ps_options_sol

# ps_options_sol - Problems 1 through 10, part a). Payoff #1...

This preview shows pages 1–4. Sign up to view the full content.

Fi8000 Practice Set #1 Check Solutions 1 Problems 1 through 10, part a). S Payoff +10 0 -10 +10 +20 +20 S Payoff +10 0 -10 +10 +20 +20 S Payoff +10 0 -10 +10 +20 +20 S Payoff +10 0 -10 +10 +20 +20 S Payoff +10 0 -10 +10 +20 +20 S Payoff +10 0 -10 +10 +20 +20 #1 #2 #3 #4 #5 #6 S Payoff +10 0 -10 +10 +20 +20 S Payoff +10 0 -10 +10 +20 +20 S Payoff +10 0 -10 +10 +20 +20 S Payoff +10 0 -10 +10 +20 +20 #7 #8 #9 #10

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Fi8000 Practice Set #1 Check Solutions 2 Problem b) Profitable Range c) Maximum Profit c) Maximum Loss 1 S < \$22.80 \$12.80 Unlimited 2 S > \$22.60 \$7.40 \$2.60 3 S > \$13.10 \$6.90 \$13.10 4 S < \$7.00 and S > \$33.00 Unlimited \$8.00 5 S > \$24.30 Unlimited \$19.30 6 S > \$30.20 Unlimited \$10.20 7 S < \$22.80 \$12.80 Unlimited 8 S < \$12.00 and S > \$38.00 Unlimited \$13.00 9 \$12.20 < S < \$27.80 \$2.80 \$2.20 10 S < \$22.00 \$2.00 \$3.00 11. Long Put, X=20; Short Bond, FV=10. 12. Long Put, X=15; Long Call, X=25; Long Bond, FV=10. 13. Short Put, X=20; Short Call, X=25; Long Bond, FV=5. 14. Long Put, X=20; Short Put, X=10; Long Call, X=20; Short Call, X=30. 15. Short Put, X=20; Long Call, X=25; Short Call, X=30; Long Bond, FV=10. 16. Short Put, X=30; Lon Put, X=15; Short Call, X=30; Long Bond, FV=5. 17. Note that the payoff to Call A is always greater than or equal to the payoff to Call B. Since Call A is cheaper, we buy Call A and write (sell) Call B, pocketing \$1.25 today. At expiration, we use any payoff from Call A to payoff anything due on Call B. At worst, we get zero at expiration and at best we get \$10 at expiration. Thus, we have an arbitrage opportunity. The table shows the transactions and cash flows. Today (time 0) Transaction Cash Flow Buy a call, X=50 - 5.25 Write a call, X=60 + 6.50 Net cash flow + 1.25 Expiration (time 1) Position S < 50 S = 50 50 < S < 60 S = 60 S > 60 Long call, X=50 0 0 S – 50 10 S – 50 Short call, X=60 0 0 0 0 60 – S Net cash flow 0 0 0 to +10 +10 +10 18. This is similar to the last problem. The payoff to Put Z is always greater than or equal to the payoff to Put Y. Since Put Z is cheaper, we buy Put Z and write (sell) Put Y, pocketing \$0.75 today. At expiration, we use any payoff from Put Z to payoff anything due on Put Y. At worst, we get zero at expiration and at best we get \$10 at expiration. Thus, we have an arbitrage opportunity. The table shows the transactions and cash flows. Today (time 0) Transaction Cash Flow Buy a put, X=40 - 2.50 Write a put, X=30 + 3.25 Net cash flow + 0.75
Fi8000 Practice Set #1 Check Solutions 3 Expiration (time 1) Position S < 30 S = 30 30 < S < 40 S = 40 S > 40 Long put, X=40 40 – S + 10 40 – S 0 0 Short put, X=30 S – 30 0 0 0 0 Net cash flow + 10 +10 0 to +10 0 0 19. We can now make general statements based on 17. and 18. Call options (on the same underlying stock and the same expiration date) with lower strike prices must have premiums greater than or equal to the premiums of call options with higher strike prices. So, C(X1) C(X2). Put options (on the same underlying stock and the same expiration date) with lower strike prices must have premiums less than or equal to the

This preview has intentionally blurred sections. Sign up to view the full version.

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

## This note was uploaded on 05/20/2011 for the course ECON 5128 taught by Professor Ram during the Spring '11 term at Cambridge College.

### Page1 / 8

ps_options_sol - Problems 1 through 10, part a). Payoff #1...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online