Rate Shifts: Exact Repricing Versus Local Approximation

Why Exact Repricing and Local Estimates Differ

Duration and convexity summarize how a present value responds to a small change in yield. For a specified shift, however, the cash flows can also be repriced directly. Comparing these two calculations is useful: exact repricing gives the value under the stated model, while the duration-convexity estimate shows how well a local approximation describes that value.

The distinction matters for immunization. Matching present value and duration makes an asset-liability surplus locally neutral to a small yield change; an asset convexity advantage can make the surplus locally positive for small changes in either direction. These local statements are not, by themselves, guarantees about the result of every finite shock.

Definition: Suppose fixed cash flows $C_t$ are paid at times $t$, measured in years, and are valued using one annual effective yield $i>-1$. Their present value is $P(i)=\sum_t C_t(1+i)^{-t}$. A parallel yield shift of $h$ replaces the yield at every payment date by $i+h$, so the exact repriced value is $P(i+h)=\sum_t C_t(1+i+h)^{-t}$, provided $i+h>-1$. The cash flows and their payment dates are assumed not to change when the yield shifts.

For a stream of positive cash flows, define modified duration $D$ and convexity $K$, both measured with respect to the annual effective yield $i$, by $D=-P'(i)/P(i)$ and $K=P''(i)/P(i)$. For dated cash flows, these are

D=\frac{\sum_t tC_t(1+i)^{-t-1}}{P(i)},\qquad K=\frac{\sum_t t(t+1)C_t(1+i)^{-t-2}}{P(i)}.

The second-order duration-convexity estimate is $P(i+h)\approx P(i)\bigl(1-Dh+\tfrac12Kh^2\bigr)$. Here $h$ is an absolute change in the annual effective rate: a rise from 5% to 6% means $h=0.01$, not $h=1\%$ of the original rate.

Why the Approximation Has This Form

For fixed cash flows, each discounted payment is a smooth function of the yield as long as the yield stays above $-1$. Taylor’s theorem around the initial yield therefore gives the local expansion below. Substituting the definitions of duration and convexity gives the familiar estimate.

Differentiate the present value.
For one payment, the first derivative of $C_t(1+i)^{-t}$ is $-tC_t(1+i)^{-t-1}$. The second derivative is $t(t+1)C_t(1+i)^{-t-2}$. Summing over payments gives $P'(i)$ and $P''(i)$.

Expand around the original yield.
For a small shift $h$, $P(i+h)=P(i)+P'(i)h+\tfrac12P''(i)h^2+O(h^3)$. The remainder notation means that the omitted terms are of order $h^3$ locally.

Express the derivatives using the measures.
Since $P'(i)=-P(i)D$ and $P''(i)=P(i)K$, the second-order estimate is $P(i+h)\approx P(i)\bigl(1-Dh+\tfrac12Kh^2\bigr)$.

This justification also explains a frequent source of error: for an annual effective yield, the second-derivative factor is $t(t+1)$, not $t^2$. The approximation becomes exact for certain cash-flow structures and shifts, but in general it omits third- and higher-order effects. Its accuracy is a question to assess, not an assumption to make.

Worked Example: Exact and Approximate Repricing of a Zero-Coupon Payment

A hypothetical security pays $10,000 exactly two years from now. The initial annual effective yield is 0%. Estimate its value after a 10-percentage-point upward shift using duration and convexity, then compare the estimate with the exact value. Also compare both methods after a 10-percentage-point downward shift.

Method. First calculate the initial present value and the two local measures. Then apply the shifted yield directly for the exact repricing. The cash flow is fixed, and both shifted yields, 10% and $-10\%$, exceed $-100\%$.

At the initial yield $i=0$, the present value is $P(0)=10{,}000$. For a single payment at $t=2$, the modified duration is $D=2/(1+0)=2$, and the convexity is $K=2(3)/(1+0)^2=6$.

For the upward shift, $h=0.10$. The local estimate is

10{,}000\left(1-2(0.10)+\frac12(6)(0.10)^2\right) =10{,}000(0.83) =\$8{,}300.00.

Exact repricing discounts the payment at 10% for two years:

P(0.10)=\frac{10{,}000}{1.10^2} =\frac{10{,}000}{1.21} =\$8{,}264.46.

The estimate is $35.54 higher than the exact value. For the downward shift, $h=-0.10$, the local factor is $1-2(-0.10)+\tfrac12(6)(-0.10)^2=1.23$, giving an estimate of $12,300.00. Exact repricing gives $10{,}000/0.90^2=10{,}000/0.81=\$12{,}345.68$, so the estimate is $45.68 lower.

Check. The exact values follow directly from discounting the same $10,000 payment at each shifted yield. The local estimate uses the initial yield’s duration and convexity and is not required to match the exact result for these finite shifts.

Repricing an Asset-Liability Surplus

Apply the same calculation separately to assets and liabilities. Write $S(i)=P_A(i)-P_L(i)$ for the surplus, where $P_A$ and $P_L$ are their present values. Exact repricing means calculating $S(i+h)=P_A(i+h)-P_L(i+h)$; the approximation means applying the duration-convexity expansion to the surplus or to each present value and then taking the difference.

If the initial present values are equal and the durations match, then $S(i)=0$ and $S'(i)=0$. The second-order estimate becomes $S(i+h)\approx \tfrac12\bigl(P_A(i)K_A-P_L(i)K_L\bigr)h^2$. This describes the local effect of the convexity difference. It does not determine exact surplus at an arbitrary shift.

Worked Example: Exact Surplus Versus a Convexity Estimate

At a 0% annual effective yield, a hypothetical portfolio holds an asset paying $50,000 at the end of year 1 and $50,000 at the end of year 3. It must meet a liability of $100,000 at the end of year 2. Reprice the surplus after shifts of $+5$ and $-5$ percentage points, and compare with the second-order estimate.

Method. Verify present-value and duration matching at the initial yield, calculate the convexities, and then reprice all three cash flows at each shifted yield. No cash flows change with the shift.

At the initial yield, the assets and liability each have present value $100,000. The asset duration is $(50{,}000(1)+50{,}000(3))/100{,}000=2$. The liability duration is $2$. Thus the initial surplus and its first derivative are both zero.

At $i=0$, the asset convexity is $\bigl(50{,}000(1)(2)+50{,}000(3)(4)\bigr)/100{,}000=7$. The liability convexity is $2(3)=6$. The second-order surplus estimate for either shift with $|h|=0.05$ is

\frac12\bigl(100{,}000(7)-100{,}000(6)\bigr)(0.05)^2 =\$125.00.

For the upward shift, exact asset value is $50{,}000/1.05+50{,}000/1.05^3$, which is approximately $47,619.05 + $43,191.88 = $90,810.93. Exact liability value is $100{,}000/1.05^2$, or $90,702.95. The exact surplus is therefore $107.98, compared with the $125.00 local estimate.

For the downward shift, exact asset value is $50{,}000/0.95+50{,}000/0.95^3$, approximately $52,631.58 + $58,317.54 = $110,949.12. Exact liability value is $100{,}000/0.95^2$, or $110,803.32. The exact surplus is $145.79. The local estimate remains $125.00 because its second-order term depends on $h^2$.

Check. Direct subtraction of the exact present values gives positive surplus for each shift. More compactly, the exact difference simplifies to $50{,}000h^2/(1+h)^3$: it is $125/1.05³ = $107.98 for $h=0.05$, and $125/0.95³ = $145.79 for $h=-0.05$. This agrees with the direct repricing.

A Nonzero Initial Yield and a Close Local Estimate

The initial yield does not have to be zero. The same formulas apply at the stated valuation yield, using its own present value, duration, and convexity. The following example shows a case where a small shift makes the second-order estimate close to exact repricing.

Worked Example: A Two-Payment Security at a 10% Yield

A hypothetical security pays $110 at the end of year 1 and $121 at the end of year 2. At an annual effective yield of 10%, estimate its value after the yield rises to 12%, then calculate the exact value. Report dollar values to the nearest cent.

Method. Find the present value and the derivative-based measures at $i=0.10$, use $h=0.02$ in the local estimate, and compare with direct discounting at 12%.

At the initial yield, the two discounted payments are $110/1.10=\$100$ and $121/1.10^2=\$100$, so $P(0.10)=\$200$. The duration is

D=\frac{100(1)+100(2)}{200(1.10)} =\frac{1.5}{1.10} =1.363636\ldots.

The convexity is

K=\frac{100(1)(2)+100(2)(3)}{200(1.10)^2} =\frac{4}{1.21} =3.305785\ldots.

Using these values, the estimated repriced value is

200\left(1-(1.363636\ldots)(0.02) +\frac12(3.305785\ldots)(0.02)^2\right) =\$194.68.

Exact repricing gives $110/1.12+121/1.12^2$. The first term is $98.2142857 and the second is $121(625/784)=\$96.4604592$, so the exact value is $194.6747449, or $194.67 to the nearest cent. The estimate is about $0.003 higher before rounding; both values round to $194.67.

Check. At the initial yield, discounting reproduces the $200 present value. At the shifted yield, discounting the two stated payments directly gives the exact comparison value; no duration approximation is involved in that calculation.

Common Traps and Practical Interpretation

Using the wrong shift units. A rise from 5% to 6% is $h=0.01$, an increase of one percentage point. Substituting $h=1$ would represent a rise of 100 percentage points.
Using $t^2$ in annual effective convexity. The second derivative of $(1+i)^{-t}$ contains $t(t+1)$. For the two-year zero-coupon payment at 0%, using $t^2=4$ instead of 6 would produce an incorrect upward-shift estimate: $10{,}000(1-0.20+0.02)=\$8{,}200$, rather than the correct duration-convexity estimate of $8,300.
Treating an approximation as an exact price. The Taylor estimate omits higher-order terms. When a shift is not small, calculate the exact present value from the cash flows whenever the model permits it.
Changing the cash flows while calling the change a yield shift. The formulas here assume fixed payment amounts and dates and one yield shift applied to every date. If the payments themselves change, the repricing must include that change too.

Exact repricing is the appropriate benchmark for a specified parallel shift under the fixed-cash-flow, single-yield model. Duration and convexity remain useful because they summarize local sensitivity and help explain differences between portfolios. Their estimate is most informative when the shift is small and the assumed cash-flow and yield model fits the question.

This distinction supports work aligned with the SOA Exam FM (Financial Mathematics) syllabus: state the valuation model, identify the shift in rate units, calculate the local measures at the original yield, and keep exact repricing separate from approximation.

Check Your Understanding

For each problem, distinguish the exact repricing from the local estimate. Round dollar amounts to the nearest cent where needed.

A $12,000 payment is due in three years. At an initial annual effective yield of 0%, find the duration-convexity estimate and the exact value after a 4-percentage-point increase. (Answer: estimate $10,675.20; exact value $10,667.96.)
A $100,000 payment is due in two years. The initial annual effective yield is 5%. After a one-percentage-point increase, calculate the duration-convexity estimate and exact value. (Answer: estimate $88,999.95; exact value $88,999.64.)
At a 0% initial yield, assets pay $20,000 at the end of years 1 and 3, while a liability of $40,000 is due at the end of year 2. For a 10-percentage-point increase, give the second-order surplus estimate and exact surplus. (Answer: estimate $200.00; exact surplus $200/(1.10)^3 = $150.26.)
A security pays $120 at the end of year 1 and $130 at the end of year 2. At a 0% initial yield, calculate its duration and convexity, then give the estimate and exact value after a 3-percentage-point increase. (Answer: duration 1.52; convexity 4.08; estimate $239.06; exact value $120/1.03 + $130/1.03² = $239.04.)
A $50,000 payment is due in three years. At a 0% initial annual effective yield, estimate and exactly reprice it after a 2-percentage-point increase. (Answer: estimate $47,120.00; exact value $50,000/1.02³ = $47,116.12.)

What You'll Learn