# 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]} |
^{0}E(^{1}P^{ k}) |
---|---|---|

0 | 52800 | |

1 | 1.360 x 10^{8} |
52800 |

2 | -1.036 x 10^{12} |
2.924 x 10^{9} |

3 | 1.057 x 10^{16} |
1.677 x 10^{14} |

4 | -1.349 x 10^{20} |
9.894 x 10^{18} |

5 | 5.975 x 10^{23} |

Exhibit s1: Results for items (c) and (d)—conditional values

*g*^{[0]}through*g*^{[4]}for^{1}*P*and conditional moments^{0}*E*(^{1}*P*) through^{0}*E*(^{1}*P*^{ 5})- Based upon its first two moments, we calculate the standard deviation of
^{1}*P*using the formula for variance derived in Exercise 3.15:[s6]

[s7]

[s8]

Since

^{0}*std*(^{1}*L*) =^{0}*std*(^{1}*P*), 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 1.36×10 ^{8}1 1 3 -1.036×10 ^{12}-.6534 -.6534 4 6.604×10 ^{16}3.5718 .5714 5 -1.544×10 ^{21}7.1597 -.6257 Exhibit s2: Results for item (f)—conditional central moments of^{1}*P*as well as the conditional central moments and cumulants of the normalization^{1}*P** of^{1}*P*. - Using a spreadsheet, we use the Cornish-Fisher expansion to calculate the .10-quantile of
^{1}*P** as –1.3367. By [3.209], the .10-quantile of^{1}*P*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

^{1}*P*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
^{1}*p*^{[1]}= 25000 and^{1}*p*^{[2]}= 50000. We use the simple program to value the CDF of^{1}*P*at each, and then apply the secant method. The resulting sequence of values^{1}*p*^{[k]}converges to the desired .10-quantile of^{1}*P*. 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^{1}*P.*The .10-quantile of

^{1}*P*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).

## Comments are closed.