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# Exercise Solution 10.2

1. 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.

2. 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.

kg[k]0E(1Pk)
052800
11.360 x 10852800
2-1.036 x 10122.924 x 109
31.057 x 10161.677 x 1014
4-1.349 x 10209.894 x 1018
55.975 x 1023
Exhibit s1: Results for items (c) and (d)—conditional values g through g for 1P and conditional moments 0E(1P ) through 0E(1P 5)
1. 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.

2. Results, calculated with a spreadsheet, are presented in Exhibit s2.
central
moments
normalized
central
moments
normalized
cumulants
1000
21.36×10811
3-1.036×1012-.6534-.6534
46.604×10163.5718.5714
5-1.544×10217.1597-.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.
3. 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.
4. Completing the squares, we obtain

[s9]

[s10]

5. 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.
6. By [3.220], the characteristic function of ( 1 + 4.5)2 is

[s11]

Applying [3.217], the characteristic function of 1P is

[s12]

7. 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 = 25000 and 1p = 50000. We use the simple program to value the CDF of 1at 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.
kp[k]Φ(p[k])
1250000.0199
2500000.3665
3307760.0444
4340930.0683
5385070.1163
6370070.0976
7372010.0998
8372140.1000
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).