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Answer Key July 2009.docx, helpful for students
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For Questions and Errors please refer to: Nowak, Matthias: [email protected] Stadtfeld, Markus: [email protected] Dilthey, Jacob: [email protected] Termeulen, Thom: [email protected] Hermanns, Ulf: [email protected] Question 1 (3 points) Consider a trader who sells 100,000 European call options on a non- dividend paying stock, when: (1) the stock price is $49, (2) the strike price is $50, (3) the risk free rate is 5%, (4) the stock price volatility is 20% per annum, (5) the time to option maturity is 20 weeks. Furthermore, suppose that the amount received for the options is $300,000 and that the trader has no other positions dependent on the stock (a) Determine the value of the option (b)Determine the delta of the option (c) What position in stocks should the trader take to make the portfolio delta neutral After 1 week the stock price decreases to $48.12. (d)What is the delta of the portfolio of options after the change (e) Take the ‘old’ hedge position as determined under (c) as point-of-departure and determine how many stocks should be bought or sold to make the portfolio delta neutral (a)
− rT
d 1 = ln ( S 0 / K )+( r + σ 2 / 2 ) T σ (^) √ T d 2 = d 1 − σ (^) √ T d 1 = ln ( 49 / 50 )+( 0.05+0. 2 / 2 )
d 2 = d 1 −0.
Looking up the values in the Normal Distribution Table yields N ( 0.054 )=0. N (−0.070) =0. C = 49 ∗0.5220− 50 e −0.05 (^2052) ∗0. C = 2. Since the trader is short in 100,000 options, the value of his portfolio is -242,261. The options have been sold for $57,739 more than their theoretical value. (b) The Delta for a Call option is calculated with: Δ = e − qT N ( d 1 ) Δ = e − (^0 ) 0.4364=0. because the trader is short in 100,000 options his delta exposure is -52, (c) To delta neutralize the portfolio the trader has to buy 52,200 shares After one week the stock price decreases to $48. (d) d 1 = ln (48.12/ 50 )+(0.05+ 0. 2 / 2 )
Δ = e 0 T 0.4582=0. because the trader is short in 100,000 options his delta exposure is now -45, (e) The ‘old’ hedge included a long position in 52,200 stocks. Since the delta has decreased for the traders portfolio only 45,820 shares are needed to delta neutralize the portfolio. Therefore 52,200-45,820= 6380 shares are sold.
The price change for each maturity can be calculated as: ∆ Pi P =− Di ∆ yi where ∆ Pi P is the relative price, Di the duration and ∆^ yi the interest rate change; for i in years of maturity. So for 1, 2, 3, 4, 5, 7 and 10 years the relative price change is: Maturi ty Relative price change 1 −2.0 × ( 3.7 %−4.0 % )=+0.6 % 2 −1.6 × ( 4.3 %−4.5 % )=+0.32 % 3 −0.6 × ( 4.7 %−4.8 %) =+ 0.06 % 4 −0.2 × ( 5.0 %−5.0 % )= 0 % 5 0.5 × ( 5.2 %−5.1 % )=−0.05 % 7 1.8 × ( 5.5 %−5.2 % )=−0.54 % 10 1.9 × ( 5.9 %−5.3 % )=−1.14 % Assuming each maturity has an equal weight, the price change of the portfolio is:
Question 3 (2 points) Consider an exponentially weighted moving average scheme
per day, and during day 0 the market variable increased by 2%. (a) What is the volatility on day 1 according to the EWMA scheme. (b)Suppose that no further shocks occur during the next 60 days. What would be the volatility according to the EWMA scheme after 60 days. Answer: (a) Equation (9.8.)
Daily volatility = = 1% = 0. Most recent daily percentage change = = 2% = 0. = 0.90 * (0.01)^2 + (1-0.90)*(0.02) 2 = 0. σ = (^) √0. σ = 0.0114 = 1.14% (b) σ (^) 2 =√0.90∗0. σ (^) 60 =√0. 59 ∗0. σ (^) 60 =√0. σ (^) 60 =0.04468=4.47 % Question 4 (3 points) Considertwo$10millionone-yearloanseachofwhichhasa1.25%chanceof defaulting.Ifadefaultoccursononeoftheloans,therecoveryoftheloanprincipleis uncertain,withallrecoveriesbetween0%and100%beingequallylikely.Iftheloan doesnotdefault,aprofitof$0.2millionismade.Supposethatifoneloandefaults,it iscertainthattheotherloanwillnotdefault. (a) Providethe4mostimportantdesirablepropertiesofariskmeasureanda shortdescriptionthereof. Artzner et al. have defined four properties for a risk measure that will equal the capital a financial institution is required to keep should satisfy.
Because $7.8 million is less than 2*$6 million, the expected shortfall measures does satisfy the subadditivity condition. Question 5 (3 points) A bank has a portfolio of options on an asset. The delta of the options is -30 and the gamma is -5. The asset price is 20 and its volatility is 1% per day. (a) Using the Quadratic model calculate the first three moments of the change in the portfolio value. The quadratic model is given by the following equation: ∆ P = S δ ∆ x +
2 γ ( ∆ X ) 2 When ∆^ x^ is assumed to be normal, we can calculate the first three moments by the following equations: E ( ∆ P ) =0.5 S 2 γ σ 2
2 δ 2 σ 2 +0.75 S 4 γ 2 σ 4
3
4 δ 2 γ σ 4 +1.875 S 6 γ 3 σ 6 Hence: ( ∆ P ) =0.5 20 2 (− 5 ) 0. 2 =−0.
3
4 (− 30 2 )(− 5 ) 0. 4 +1.875 20 6 (− 5 ¿¿ 3 )0. 6 =−32.415 ¿ (b) Calculate a 1-day 99% VaR using the first two moments. Using the first two moments, the 1-day 99% VaR for the portfolio can be calculated as follows:
μp = E ( ∆ P )=−0. σ (^) p 2 = E (^) [( ∆ P ) 2 ] −[ E^ (^ ∆^ P )^ ] 2 =36.03−(−0.01) 2 =36. σ (^) p =6. Assuming that the probability distribution of ∆ P is normal, the 0.01 quantile is -0.1 – 2.33 x 6.0025 = -14. Therefore, we are 99% certain that ∆ P >−14. ) Calculate a 1-day 99% VaR using the first three moments. In order to calculate the first three moments, the skewness of the Cornish fisher expansions has to be taken into account, which can be calculated as follows: ξP =
σ (^) P 3 E^ [(^ ∆^ P − μP^ ) 3 ]=^ [ E ( ΔP ) 3 ]− 3 [ E ( ΔP ) 2 ] × μP + 2 μP 3 σ (^) P 3 ξP =−0. Using the Cornish-Fisher expansion to adjust for skewness and setting q = 0.01, we obtain ωq = zq +
( zq 2 − (^1) ) × ξP ωq =−2.33+
((−2.33 ) 2 − 1 ) × (−0.0999) =−2. so that the 0.01 quantile of the distribution is VaR = μ (^) p + ωq σ (^) p =−0.1+ (−2.40374 × 6.0025) =14.
Insurance company cannot distinguish between good and bad risks. It offers the same price to everyone and inadvertedly attracts more of the bad risks (b)