Exercise Solution 10.2

Exercise Solution 10.2

  1. Define  1 conditional on information available at time 0, has distribution 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.

k g[k] 0E(1Pk)
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)
  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
    1 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.
  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[1] = 25000 and 1p[2] = 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.
    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).

 
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