Exercise Solution 10.2
- Define
1 
N(0,1) such that[s1]
Substituting this into [10.39], we obtain
[s2]
[s3]
[s4]
Values for
, 
and 
are, respectively, –1280, –11520, and 54080. - Because it entailed no approximation, our work in part (a) comprised a mapping:
[s5]
c, d. Results, calculated with a spreadsheet, are presented in Exhibit s1.
| k | g[k] | 0E(1P k) |
|---|---|---|
| 0 | 52800 | |
| 1 | 1.360 x 108 | 52800 |
| 2 | -1.036 x 1012 | 2.924 x 109 |
| 3 | 1.057 x 1016 | 1.677 x 1014 |
| 4 | -1.349 x 1020 | 9.894 x 1018 |
| 5 | 5.975 x 1023 |
Exhibit s1: Results for items (c) and (d)—conditional values g[0] through g[4] for 1P and conditional moments 0E(1P ) through 0E(1P 5)
- Based upon its first two moments, we calculate the standard deviation of 1P using the formula for variance derived in Exercise 3.15:
[s6]
[s7]
[s8]
Since 0std(1L) = 0std(1P), the portfolio’s standard deviation of loss is USD 11661.
- Results, calculated with a spreadsheet, are presented in Exhibit s2.
central
momentsnormalized
central
momentsnormalized
cumulants1 0 0 0 2 1.36×108 1 1 3 -1.036×1012 -.6534 -.6534 4 6.604×1016 3.5718 .5714 5 -1.544×1021 7.1597 -.6257 Exhibit s2: Results for item (f)—conditional central moments of 1P as well as the conditional central moments and cumulants of the normalization 1P * of 1P. - Using a spreadsheet, we use the Cornish-Fisher expansion to calculate the .10-quantile of 1P* as –1.3367. By [3.209], the .10-quantile of 1P is 37213. Because the portfolio’s current value is 53600, the .90-quantile of loss is 53600 – 37213 = 16387.
- Completing the squares, we obtain
[s9]
[s10]
- Because 1 is conditionally standard normal, the random variable (1 + 4.5)2 is conditionally chi-squared with degrees of freedom ν = 1 and non-centrality parameter
= 20.25. - By [3.220], the characteristic function of (1 + 4.5)2 is
[s11]
Applying [3.217], the characteristic function of 1P is
[s12]
- This item is best performed by coding a simple program that applies the inversion theorem by using the trapezoidal rule. We start with seed values 1p[1] = 25000 and 1p[2] = 50000. We use the simple program to value the CDF of 1P at each, and then apply the secant method. The resulting sequence of values 1p[k] converges to the desired .10-quantile of 1P. See Exhibit s3.
k p[k] Φ(p[k]) 1 25000 0.0199 2 50000 0.3665 3 30776 0.0444 4 34093 0.0683 5 38507 0.1163 6 37007 0.0976 7 37201 0.0998 8 37214 0.1000 Exhibit s3: Results of item (k)—Within eight steps, the secant method finds the .10-quantile of 1P.The .10-quantile of 1P is obtained as 37214. Because the portfolio’s current value is 53600, the .90-quantile of loss is 53600 – 37214 = 16386. This compares favorably with the approximate result of 16387 obtained in part (g).