An insurance company must pay liabilities of 99 at the end of one year, 102 at the end of two years and 100 at the end of three years. The only investments available to the company are the following three bonds. Bond A and Bond C are annual coupon bonds. Bond B is a zero-coupon bond.

Bond	Maturity (in years)	Yield-to-Maturity (Annualized)	Coupon Rate
A	1	6%	7%
B	2	7%	0%
C	3	9%	5%

All three bonds have a par value of 100 and will be redeemed at par.

Calculate the number of units of Bond A that must be purchased to match the liabilities exactly.

• 0.8807
• 0.8901
• 0.8975
• 0.9524
• 0.9724

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

Step 1: Understand the Liabilities and Objective

The objective is to exactly match the given liabilities at the end of each year using the available bonds. This involves determining the number of units of each bond to purchase such that the total cash inflows from these bonds at each time [math]t[/math] precisely cover the liabilities at that time. The liabilities are:

• Liability at Year 1: $99
• Liability at Year 2: $102
• Liability at Year 3: $100

Step 2: Analyze Available Investments

We are provided with three bonds, each with a par value of $100 and redeemable at par. We need to understand the cash flows generated by one unit of each bond.

Bond	Maturity (in years)	Yield-to-Maturity (Annualized)	Coupon Rate	Cash Flow per Unit
A	1	6%	7%	At [math]t=1[/math]: Coupon + Par = [math]$7 + $100 = $107[/math]
B	2	7%	0%	At [math]t=2[/math]: Par = [math]$100[/math]
C	3	9%	5%	At [math]t=1[/math]: Coupon = [math]$5[/math] At [math]t=2[/math]: Coupon = [math]$5[/math] At [math]t=3[/math]: Coupon + Par = [math]$5 + $100 = $105[/math]

We will use a cash flow matching strategy, working backward from the longest maturity liability (Year 3) to the shortest (Year 1).

Step 3: Match Liability at Year 3

The liability at Year 3 is $100. Bond C is the only bond that matures at Year 3 and provides a cash flow at this time. Let [math]N_C[/math] be the number of units of Bond C purchased. One unit of Bond C provides a cash flow of $105 at Year 3 (a $5 coupon plus $100 par value). To match the liability at Year 3, the cash flow from [math]N_C[/math] units of Bond C must equal the $100 liability.

[[math]]N_C \times 105 = 100[[/math]]

[[math]]N_C = \frac{100}{105} \approx 0.95238[[/math]]

For calculations, we will use the more precise value [math]N_C = \frac{100}{105}[/math].

Step 4: Match Liability at Year 2

The liability at Year 2 is $102. First, we account for any cash flow from Bond C at Year 2. Since [math]N_C[/math] units of Bond C were purchased, they will provide a coupon payment at Year 2. Cash flow from Bond C at Year 2 = [math]N_C \times $5 = \frac{100}{105} \times 5 = \frac{500}{105} = \frac{100}{21} \approx $4.7619[/math] Remaining liability at Year 2 = Total liability at Year 2 - Cash flow from Bond C at Year 2 Remaining liability at Year 2 = [math]$102 - \frac{100}{21}[/math] Let [math]N_B[/math] be the number of units of Bond B purchased. Bond B is a zero-coupon bond, providing $100 per unit at Year 2. It is the only bond remaining that provides cash flow at Year 2 to cover the remaining liability. So, the cash flow from [math]N_B[/math] units of Bond B must cover the remaining liability at Year 2.

[[math]]N_B \times 100 = 102 - \frac{100}{21}[[/math]]

[[math]]N_B = \frac{1}{100} \left(102 - \frac{100}{21}\right) = \frac{1}{100} \left(\frac{102 \times 21 - 100}{21}\right) = \frac{1}{100} \left(\frac{2142 - 100}{21}\right)[[/math]]

[[math]]N_B = \frac{1}{100} \left(\frac{2042}{21}\right) = \frac{2042}{2100} \approx 0.97238[[/math]]

For calculations, we will use the more precise value [math]N_B = \frac{2042}{2100}[/math].

Step 5: Match Liability at Year 1

The liability at Year 1 is $99. First, we account for any cash flow from Bond C at Year 1. Cash flow from Bond C at Year 1 = [math]N_C \times $5 = \frac{100}{105} \times 5 = \frac{100}{21} \approx $4.7619[/math] Remaining liability at Year 1 = Total liability at Year 1 - Cash flow from Bond C at Year 1 Remaining liability at Year 1 = [math]$99 - \frac{100}{21}[/math] Let [math]N_A[/math] be the number of units of Bond A purchased. Bond A provides a cash flow of $107 per unit at Year 1 (a $7 coupon plus $100 par value). It is the only bond remaining that provides cash flow at Year 1 to cover the remaining liability. So, the cash flow from [math]N_A[/math] units of Bond A must cover the remaining liability at Year 1.

[[math]]N_A \times 107 = 99 - \frac{100}{21}[[/math]]

[[math]]N_A = \frac{1}{107} \left(99 - \frac{100}{21}\right) = \frac{1}{107} \left(\frac{99 \times 21 - 100}{21}\right) = \frac{1}{107} \left(\frac{2079 - 100}{21}\right)[[/math]]

[[math]]N_A = \frac{1}{107} \left(\frac{1979}{21}\right) = \frac{1979}{107 \times 21} = \frac{1979}{2247} \approx 0.88073[[/math]]

Step 6: Final Answer

The number of units of Bond A that must be purchased to match the liabilities exactly is approximately [math]0.8807[/math].

Key Insights

• Cash flow matching (exact immunization) is achieved by ensuring that the present value of assets equals the present value of liabilities, and cash inflows at each time point meet the outflows.
• When performing exact cash flow matching, it's generally most efficient to start with the longest maturity liability and work backward.
• Coupon payments from bonds purchased for longer-term liabilities must be considered as cash inflows for earlier liabilities.
• For each time point, identify all bonds that provide cash flow and determine the required number of units to cover the net liability (total liability minus cash flows from bonds already purchased for later liabilities).

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

Solution: A

Let N be the number of shares bought of the bond as indicated by the subscript.

[[math]] \begin{array}{l}{{N_{C}(105)=100,N_{C}=0.9524}}\\ {{N_{B}(100)=102-0.9524(5),N_{B}=0.9724}}\\ {{N_{A}(107)=99-0.9524(5),N_{A}=0.8807}}\end{array} [[/math]]

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