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The answers to three financial economics questions. The first question involves estimating the probability of a daily return on the s&p 500 being less than 0.04% and calculating the one-day 99% var for a portfolio invested in the s&p 500. The second question deals with calculating the value of a european call option on a non-dividend-paying stock, assuming no possibility of default and considering a 2% chance of default at maturity. The third question focuses on calculating the total economic capital for two business units and determining the incremental effect of each business unit on the total economic capital.
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Question 7 (2 points) During the period July 11, 1988 and July 10, 1998 a total of 2,256 daily return observations on the S&P 500 have been observed, ranging from -6.87% to +5.12%. Consider the left tail of the distribution and apply extreme value theory with u=0.02, there were in total 28 returns less than -2%. Furthermore use as parameters: ξ =^0.^3232 , and β =^0.^0055. a) Estimate the probability that the return will be less than 0.04. b) What is the value of the one-day 99% VaR for a portfolio where $1 million is invested in the S&P 500? Answer: We assume that the distribution is symmetric! (12.5) Prob ( v > x )= nu n (
x − u ß ) − 1 / ¿¿ a) Prob (^ v > 0.04)^ =
2256 (
0.0055 ) − 1 /0. =0. Prob ( v < 0.04) = 1 − Prob ( v > 0.04) = 1 −0.001121=0. b) (12.6) VaR = u +^ ß ❑
( n nu ( 1 − q ) ) −¿− 1 ⌉ ¿ q = F(VaR) VaR =0.02+
⌈ (^) (
( 1 −0.99)) −0. − 1 ⌉ =0. For 1 Million $ is invested: 0.02121x1,000,000 = 21200
Question 8 (2 points) Consider a European call option on a non-dividend-paying stock where the stock price is $52, the strike price $50, the risk-free rate is 5%, the volatility is 30%, and the time-to-maturity is 1 year. (a) What is the value of the option assuming no possibility of a default? (b) What is the value of the option to the buyer if there is a 2% chance that the option seller will default at maturity? Answer: See appendix E (p.463) for the formulas a) c = S 0 N(d 1 ) - Ke-rT^ N(d 2 ) d 1 = ln(
K ) +( r + o 2 2 )
o (^) √ T d 2 = ln(
K ) −( r + o 2 2 )
o (^) √ T d 1 = ln(
50 ) +(0.05+
2 2 )
0.3 (^) √ 1 =0.4474 d 2 =0. c = 52 N ( 0.4474) − e −0.05∗ 1 N ( 0.1474 )=8. b)