Solution of Derivative

# 081656 w 2 1002 32 p 100000 1 39972 16331 2 w 3 rp

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Unformatted text preview: 77625 0.077682 0.077059 0.086744 0.089484 0.090998 0.101981 0.107003 0.125192 Year 5 European Calculations 0.07614931 0.076312 0.07572 0.074778 0.08817 0.089286 0.089912 0.104526 0.107746 0.128608 0.11307958 0.11298736 0.11040541 0.11048209 a) Note the forward rates only depend on the initial bond prices; for example, rf (2) = B (0, 1) /B (0, 2) − 1 = .925926/.849454 − 1 = 9.0025%. This immediately implies the yield volatilities do not affect these forward rates. b) These rates were computed by formulas such as re (2) = ru + rd 2 145 (30) Part 5 Advanced Pricing Theory and re (3) = 1 B (0, 2) rdd + rdu rdu + ruu 1 1 + . 2 (1 + rd ) 2 2 (1 + ru ) 2 (31) c) From Table Four, we see that the difference between the two settlement styles is larger for the high volatility tree (#1) for the year 2 and 3 forward rates. In addition, the difference is larger for the later years. In looking at the short rate trees we see that the short rate tree #1 has a lower dispersion in year 4 (ranging from 7.55% to 15.81%) than it does in tree #2 (ranging from 7.28% to 16.28%). This causes the difference for the 5 year rates to be more pronounced for tree #2. 146 Chapter 24 Risk Assessment Question 24.2. A 95% VaR uses Z1 = −1.645 and the 99% VaR uses Z2 = −2.326. Given the horizon h (in years), √ 2 the value of 10 million will be 10e α −σ /2 h+σ hZi million. Table One shows these values as well as the loss (VaR). Table One (Problem 24.2) 95% Values A B Loss (VaR) A B 99% Values A B Loss (VaR) A B 1 day 9,747,824 9,622,055 1 day 252,176 377,945 10 day 9,242,241 8,866,025 10 day 757,759 1,133,975 20 day 8,960,529 8,445,521 20 day 1,039,471 1,554,479 1 day 9,644,067 9,468,836 355,933 531,164 10 day 8,934,714 8,427,213 1,065,286 1,572,787 20 day 8,541,801 7,860,500 1,458,199 2,139,500 Question 24.4. The portfolio mean is αp = 16.3% and the standard deviation is σp = 28.65%. Letting h be the holding period, there is a 95% (or 99%) chance the value of the portfolio will exceed √ \$10m × 1 + αp h + σp hZi (1) where Z1 = −1.645 (95%) and Z2 = −2.326 (99%). See Table Three for the numerical answers. Table Three (Problem 24.4) Values 95% 99% Loss (VaR) 95% 99% 1 day 9,757,785 9,655,581 1 day 242,215 344,419 147 10 day 9,264,584 8,941,387 10 day 735,416 1,058,613 20 day 8,986,124 8,529,054 20 day 1,013,876 1,470,946 Part 5 Advanced Pricing Theory Question 24.6. The 100,000 105-strike one year put options have a premium of \$1,026,694.90, hence W = 11, 026, 694. The delta (per share) is −0.3997. Using equations (24.10) and (24.11), we obtain .15 × 100 (100000 (1 − .3997)) = .081656 W (2) 1002 × .32 σp = (100000 (1 − .3997))2 = .16331 2 W (3) Rp = and Table Five shows the 6 VaR values using the normal approximation. Table Five (Problem 24.6) 95% 99% 1 day 10 day 20 day 1 day 10 day 20 day Value 10,874,121 10,561,083 10,382,670 10,809,886 10,357,951 10,095,399 Loss (VaR) 152,574 465,612 644,025 216,809 668,743 931,296 Question 24.8. We ﬁrst do the problem analytically; since we have written options we receive the premiums which is \$1,571,210. It is virtually impossible for S to fall enough in a week for there to be a loss (Z ≈ −5.64). The 95% 10-day VaR is therefore the loss when Z = 1.645; with this value the option position will be \$1,901,066 which is a loss of \$329,856. Monte Carlo simulations should conﬁrm this. The delta of our position is −100000 (−.22128 + .539417) = −31814. If we use the delta approximation, we could use a normal approximation with return Rp = 100000 .15 (100) (.318135) = 30.372% 1571210 (4) 1002 (31814)2 .32 = 36.899% 15712102 (5) and variance 2 σp = Using a normal approximation, the value of our portfolio would be Wh = −1571210 1 + Rp 148 10 10 + σp Z. 365 365 (6) Chapter 24 Risk Assessment Our proﬁt is Wh + 1571210. If Z = 1.645, the proﬁt will be negative; this amounts to a 95% 10day VaR of \$272,947 which is much less than the true VaR of \$329,856. This large error is due to the linear approximation of a highly non-linear payoff. The payoff is similar to Figure 24.3. The negative delta of −31814 underestimates the true loss due to a negative gamma. Question 24.10. As in problem 24.8, the loss occurs when the stock price rises, i.e. Z = 1.645. At this value the portfolio is worth −1, 901, 066. The conditional expected value of the portfolio should be approximately 2.025m leading to tail VaR that is approximately \$454,000. The simulation method is to sort the portfolio values and then take the average only of those simulations where the value is greater than 1.9m. Question 24.12. Since 8 = 2 7 + 1 10, interpolation implies 3 3 2 (.06) + 3 2 σ8 = (.10) + 3 y8 = 1 (.065) = .06167. 3 1 (.095) = .09833. 3 (7) (8) The yield being 6.167% implies the bond position is worth 10e−.06167(8) = \$6.1059m. Using these values, the 10-day 95% VaR is \$6.1059m × 1 + .09833 × 8 × 10 × (−1.645) − \$6.1059m = −\$1.31m 365 (9) 10 × (−2.326) − \$6.1509m = −\$1.85m 365 (10) and a 10-day 99% VaR of \$6.1509m × 1 + .09833 × 8 × Question 24.14. Table Eight shows the transition probabilities for years 2, 3, and 4. These are determined by calculations similar to (24.21). As an example, the probability of going from an F rating to an FF rating in two years is .9 × .07 + .07 × .8 + .03 × .3 = .128. (11) The ﬁrst term is the probability of keeping an F rating in one year and then going from F to FF in year 2. The...
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## This document was uploaded on 03/11/2014 for the course FIN 402 at FPT University.

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