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midterm3_fall101

# midterm3_fall101 - E xam 3(FINA 4320/6320 Dec 2 2010...

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Unformatted text preview: E xam 3 (FINA 4320/6320) Dec 2, 2010, 5-6pm (60 min) I. Multiple Choice Questions: (21 points) 1. Assume that there is a forward market for a commodity. The forward price of the commodity is \$47. The contract expires in one year. The r isk-free rate is 10%. Now commodity six months later, the spot price is \$54. What is the forward contract worth at this six t ime? A . 9.09 B. 9.19 C. 11.27 D. 7.00 2. On a particular day the S&P 500 futures settlement price was 899.2. You buy one contract at around the close of the market at the settlement price. The next day, the contract contract opens at 899.90 and the settlement price at the close of the day is 899.08. contract Determine the value of the futures contract at the opening, an instant before the close. Determine Remember that the S&P futures contract has a \$250 multiplier. Remember A. \$205, -\$30 B. \$175, -\$30 C. \$175, -\$205 D. \$205, -\$205 3. On a particular day, the September S&P 500 stock index futures was priced at 940.50. The S&P 500 index was at 936.42. The contract expires 70 days later. 940.50. a). Assume continuous compounding, suppose the risk-free rate is 5.26%, and the dividend yield on the index is 2.65%. b) Assuming annual compounding, suppose the risk-free rate is 5.26%, Assuming and the future value of the dividends on the index is \$4.27. and What are the futures’ theoretical prices under assumption (a) and (b) respectively? A. 941.08 , 945.59 B. 945.59, 945.83 C. 945.59, 941.32 D. 941.08, 941.32 I I. Short answer questions: (79 points) 1. On September 26 the spot price of wheat was \$4.5225 per bushel and the price of a December wheat futures was \$4.66 per bushel. The interest forgone on money December tied up in a bushel until expiration is 0.04, and the cost of storing the wheat is tied 0.0975 per bushel. The risk premium is 0.045 per bushel. (24 points) 0.0975 a. what is the expected price of wheat on the spot market in December? S0=\$4.5225 , f0(T)= \$4.66 ,θ =0.04+ 0.0975=0.1375 , E(Φ)=0.045 E(ST )= E(f T (T)) =f 0(T) + E(Φ)= \$4.66 + 0.045=\$4.705 E(S b. Show how the futures price is related to the spot price. b. f 0(T) = S0 + θ =\$4.5225 +0.1375 =\$4.66 =\$4. c. Show how the expected futures price at expiration is related to the futures price c. today. t oday. E(f T(T)) =f 0(T) + E(Φ)= \$4.66 + 0.045=\$4.705 d. Explain who earns the r isk premium and why. d. 2. Use the following data from January 31 of a particular year for a group of March 470 options on futures contracts to answer parts a through g. (35 points) 470 Futures price:473.5, expiration: Mach 18, risk-free rate:0.0284% , call price:7.25, put price:5.45 price:5.45 a. Intrinsic value of the call? Max(0, f 0 – X)= Max(0, 473.50 – 470)= 3.50 b. Time value of the call? =Call Price – Intrinsic Value= 7.25– 3.50= 3.75 =Call c. Lower bound of the call? = Max[0, (f0 – X)(1 + r)-T]= Max[0, (473.50 -470)(1.0284)-0.1260]= 3.49 d. Intrinsic value of the put ? Max(0, X – f0)= Max(0, 470 – 473.50)= 0 )= e. Time value of the put ? Put Price – Intrinsic Value= 5.45 – 0= 5.45 f. Lower bound of the put? Max[0, (X – f0)(1 + r)-T]= Max[(0, (470 – 473.50)(1.0284)-0.1260]= 0 g. Does the put-call parity hold? C= P + (f0 – X)(1 + r) -T= 5.45 + (473.50 – 470)(1.0284)-0.1260 C= = 8.94 8.94 The actual call price is 7.25, so put-call parity does not hold. T he 5 points each. 3. Suppose you are a dealer in sugar. It is September 26, and you hold 112,000 pounds of sugar worth \$0.0469 per pound. The price of a futures contract expiring pounds in January is \$0.050 per pound. Each contract is for 112,000 pounds. (20 points) in a). Determine the original basis. Then calculate the profit from a hedge if it is held to expiration and the basis converges to zero. The dealer is long sugar in the spot market and should sell sugar futures to set up a hedge The S0 = 0.0469 f0 = 0.050 b0 = S0 – f0 = 0.0469 – 0.0500 = –0.0031 (5points) In terms of the basis, In π = – b0 + bt = – (–0.0071) + 0 = 0.0031 (5 points) In dollars, π = 112,000(\$0.0031) = \$347.20 In Thus, the profit on the hedge is –1 times the original basis times the number of pounds. Thus, b) Rework this problem, but assume the hedge is closed on December 10, when the spot price is \$0.0574 Rework and the January futures price is \$0.0590. and bt = St – ft = 0.0574 – 0.0590 = –0.0016 In terms of the basis, π = – b0 + bt = – (–0.0031) + (–0.0016) = 0.0015 The basis went from –0.0071 to –0.0016, a profit of 0.0015. In dollars, The π = 112,000(\$0.0055) = \$616 Thus, the basis strengthened so the hedger gained, though not as much as if the hedge had been held to expiration. held V T(0,T) = ST - F (0,T). Vt(0,T) = St – F(0,T)(1+r)-(T-t) f0(T) = S0(1+r)T - DT f 0(T) = (S0 – D 0)(1+r) T V t (0,T) = St – D t,T – F (0,T)(1 + r) -(T-t) S0 = E (ST) - s - iS 0 - E (φ ) f 0(T) = S0 + θ E(ST ) = E(f T (T)) = f 0(T) + E(φ ) -T Pe(f 0(T),T,X) = Ce(f 0(T),T,X) + (X - f 0(T))(1+r) . Pe(S0,T,X) = Ce(S0,T,X) + (X - f0(T))(1+r)-T ...
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