricing result.

Let $x=1.01$. The exact surplus is

S(0.01)=\frac{50}{x}+\frac{50}{x^3}-\frac{100}{x^2} =\frac{50x^2+50-100x}{x^3} =\frac{50(x-1)^2}{x^3}.

Since $x-1=0.01$, the numerator is $50(0.01)^2=0.005$, and $x^3=1.01^3=1.030301$. Therefore, $S(0.01)=0.005/1.030301\approx \$0.00485295075$. Rounded to six decimal places, the exact surplus is \$0.004853.

From the first example, the second-order approximation is $50h^2$. At $h=0.01$, it is $50(0.01)^2=\$0.005000$. The remainder is exact surplus minus approximation: $R_3(0.01)\approx0.00485295075-0.005000=-\$0.00014704925$. Rounded to six decimal places, the remainder is $-\$0.000147$. The exact surplus is slightly below the quadratic estimate.

Why the Remainder Matters

The Taylor expansion separates the local response into terms with distinct roles. $S(i_0)$ is the starting surplus. $S'(i_0)h$ is the first-order response. $\tfrac12 S''(i_0)h^2$ is the second-order contribution, and $R_3(h)$ collects the terms not represented by that quadratic approximation. In a value- and first-derivative-matched portfolio, only the first two terms are eliminated; the remainder is not automatically zero.

The remainder is often small for sufficiently small yield changes because it is of third order in $h$ when the third derivative is bounded near the base yield. That does not mean it can be ignored for any shift size. Its sign is not determined solely by $S''(i_0)$, and an approximation should not be reported as an exact surplus unless the remainder is known to be zero.

This distinction matters in exam calculations aligned with the SOA Exam FM (Financial Mathematics) syllabus. A question may ask for the second-order approximation, the coefficient of $h^2$, or exact repricing; these are different requests. State which quantity is being calculated, keep the factor of one-half, and use the exact cash-flow present values when exact repricing is requested.

Set the base and shift.
Write the new yield as $i_0+h$, and confirm it remains greater than $-1$.

Form surplus.
Use $S(i)=A(i)-L(i)$, with both present values based on the same yield convention.

Check the matches.
Evaluate $S(i_0)$ and $S'(i_0)$; do not assume either is zero.

Build the quadratic approximation.
Use $S(i_0)+S'(i_0)h+\tfrac12S''(i_0)h^2$, retaining every nonzero term.

Separate exact value from approximation.
If exact surplus is requested, reprice the cash flows at $i_0+h$; otherwise identify the omitted remainder.

Common Mistakes

Omitting the factor of one-half. The second derivative is not the coefficient of $h^2$; Taylor’s formula uses $S''(i_0)/2$.
Assuming value matching also matches the derivative. Equal present values give $S(i_0)=0$, but $S'(i_0)$ must be checked separately.
Forgetting the sign of liabilities. Differentiate asset and liability present values consistently, then subtract the liability derivative from the asset derivative.
Calling the quadratic approximation exact. The third-order remainder is generally present for a finite yield change.
Mixing the shift with the new yield. In the expansion, $h=i-i_0$, not the new yield itself.

Check Your Understanding

For each problem, calculate the requested expansion or surplus from the stated cash flows. Treat rates as annual effective yields and amounts as dollars.

At $i_0=0$, assets pay \$40 at time 1 and \$60 at time 3; the liability pays \$100 at time 2. Find $S(0)$, $S'(0)$, $S''(0)$, and the coefficient of $h^2$. (Answer: $S(0)=0$, $S'(0)=-20$, $S''(0)=200$, coefficient $=100$.)
At $i_0=0$, assets pay \$50 at time 0 and \$50 at time 4; the liability pays \$100 at time 2. Verify value and first-derivative matching, then find the coefficient of $h^2$. (Answer: $S(0)=0$, $S'(0)=0$, $S''(0)=400$, coefficient $=200$.)
At base yield $0$, assets pay \$40 at times 1 and 3, while the liability pays \$80 at time 2. Find the exact surplus at yield $0.02$ and the second-order approximation. Round the exact surplus to six decimal places. (Answer: exact surplus $=40(0.02)^2/(1.02)^3\approx \$0.015077$; second-order approximation $=\$0.016000$.)
At $i_0=0$, assets pay \$60 at time 1 and \$40 at time 3; the liability pays \$100 at time 2. Write the expansion through second order and identify which term dominates for sufficiently small nonzero $h$. (Answer: $S(0)=0$, $S'(0)=20$, $S''(0)=0$, so $S(h)=20h+R_3(h)$; the linear term is the leading term.)

What You'll Learn