6 Options II

# Fig 9 1 let z n0 1 be a standard normal random

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: OPTIONS II p. 10 of 18 D. Binomial (Two- State) Option Pricing Trees {Hull Ch. 12 (omit §12.9); BKM §21.3} 1. Call Option Pricing (1- Period): [Fig. 3] uS0 Fig. 3 ST−K \$200 a) A no- dividend stock sells for S0 = \$100 \$75 today. At the end of one year, its price will S0 c either: dS0 \$100 • rise by factor u = 2 to uS0 = \$200, or \$0 \$50 • decline by factor d = ½ to dS0 = \$50. b) Consider a call option on this stock with K = \$125 and T = 1 year. In one year, it will be worth its payoff ST − K = 200 − 125 = \$75 (if ST = 200 > K) or \$0 (if ST = \$50 < K). c) Now construct the following perfect hedge portfolio H: • Buy ½ share of stock (pay \$50) • Write/sell one call option (receive \$c) 1) Cost today H0 = \$50 − c 2) Value in 1 year {HT}: • If ST = \$200, buy ½ share for 100 and sell 1 share to option holder for K = \$125. You are left with HT = \$125 − \$100 = \$25. • If ST = \$50, option not exercised. You own stock worth HT = 50(1/2) = \$25. 3) By replication principle #3, H0 = HTe−rT. Let rf = 10%/year. • 50 – c = 25e−0.10(1) = \$22.62 ⇒ c = 50 − 22.62 = \$27.38 d) Summarizing the above result. 1) We know the current stock price S0 and option strike price K and time to maturity T. 2) If we also know the only two possible values of the stock price ST, we can use replica- tion to deduce the value of the call option and its premium c. 2. The Way it Works – Arbitrage: {Hull §12.1 - 12.2} a) By constructing the perfect hedge portfolio, we invest H0 = ½S0 – c = \$50 – c at the beginning, and get \$25 back for certain (i.e., with no risk) at the end of one year. 1) This means that our investment grew at a risk- free rate of return rf, and there can’t be more than one risk- free rate at a given time. 2) So if rf = 10%, then (50 – c)e0.10(1) = \$25, so 50 – c = 25e- 0.10(1) = \$22.62, and c = \$27.38. 3) That is, ½S0 – c = \$22.62 invested now grows at rf = 10% to \$25 in one year. b) Suppose that c = \$30 > \$27.38 (and hold S0 = \$100 and rf = 10% constant)....
View Full Document

## This document was uploaded on 02/18/2014 for the course ECON 174 at UCSD.

Ask a homework question - tutors are online