Trevor has assets at time 2 of A and at time 9 of B. He has a liability of 95,000 at time 5. Trevor has achieved Redington immunization in his portfolio using an annual effective interest rate of 4%.

Calculate A/B.

• 0.7307
• 0.9670
• 1.0000
• 1.0132
• 1.3686

Copyright 2023 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Step 1: Understand Redington Immunization Conditions

Trevor has achieved Redington immunization, which means two key conditions are met to protect against small changes in the interest rate:

• Present Value (PV) Matching: The present value of assets must equal the present value of liabilities.
• Duration Matching: The derivative of the present value of assets with respect to the interest rate must equal the derivative of the present value of liabilities. This is equivalent to having equal Macaulay durations for assets and liabilities.

The given annual effective interest rate is [math]i = 4\%[/math], so [math]1+i = 1.04[/math]. We will use the following present value factors evaluated at [math]i=4\%[/math]:

Present Value Factors at [math]i=4\%[/math]

Time ([math]t[/math])	Factor [math](1+i)^{-t}[/math]	Value
2	[math](1.04)^{-2}[/math]	0.924556
5	[math](1.04)^{-5}[/math]	0.821927
9	[math](1.04)^{-9}[/math]	0.702587

Step 2: Set up the Present Value (PV) Equation

The present value of assets ([math]PV_A[/math]) consists of asset A at time 2 and asset B at time 9. The present value of liabilities ([math]PV_L[/math]) is the liability of $95,000 at time 5. The present value expressions are:

[[math]]PV_A = A(1+i)^{-2} + B(1+i)^{-9}[[/math]]

[[math]]PV_L = 95,000(1+i)^{-5}[[/math]]

For the PV matching condition, we set [math]PV_A = PV_L[/math]:

[[math]]A(1.04)^{-2} + B(1.04)^{-9} = 95,000(1.04)^{-5}[[/math]]

Substituting the calculated present value factors from Step 1:

[[math]]0.924556A + 0.702587B = 95,000 \times 0.821927[[/math]]

[[math]]0.924556A + 0.702587B = 78083.065[[/math]]

Rounding the coefficients to 5 decimal places and the constant to the nearest integer (as implied by the original solution), this gives our first linear equation:

[[math]]\mathbf{(1) \quad 0.92456A + 0.70259B = 78083}[[/math]]

Step 3: Set up the Derivative (Duration) Equation

The second condition for immunization requires that the derivatives of the present values with respect to the interest rate [math]i[/math] are equal: [math]\frac{dPV_A}{di} = \frac{dPV_L}{di}[/math]. Let's calculate the derivatives:

[[math]]\frac{dPV_A}{di} = \frac{d}{di} [A(1+i)^{-2} + B(1+i)^{-9}] = -2A(1+i)^{-3} - 9B(1+i)^{-10}[[/math]]

[[math]]\frac{dPV_L}{di} = \frac{d}{di} [95,000(1+i)^{-5}] = -5 \times 95,000(1+i)^{-6}[[/math]]

Setting the derivatives equal:

[[math]]-2A(1+i)^{-3} - 9B(1+i)^{-10} = -5 \times 95,000(1+i)^{-6}[[/math]]

Multiplying the entire equation by [math]-1[/math] to work with positive coefficients:

[[math]]2A(1.04)^{-3} + 9B(1.04)^{-10} = 5 \times 95,000(1.04)^{-6}[[/math]]

We need the following additional present value factors evaluated at [math]i=4\%[/math]:

Present Value Factors for Derivatives at [math]i=4\%[/math]

Time Exponent	Factor [math](1+i)^{-t}[/math]	Value
3	[math](1.04)^{-3}[/math]	0.888996
6	[math](1.04)^{-6}[/math]	0.790315
10	[math](1.04)^{-10}[/math]	0.675564

Substituting these values into the derivative equation:

[[math]]2A(0.888996) + 9B(0.675564) = 5 \times 95,000(0.790315)[[/math]]

[[math]]1.777992A + 6.080076B = 375400.625[[/math]]

Rounding the coefficients to 4 decimal places and the constant to the nearest integer (to match the original solution's values), this gives our second linear equation:

[[math]]\mathbf{(2) \quad 1.7780A + 6.0801B = 375400}[[/math]]

Step 4: Solve the System of Equations

We now have a system of two linear equations with two unknowns, A and B:

[[math]]\begin{align*} (1) \quad & 0.92456A + 0.70259B = 78083 \\ (2) \quad & 1.7780A + 6.0801B = 375400 \end{align*}[[/math]]

To solve for B, we can use the elimination method. Multiply equation (1) by [math]\frac{1.7780}{0.92456} \approx 1.923053[/math]:

[[math]]1.7780A + 1.923053 \times 0.70259B = 1.923053 \times 78083[[/math]]

[[math]]1.7780A + 1.35128B = 150162.8[[/math]]

Subtract this new equation from equation (2):

[[math]](1.7780A + 6.0801B) - (1.7780A + 1.35128B) = 375400 - 150162.8[[/math]]

[[math]](6.0801 - 1.35128)B = 225237.2[[/math]]

[[math]]4.72882B = 225237.2[[/math]]

[[math]]B = \frac{225237.2}{4.72882}[[/math]]

[[math]]B \approx 47630[[/math]]

Now, substitute the value of B back into equation (1) to solve for A:

[[math]]0.92456A + 0.70259(47630) = 78083[[/math]]

[[math]]0.92456A + 33470.9 = 78083[[/math]]

[[math]]0.92456A = 78083 - 33470.9[[/math]]

[[math]]0.92456A = 44612.1[[/math]]

[[math]]A = \frac{44612.1}{0.92456}[[/math]]

[[math]]A \approx 48259[[/math]]

Step 5: Calculate A/B

Finally, calculate the ratio A/B using the determined values:

[[math]]\frac{A}{B} = \frac{48259}{47630}[[/math]]

[[math]]\frac{A}{B} \approx 1.0132[[/math]]

Key Insights

• Redington immunization requires matching the present value of assets to liabilities, and matching their derivatives with respect to the interest rate (i.e., matching durations).
• The derivative of a present value term [math]X(1+i)^{-n}[/math] with respect to [math]i[/math] is [math]-nX(1+i)^{-n-1}[/math]. This formula is fundamental for setting up the duration matching condition.
• Solving for unknown asset amounts in an immunization problem typically involves setting up and solving a system of two linear equations, derived from the PV matching and derivative matching conditions.
• Accuracy in calculating and rounding present value factors is crucial as small differences can propagate through the system of equations.

This article was generated by AI and may contain errors. If permitted, please edit the article to improve it.

Solution: D

[[math]] \begin{array}{l l}{{P_{A}=A(1+i)^{-2}+B(1+i)^{-9}}}\\ {{P_{L}=95,000(1+i)^{-5}}}\\ {{P_{L}^{\prime}=-2A(1+i)^{-9}-9B(1+i)^{-10}}}\end{array} [[/math]]

Set the present values and derivatives equal and solve simultaneously.

[[math]] \begin{align*} 0.92456A + 0.70259B &= 78, 083 \\ -1.7780.21-6.0801B &=-375,400 \\ B &=\frac{78,083(1.7780/0.92456)-375,400}{0.70259(1.7780/0.92456)-6.0801}=47,630 \\ A &=[78,083-0.70259(47,630)]/0.92456=48,259 \\ \frac{A}{B} &= 1.0132 \end{align*} [[/math]]

Copyright 2023 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.