1 < = α σ +σ < 0. Using the parameters and h = 1/365 this is N ( ) = If we use h = 1/252, the value would be N ( ) =
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1 Chater 6 Value at Risk Question 6.1 Since the rice of stock A in h years (S h ) is lognormal, 1 < = α σ +σ < 0 ( ) P Sh S0 P h hz σ α σ α = P Z < h = N h. σ σ (1) () Using the arameters and h = 1/365 this is N ( ) = If we use h = 1/5, the value would be N ( 0.05) = Question 6. A 95 ercent, VaR uses Z 1 = 1.645, and the 99 ercent VaR uses Z =.36. Given the horizon h (in years), the value of 10 million will be ( ) h hz 10e α σ +σ i million. Table 1 shows these values as well as the loss (VaR). Table 1 (Problem 6.) 34
2 Chater 6/Value at Risk 343 Question 6.3 Using the normal aroximation, the ortfolio will have a mean return of α = 16.8 ercent and standard deviation of σ = 3.17 ercent. Letting h be the holding eriod, there is a 95 ercent (or 99 ercent) chance the value of the ortfolio will exceed $10m 1 h hz +α +σ i (3) where Z 1 = (95%) and Z =.36. See Table for the numerical answers. Table (Problem 6.3) Question 6.4 The ortfolio mean is α = 16.3 ercent and the standard deviation is σ = 8.65 ercent. Letting h be the holding eriod, there is a 95 ercent (or 99 ercent) chance the value of the ortfolio will exceed $10m 1 h hz +α +σ i (4) where Z 1 = (95 ercent) and Z =.36 (99 ercent). See Table 3 for the numerical answers. Table 3 (Problem 6.4)
3 344 Part Five/Advanced Pricing Theory Question 6.5 See Table 4 for the numerical answers. Risk is not eliminated at ρ = 1 for the ortfolio volatility is σ = 15 ercent. If we had 60 ercent in A and 40 ercent in B, then Table 4 (Problem 6.5) ( ) ( ) ( )( )( )( ) σ = + = (5)
4 Chater 6/Value at Risk 345 Question 6.6 The 100, strike, one-year ut otions have a remium of $1,06,694.90; hence, W = 11,06,694. The delta (er share) is Using equations (6.10) and (6.11), we obtain R 0.15 = 100 ( ( )) = W (6) and ( 100,000 ( )) σ = = W Table 5 shows the six VaR values using the normal aroximation. Table 5 (Problem 6.6). (7) Question 6.7 Monte Carlo simulations should give answers close to the analytical values that are shown in Table 6. These are derived by using the Z = and Z =.36 to find the future stock rice, which will imly Black-Scholes values for the otion, which determines the ortfolio value. Table 6 (Problem 6.7) Question 6.8 We first do the roblem analytically; since we have written otions, we receive the remiums, which is $1,571,10. It is virtually imossible for S to fall enough in a week for there to be a loss (Z 5.64). The 95 ercent, 10-day VaR is, therefore, the loss when Z = 1.645; with this value, the otion osition will be $1,901,066, which is a loss of $39,856. Monte Carlo simulations should confirm this.
5 346 Part Five/Advanced Pricing Theory The delta of our osition is 100,000 ( ) = 31,814. If we use the delta aroximation, we could use a normal aroximation with return and variance 100,000 R = 0.15 ( 100 )( ) = 30.37% (8) 1,571,10 ( ) , %. (9) 1,571,10 σ = = Using a normal aroximation, the value of our ortfolio would be Wh = 1,571,10 1+ R +σ Z (10) Our rofit is W h + 1,571,10. If Z = 1.645, the rofit will be negative; this amounts to a 95 ercent, 10- day VaR of $7,947, which is much less than the true VaR of $39,856. This large error is due to the linear aroximation of a highly nonlinear ayoff. The ayoff is similar to Figure 6.3. The negative delta of 31,814 underestimates the true loss due to a negative gamma. Question 6.9 See Table 7 for the tail VaRs, as well as the conditional exected values and the terms d 1 and d (as in Examle 6.11). Note the numerical tail VaR is stated as a loss. For examle, the 95 ercent, one-day tail VaR for stock B is a loss of $471,8. Table 7 (Problem 6.9)
6 Chater 6/Value at Risk 347 Question 6.10 As in roblem 6.8, the loss occurs when the stock rice rises, i.e., Z = At this value the ortfolio is worth 1,901,066. The conditional exected value of the ortfolio should be aroximately.05m leading to tail VaR that is aroximately $454,000. The simulation method is to sort the ortfolio values and then take the average only of those simulations where the value is greater than 1.9m. Question 6.11 The one-year cash or nothing 50-strike ut has a remium of The cash or nothing 15- strike call has a remium of a) If we write the ut,there is only a 0.39 ercent chance of having to ay $1. Hence the 99 ercent, one-year VaR is a gain of the remium If we write the call, there is a 1.38 ercent chance of aying $1. Hence the 99 ercent, one-year VaR is a loss of b) If we write both otions, the lower 1 ercent of the return distribution has one of the otions being exercised, which imlies a 99 ercent, one-year VaR equal to a loss of = Note the VaR measure is not subadditive since ρ ( A B ) = >= = ρ ( A) + ρ ( B ). (11) c) The tail VaRs for writing the call ( ) and writing both otions (0.9861) are the same since the loss is indeendent of S (conditional on the 1 ercent event). For the ut, we have to look at the exected loss, conditional on the 99 ercent S value. The ut osition has a 0.39 ercent chance of a loss of = The conditional robability of this loss is /0.01 = Hence, the tail VaR (loss) for the written ut is 0.39 (0.9998) ( ) = (1) Note we now have subadditivity as Question 6.1 Since 8 = , interolation imlies ρ ( A B ) = < = ρ ( A) + ρ ( B ). (13) 1 y 8 = ( 0.06) + ( 0.065) = (14) σ 8 = ( 0.10 ) + ( 0.095) = (15) 3 3
7 348 Part Five/Advanced Pricing Theory The yield being ercent imlies the bond osition is worth 10e (8) = $6.1059m. Using these values, the 10-day, 95 ercent VaR is 10 $6.1059m ( ) $6.1059m = $1.31m 365 (16) and a 10-day, 99 ercent VaR of 10 $6.1059m (.36 ) $6.1509m = $1.85m. 365 (17) Question 6.13 We must account for the correlation of the eight-year bond with the seven- and ten-year bonds. Note that the eight-year bond is not matched to a /3,1/3 ortfolio of the seven- and ten-year bonds. As in equation (6.16), the cash flow maing ortfolio must solve ( ) ( ) ( ) ( )( )( ) ( ) , =ω + ω + ω ω (18) which yields two solutions ω = or ω = We choose the more sensible solution ω = We also need to look at the nine-year zero couon bond. Similar analysis finds and a cash flow maing ortfolio that solves 1 y 9 = ( 0.06) + ( 0.065) = (19) σ 9 = ( 0.10 ) + ( 0.095) = (0) 3 3 ( ) ( ) ( ) ( )( )( ) ( ) , =ω + ω + ω ω (1) which has a solution of ω = The $3m cash flows in years seven through ten will have an initial value of W 0.06(7) (8) (9) 0.065(10) = $3m e + e + e + e = $7, 065, 617m. () In terms of ercentage of the total value invested in each of the four bonds, we have the ortfolio [0.7898, 0.595, 0.401, 0.166]. Hence, the ortfolio of four bonds in terms of the sevenand ten-year bonds is ω = (0.7963) (0.5891) = (3)
8 Chater 6/Value at Risk 349 in the seven-year bond and in the ten-year bond. This ortfolio has a variance of ( ) + ( ) (4) for a volatility of = = for a 10-day, 95 ercent VaR of 10 $7.0656m ( ) $7.0656m = $1,518, 16. (5) 365
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