Why a Second-Order Estimate Has an Error
Duration and convexity give a useful estimate of how a bond’s price changes when its yield changes. The estimate includes the first two terms of a Taylor expansion: the local slope of the price-yield curve and its local curvature. It does not, in general, reproduce the exact price at the new yield. The omitted third- and higher-order terms account for the difference.
To study that difference precisely, consider a bond with fixed cash flows at times \(t=1,2,\ldots,n\), measured in years, and a single annual effective yield \(y>-1\). Assume the cash flows are certain, do not change with yield, and are discounted using annual compounding. There are no embedded options or other features that alter the cash flows. Its price is
where \(CF_t\) is the payment at time \(t\). Write \(y_0\) for the initial yield and \(\Delta y\) for the yield change, so the new yield is \(y_0+\Delta y\). The second-order estimate uses the price and its first two derivatives at \(y_0\).
For annual compounding, modified duration \(D\) and convexity \(C\) at \(y_0\) are \(D=-P'(y_0)/P(y_0)\) and \(C=P''(y_0)/P(y_0)\). Dividing the Taylor estimate’s change by the initial price gives the familiar relative-price estimate:
The yield change in this expression is a decimal. For example, a rise of 25 basis points is \(\Delta y=0.0025\), not \(0.25\) or \(25\). The formula is an estimate because it stops after the second-order term.
Where the Omitted Terms Come From
Differentiate the cash-flow price formula with respect to yield. For any integer \(k\geq 1\),
Each differentiation adds a factor involving the payment time and one more power of \(1+y\) in the denominator. Taylor’s theorem then expresses the exact price as the second-order estimate plus a remainder:
for some \(\xi\) between \(y_0\) and \(y_0+\Delta y\), provided the price function has three continuous derivatives throughout that interval. In other words, the error is the third-order remainder, not a failure of the duration or convexity calculations.
If all cash flows are nonnegative, then \(P'''(y)\leq 0\) for \(y>-1\). This follows directly from the derivative formula: each term in the sum for \(P'''(y)\) is nonnegative before the leading minus sign. Consequently, for a positive yield change the third-order remainder is nonpositive; for a negative yield change it is nonnegative. These are statements about the second-order estimate’s error, not guarantees that the estimate is exact. With signed cash flows, the derivative’s sign need not follow this rule.
Worked Example: A Sequence of Shocks to a Three-Year Zero-Coupon Bond
Worked Example: A Sequence of Shocks to a Three-Year Zero-Coupon Bond
A hypothetical three-year zero-coupon bond pays $1,000 at time 3. Its initial annual effective yield is \(0\%\). Find the second-order estimated price and the exact price after yield increases of 1, 2, and 4 percentage points. Measure each error as exact price minus estimated price.
Method and initial derivatives. The price function is \(P(y)=1000(1+y)^{-3}\), so \(P(0)=1000\). Differentiating gives \(P'(y)=-3000(1+y)^{-4}\) and \(P''(y)=12000(1+y)^{-5}\). At the initial yield, \(P'(0)=-3000\) and \(P''(0)=12000\). Thus the second-order change estimate for any shock \(\Delta y\) is
Shock of 1 percentage point. Here \(\Delta y=0.01\). The estimated change is \(-3000(0.01)+6000(0.01)^2=-30+0.60=-29.40\), so the estimated price is \(1000-29.40=\$970.600\). Exact repricing gives \(1000/(1.01)^3=1000/1.030301=\$970.590\). The error is \(970.590-970.600=-\$0.010\), to three decimal places.
Shock of 2 percentage points. Here \(\Delta y=0.02\). The estimated change is \(-3000(0.02)+6000(0.02)^2=-60+2.40=-57.60\), giving an estimated price of \(\$942.400\). Exact repricing gives \(1000/(1.02)^3=1000/1.061208=\$942.322\). The error is \(-\$0.078\), to three decimal places.
Shock of 4 percentage points. Here \(\Delta y=0.04\). The estimated change is \(-3000(0.04)+6000(0.04)^2=-120+9.60=-110.40\), giving an estimated price of \(\$889.600\). Exact repricing gives \(1000/(1.04)^3=1000/1.124864=\$888.996\). The error is \(-\$0.604\), to three decimal places.
Check and interpretation. Each estimate and exact price starts from the same initial price of $1,000. The exact price is below the second-order estimate for all three positive shocks, consistent with the nonpositive third-order remainder for positive fixed cash flows. The error grows in magnitude across this sequence. Rounding was applied only to the final reported prices and errors, to three decimal places.
Common mistake avoided: Treating the second-order result as an exact repricing would report $889.600 as the exact price after the 4-point yield increase. Exact repricing instead gives $888.996; the difference is the omitted remainder and higher-order effects.
Using the Third-Order Term to Interpret the Error
The third derivative at the initial yield gives a useful local indication of the omitted error. It is not generally the exact error, because the Lagrange remainder uses the derivative at an intermediate yield \(\xi\), rather than necessarily at \(y_0\). The third-order Taylor term at the initial yield is \(P'''(y_0)(\Delta y)^3/6\); for a sufficiently small shock, this term often indicates the error’s approximate size and direction.
Worked Example: Adding the Local Third-Order Term
A hypothetical bond pays $100 at time 1, $100 at time 2, and $1,100 at time 3. Its initial annual effective yield is \(0\%\). For a yield increase of 1 percentage point, calculate the second-order estimate, exact repricing, and local third-order term.
Method. At zero yield, the initial price is \(P(0)=100+100+1100=\$1,300\). The first derivative is \(P'(0)=-[1(100)+2(100)+3(1100)]=-\$3,600\). The second derivative is \(P''(0)=2(100)+6(100)+12(1100)=\$14,000\). For \(\Delta y=0.01\), the estimated change is \(-3600(0.01)+\frac12(14000)(0.01)^2=-36+0.70=-\$35.30\). The estimated price is therefore \(\$1,264.70\).
Exact repricing. At the new yield of \(1\%\), the price is \(100/1.01+100/(1.01)^2+1100/(1.01)^3\). The three discounted cash flows are, to six decimal places, \(\$99.009901\), \(\$98.029605\), and \(\$1,067.649163\). Their sum is \(\$1,264.688669\). Thus the error, exact minus estimated, is \(1264.688669-1264.70=-\$0.011331\), or \(-\$0.011\) to three decimal places.
Local third-order term. The third derivative at zero is \(-[6(100)+24(100)+60(1100)]=-\$69,000\). The third-order term is \(-69000(0.01)^3/6=-\$0.0115\). This is close to, but not exactly, the full error of \(-\$0.011331\). Higher-order terms account for the remaining difference. The exact price and both estimates are consistent when rounded to cents; intermediate values above are shown to make the comparison transparent.
Common mistake avoided: Calling the local third-order term the exact remainder confuses \(P'''(y_0)\) with \(P'''(\xi)\). The term is a useful further approximation, while the exact error is determined by the full repricing.
Negative Shocks and the Error Sign
The sign of the shock matters because the third-order factor is \((\Delta y)^3\). For positive fixed cash flows, the third derivative is nonpositive, so a negative yield change gives a nonnegative second-order error under the stated convention. A concise calculation illustrates how small that error can be when the shock is small.
Worked Example: A Small Yield Decrease
A hypothetical two-year zero-coupon bond pays $110.25 at time 2. Its initial annual effective yield is \(5\%\). Estimate its price after a 1-percentage-point yield decrease using duration and convexity, then compare with exact repricing.
Method and derivatives. The price function is \(P(y)=110.25(1+y)^{-2}\), and \(P(0.05)=110.25/(1.05)^2=\$100\). Its derivatives are \(P'(y)=-2(110.25)(1+y)^{-3}\) and \(P''(y)=6(110.25)(1+y)^{-4}\). At \(y=0.05\), these are \(P'(0.05)=-200/1.05=-190.476190\) and \(P''(0.05)=600/1.05^2=544.217687\).
Second-order estimate. The change is \(\Delta y=-0.01\). The estimated price change is \((-190.476190)(-0.01)+\frac12(544.217687)(0.01)^2 =1.904762+0.027211=\$1.931973\). The estimated price is \(\$101.931973\).
Exact repricing and check. The new yield is \(4\%\), so the exact price is \(110.25/(1.04)^2=\$101.932322\). The error is \(101.932322-101.931973=\$0.000350\), to six decimal places. It is positive, as expected for a negative shock and nonnegative fixed cash flows. The price estimate is close, but the small nonzero error remains.
Common mistake avoided: A negative yield change must be entered as \(-0.01\), not \(0.01\). The first-order term changes sign with the shock, while the second-order term remains positive because the shock is squared.
Reading a Sequence of Errors
A sequence of shocks is useful because it separates a model’s local accuracy from exact repricing. In the three-year zero-coupon example, the second-order error moved from about one cent at a 1-point shock to about sixty cents at a 4-point shock. The third-order term scales with the cube of the shock when measured locally: multiplying a shock by a factor of two multiplies that term by a factor of eight. The total error need not follow that exact scaling, since the remainder also reflects changes in derivatives over the yield interval and includes higher-order terms.
When interpreting such a sequence, keep the comparison consistent: use the same initial price and yield, apply each shock to that initial yield, and calculate the error using the same sign convention. Exact repricing evaluates the original cash-flow formula at the new yield. The second-order estimate uses derivatives at the initial yield. Recalculating duration or convexity at each shocked yield would instead create a different local estimate at each point.
This distinction matters in practical valuation work. The duration-convexity estimate is efficient for assessing small yield movements, while exact repricing is available when the cash flows and yield are specified. A reported estimate should not be presented as an exact price, and an apparent discrepancy should first be checked against the omitted Taylor terms before attributing it to an arithmetic error.
Common Mistakes
- Using percentage points instead of decimal yield changes. A 1-percentage-point change is \(0.01\). Entering \(1\) in place of \(0.01\) changes the scale of both Taylor terms dramatically.
- Dropping the square on the convexity term. The second-order contribution is \(\frac12P''(y_0)(\Delta y)^2\), so it is nonnegative when convexity is positive, whether the yield rises or falls.
- Reversing the error definition. Here \(E_2\) is exact price minus estimated price. If the estimate is above the exact price, the error is negative.
- Assuming the third-order term is always the exact error. Taylor’s theorem evaluates the remainder derivative at an intermediate yield. The derivative at the initial yield only gives the local third-order term.
- Applying the sign conclusion to arbitrary cash flows. The stated sign rule relies on nonnegative fixed cash flows. Signed cash flows can produce a third derivative with a different sign.
Check Your Understanding
For each problem, use annual compounding and fixed cash flows. Unless stated otherwise, measure approximation error as exact price minus second-order estimated price.
- A four-year zero-coupon bond has a $1,000 maturity payment and initial yield \(0\%\). Find its estimated price after a 2-percentage-point yield increase using the second-order Taylor formula. (Answer: \(\$924.00\))
- A bond pays $100 at time 1 and $200 at time 2. Its initial yield is \(0\%\). Find the second-order estimated price after a 1-percentage-point yield increase. (Answer: \(\$295.07\))
- A two-year zero-coupon bond is priced at $500 when its annual effective yield is \(4\%\). Estimate its price after a 0.5-percentage-point yield increase using its first two price derivatives at the initial yield. (Answer: \(\$495.23\))
- A three-year zero-coupon bond pays $800 at maturity and has initial yield \(0\%\). For a 1-percentage-point yield increase, find the exact-minus-estimated error, rounded to four decimal places. (Answer: \(-\$0.0079\))
- A bond pays $100 at time 1, $100 at time 2, and $1,100 at time 3, with initial yield \(0\%\). For a 1-percentage-point yield decrease, calculate the second-order estimated price and the local third-order Taylor term. (Answer: estimated price \(\$1,336.70\); local third-order term \(+\$0.0115\))