Question

7. Below is a list of prices for $1,000 par zero-coupon bonds of various maturities. Maturity (Years) Bond AWNA Price $930 $850 $770 $700 1.4 a. Compute the zero-coupon rates for years 1, 2, 3 and 4. b. Consider an 8% coupon $1,000 par bond (denoted by B) paying annual coupons and expiring in 4 years. Compute the no-arbitrage price of the bond and its yield-to-maturity. c. If the expectations hypothesis holds, what is your forecast for the 3-year interest rate expected next year? d. If bond B was trading today for $985, is there an arbitrage opportunity that can be exploited? If so, explain how an investor would exploit such a strategy, i.e. indicate which securities the investor would buy or sell, as well as the quantities.

Help with finance question please.

Answer 1

Price of the zero coupon bond (ZCB) maturing in n years = P_n = 1000 / (1 + z_n)ⁿ

Hence, the zero coupon rate for n years = z_n = (1,000 / P_n)^1/n - 1

Part (a)

Bond	Maturity (years)	Price	Part (a) Zero coupon rate
	n	Pn	zn
Z1	1	930	7.53%
Z2	2	850	8.47%
Z3	3	770	9.10%
Z4	4	700	9.33%

Part (b)

Price of the coupon bond = Sum of PV of all the coupons + PV of principal repayment

Coupon = 8% = 8% x 1,000 = 80

So, the cash flows for this bond are $ 80 in first three years and $ 80 + 1,000 = 1,080 in year 4. This bond can be seen as a combination of 80/1000 = 0.08 number of Z1, Z2, Z3 and 1,080 / 1,000 = 1.080 number of Z4. This portfolio will produce exactly the same cash flows as B.

Hence, B = 0.08 x (Z₁ + Z₂ + Z₃) + 1.08 x Z₄

Hence, its no arbitrage price should be = 0.08 x (P₁ + P₂ + P₃) + 1.08 x P₄ = 0.08 x (930 + 850 + 770) + 1.08 x 700 = 960.00

Part (c)

Forecast for 3 year interest rate next year = ₁F₄ will be given by:

(1 + z₁)(1 +  ₁F₄)³ = (1 + z₄)⁴

Hence, ₁F₄ = [(1 + z₄)⁴ / (1 + z₁)]^1/3 - 1 = [(1 + 9.33%)⁴ / (1 + 7.53%)]^1/3 - 1 = 9.93%

Part (d)

If Bond B is actually trading at 985, i.e. at a price higher than the no arbitrage price determined in part (b), then there exists an arbitrage profit.

• Short (sell) the bond B
• Buy (long) the synthetic bond i.e
  • Buy 0.08 number of Z1, Z2, Z3 and
  • Buy 1.08 number of Z4