Three-Instrument Portfolios for Redington Matching

Why Three Matching Conditions Require Three Instruments

A portfolio can match a liability’s present value and duration with two independent asset instruments, but those two conditions do not generally determine the portfolio’s convexity. To match all three quantities—present value, duration, and convexity—use three instruments whose values, durations, and convexities provide three independent equations. The calculation is a linear system once the present-value contribution of each asset is used as the unknown.

The valuation model matters. In this tutorial, cash flows are deterministic, occur at specified integer times, and are valued using a single effective annual yield $i$. A yield change is a parallel change in this rate. Instrument holdings are fixed as the yield changes. The formulas do not directly apply to cash flows that change with the yield, or to an embedded option whose exercise behavior changes with the yield.

Definition: For cash flows $C_t$ paid at times $t=1,2,\ldots,n$, present value at effective annual yield $i$ is $P(i)=\sum_{t=1}^{n} C_t(1+i)^{-t}$. Define duration and convexity at that yield by $D(i)=-P'(i)/P(i)$ and $C(i)=P''(i)/P(i)$, respectively. Here $P'$ and $P''$ are derivatives with respect to the annual effective yield.

Under this definition, duration measures the negative first derivative of present value per dollar of present value, and convexity measures the second derivative per dollar of present value. They are yield sensitivities, so the exact values depend on the yield at which they are calculated. In particular, the duration convention here is based on differentiation with respect to $i$; it is not the time-weighted Macaulay duration.

The Three-Equation System

Suppose there are three asset instruments. Let $q_j$ be the quantity held in instrument $j$, $P_j$ its present value per unit at the valuation yield, $D_j$ its duration, and $C_j$ its convexity, for $j=1,2,3$. Let the liability have present value $L$, duration $D_L$, and convexity $C_L$, all evaluated at the same yield.

It is convenient to define $x_j=q_jP_j$, the present-value contribution of asset $j$. Matching the liability then requires:

\begin{aligned} x_1+x_2+x_3 &= L,\\ x_1D_1+x_2D_2+x_3D_3 &= LD_L,\\ x_1C_1+x_2C_2+x_3C_3 &= LC_L. \end{aligned}

The first equation matches present value. The second matches dollar duration, $-P'$, and the third matches dollar convexity, $P''$. The unknowns $x_j$ are present-value amounts, not necessarily the quantities $q_j$ of the instruments. After solving for the $x_j$, recover holdings by $q_j=x_j/P_j$.

Why the Equations Match the Derivatives

For fixed cash flows, differentiating each discounted payment gives:

\begin{aligned} P'(i)&=-\sum_t tC_t(1+i)^{-t-1},\\ P''(i)&=\sum_t t(t+1)C_t(1+i)^{-t-2}. \end{aligned}

For a portfolio with fixed holdings, its present value is $P_A(i)=\sum_j q_jP_j(i)$. Differentiation is linear, so $P_A'(i)=\sum_j q_jP_j'(i)$ and $P_A''(i)=\sum_j q_jP_j''(i)$. Since $P_j'=-P_jD_j$ and $P_j''=P_jC_j$, matching present value and the first two derivatives is equivalent to the three equations above.

This also explains the scope of the result. If the three equations hold, the asset and liability present values, first derivatives, and second derivatives agree at the valuation yield. That is an exact second-order match at that yield. It does not, by itself, guarantee that the asset value will exceed the liability value for every nonzero yield change. In the usual Redington condition, asset convexity is required to be at least liability convexity after present value and duration have been matched; equality is a boundary case, and higher-order behavior can matter.

Evaluate consistently.
At the stated valuation yield, calculate each instrument’s unit present value, duration, and convexity, as well as the corresponding liability quantities.

Use present-value contributions as unknowns.
Write the value, dollar-duration, and dollar-convexity equations in $x_j$, not in unadjusted instrument quantities.

Solve and convert to holdings.
Solve the three linear equations. If actual instrument quantities are requested, divide each $x_j$ by its unit present value $P_j$.

Check all three matches.
Substitute the solution into each equation. A portfolio that matches present value and duration but fails the convexity equation is not a three-condition match.

Worked Construction with Three Zero-Coupon Instruments

Worked Example: Matching a Three-Payment Liability

A hypothetical liability pays $2{,}756.25$ at the end of year 2, $5{,}788.125$ at the end of year 3, and $3{,}038.765625$ at the end of year 4. The valuation yield is $i=5\%$ effective annually. Construct a matching portfolio using zero-coupon instruments that each pay $1$ per unit at the end of year 1, 3, and 5, respectively. Find both their present-value contributions and the amounts held.

Method. First calculate the present value of the liability cash flows. Then solve for the asset present-value contributions using the three matching equations. A zero-coupon payment at time $t$ has $D=t/(1+i)$ and $C=t(t+1)/(1+i)^2$.

The discount factors needed for the liability are $v^2=1/1.05^2$, $v^3=1/1.05^3$, and $v^4=1/1.05^4$. The present values are:

\begin{aligned} 2{,}756.25v^2&=2{,}500,\\ 5{,}788.125v^3&=5{,}000,\\ 3{,}038.765625v^4&=2{,}500. \end{aligned}

Thus $L=10{,}000$. The present-value proportions of the liability cash flows are $0.25,0.50,0.25$ at times 2, 3, and 4. For zero-coupon cash flows, the factors $1/(1+i)$ and $1/(1+i)^2$ are common to every duration and convexity. The matching equations can therefore be solved using the time moments $\sum x_tt$ and $\sum x_tt(t+1)$, where $x_t$ is the proportion of total present value assigned to a payment time.

The liability’s time moments are:

\begin{aligned} \text{Mean time}&=0.25(2)+0.50(3)+0.25(4)=3,\\ \text{Second time moment}&=0.25(2\cdot3)+0.50(3\cdot4)+0.25(4\cdot5)=12.5. \end{aligned}

Let $x_1,x_3,x_5$ be the proportions of asset present value in the year-1, year-3, and year-5 zeroes. The three equations are:

\begin{aligned} x_1+x_3+x_5&=1,\\ x_1+3x_3+5x_5&=3,\\ 2x_1+12x_3+30x_5&=12.5. \end{aligned}

From the first two equations, $x_5=x_1$: substituting $x_3=1-x_1-x_5$ into the time equation gives $3-2x_1+2x_5=3$. Substituting $x_3=1-2x_1$ and $x_5=x_1$ into the third equation gives $12+8x_1=12.5$, so $x_1=x_5=0.0625=1/16$, and $x_3=0.875=7/8$.

The asset present-value contributions are therefore $625$, $8{,}750$, and $625$ dollars. Each unit of the respective zero pays $1$ at maturity, so its unit present value is $v^t$, and the amount held is $x_t/v^t=x_t(1.05)^t$. This gives:

\begin{aligned} q_1&=625(1.05)=656.25,\\ q_3&=8{,}750(1.05)^3=10{,}129.21875,\\ q_5&=625(1.05)^5=797.6759765625. \end{aligned}

Rounded to the nearest cent, hold $656.25$, $10{,}129.22$, and $797.68$ units of the year-1, year-3, and year-5 zeroes. As a check, their present-value contributions sum to $625+8{,}750+625=10{,}000$; their weighted time moment is $3$; and their weighted $t(t+1)$ moment is $12.5$. The duration and convexity therefore match at the stated yield. The common mistake avoided here is treating the face amounts as the $x_t$: the equations use present-value contributions, so the face amounts must be converted using the unit prices.

A Second Solution with Different Maturity Dates

Worked Example: Solving with Year-1, Year-2, and Year-4 Zeroes

A hypothetical liability pays $5{,}250$ at the end of year 1, $2{,}756.25$ at the end of year 2, and $2{,}894.0625$ at the end of year 3. At a $5\%$ effective annual valuation yield, match it using zero-coupon instruments maturing in years 1, 2, and 4. Find the share of total asset present value in each instrument and the corresponding face amounts.

Discounting the liability payments gives $5{,}000$, $2{,}500$, and $2{,}500$, respectively, so $L=10{,}000$. Its present-value proportions are $0.50,0.25,0.25$, at times 1, 2, and 3. The time and second time moments are:

0.50(1)+0.25(2)+0.25(3)=1.75,\qquad 0.50(2)+0.25(6)+0.25(12)=5.5.

Let $x_1,x_2,x_4$ be the fractions of asset present value in the zeroes maturing in years 1, 2, and 4. Since all zeroes are evaluated at the same yield, matching duration and convexity is equivalent to matching these time moments:

\begin{aligned} x_1+x_2+x_4&=1,\\ x_1+2x_2+4x_4&=1.75,\\ 2x_1+6x_2+20x_4&=5.5. \end{aligned}

Subtracting the first equation from the second gives $x_2+3x_4=0.75$, so $x_2=0.75-3x_4$ and $x_1=0.25+2x_4$. Substitution in the third equation gives $5+6x_4=5.5$, hence $x_4=1/12$, $x_2=1/2$, and $x_1=5/12$.

The dollar present-value contributions are $10{,}000(5/12)=4{,}166.666\ldots$, $5{,}000$, and $833.333\ldots$. Converting to face amounts gives $x_t(1.05)^t$: the year-1 zero amount is $4{,}375$, the year-2 amount is $5{,}512.50$, and the year-4 amount is $1{,}012.921875$. Rounded to the nearest cent, these are $4{,}375.00$, $5{,}512.50$, and $1{,}012.92$. Substitution verifies the proportions sum to 1, the time moment is $1.75$, and the second time moment is $5.5$. The trap is to use the liability’s payment amounts as weights; the appropriate weights are its discounted present values.

Scaling the Match and Checking the Result

Worked Example: Matching a Larger Liability at a Different Yield

A hypothetical liability has present-value contributions of $7{,}000$, $70{,}000$, and $13{,}000$ dollars at years 1, 4, and 7, respectively, when valued at $i=4\%$ effective annually. Its total present value is $90{,}000$. Match it using zero-coupon instruments maturing in years 2, 4, and 6. Find the fractions of asset present value allocated to the instruments and the face amounts held.

The liability proportions are $7/90,70/90,13/90$. Its time and second time moments are:

\begin{aligned} \frac{7(1)+70(4)+13(7)}{90}&=\frac{378}{90}=4.2,\\ \frac{7(2)+70(20)+13(56)}{90}&=\frac{2{,}142}{90}=23.8. \end{aligned}

Let $x_2,x_4,x_6$ be the fractions of asset present value in the three zeroes. The equations are:

\begin{aligned} x_2+x_4+x_6&=1,\\ 2x_2+4x_4+6x_6&=4.2,\\ 6x_2+20x_4+42x_6&=23.8. \end{aligned}

Subtract four times the first equation from the second to get $x_6-x_2=0.1$. Subtract twenty times the first equation from the third to get $-14x_2+22x_6=3.8$. Substituting $x_6=x_2+0.1$ gives $8x_2+2.2=3.8$, so $x_2=0.2$, $x_6=0.3$, and $x_4=0.5$.

The present-value allocations are $18{,}000$, $45{,}000$, and $27{,}000$ dollars. A unit of each zero costs $v^t$, so the face amounts are:

\begin{aligned} q_2&=18{,}000(1.04)^2=19{,}468.80,\\ q_4&=45{,}000(1.04)^4=52{,}643.6352,\\ q_6&=27{,}000(1.04)^6=34{,}163.613499392. \end{aligned}

Rounded to the nearest cent, hold $19{,}468.80$, $52{,}643.64$, and $34{,}163.61$ dollars of face value. Their present-value allocations sum to $90{,}000$. The asset time moment is $0.2(2)+0.5(4)+0.3(6)=4.2$, and the second moment is $0.2(6)+0.5(20)+0.3(42)=23.8$, verifying the match. This calculation also illustrates why the valuation yield must be consistent: duration and convexity factors are computed at the same rate used to value both sides.

Common Errors and Interpretation

Weighting by quantities instead of present values. A holding of one unit in two different instruments need not represent the same dollar contribution. Use $x_j=q_jP_j$ in the matching equations.
Mixing valuation rates. Calculate asset and liability values, durations, and convexities at the same effective annual yield and under the same yield-change convention.
Matching duration but not dollar duration. The equation is $\sum_j x_jD_j=LD_L$, not an unweighted average of the asset durations.
Stopping after two equations. Present-value and duration matching leave convexity untested. A three-instrument portfolio must also satisfy the third equation to achieve exact second-order matching.
Interpreting an exact convexity match as a global guarantee. Matching the first two derivatives at one yield is a local condition. It does not establish protection against every size or shape of yield change.

The system has a unique solution only when its three equations are independent. If the instrument duration-convexity combinations make one equation a linear combination of the others, the system may have no solution or multiple solutions. Also, an algebraic solution need not require all holdings to be positive. Whether a portfolio is feasible under long-only or other investment constraints is a separate question from whether the three matching equations can be solved.

Check Your Understanding

For zero-coupon instruments valued at one common effective annual yield, use present-value proportions and the time moments $t$ and $t(t+1)$. In each problem, report the asset present-value fractions unless face amounts are requested.

A liability has present-value proportions $0.25,0.50,0.25$ at years 2, 3, and 4. Match it with zeroes maturing in years 1, 3, and 5. (Answer: $1/16,\ 7/8,\ 1/16$, respectively.)
A liability has present-value proportions $0.50,0.25,0.25$ at years 1, 2, and 3. Match it with zeroes maturing in years 1, 2, and 4. (Answer: $5/12,\ 1/2,\ 1/12$, respectively.)
A liability has present-value proportions $0.50,0.25,0.25$ at years 2, 4, and 6. Match it with zeroes maturing in years 1, 3, and 5. (Answer: $1/4,\ 1/4,\ 1/2$, respectively.)
A liability has present-value proportions $7/90,70/90,13/90$ at years 1, 4, and 7. Match it with zeroes maturing in years 2, 4, and 6. (Answer: $1/5,\ 1/2,\ 3/10$, respectively.)
At a $5\%$ effective annual yield, the matching present-value fractions for zeroes maturing in years 1, 3, and 5 are $0.10,0.60,0.30$. The liability present value is $\$100{,}000$. Find the face amounts of the zeroes. (Answer: $\$10{,}500.00,\ \$69{,}457.50,\ \$38{,}288.45$, respectively.)

What You'll Learn