Exercise Solution 10.3

1. We calculate ⁰p as follows

   [s1]

   

   [s2]

   

   [s3]

   
2. We can easily write out a portfolio mapping off the top of our heads, much as we wrote out a formula for ⁰p in part (a). Formally, though, we first define assets:

   [s4]

   

   and portfolio holdings

   [s5]

   

   We set

   [s6]

   

   and define a mapping ¹S = φ(¹R) with

   [s7]

   

   Since θ = ωφ, we obtain

   [s8]

   

   Our mapping function is a quadratic polynomial.

3. We express our mapping from part (b) as

   [10.42]

   

   where

   [s9]

   

   [s10]

   

   [s11]

   
4. As described in Section 2.7.3, we construct the Cholesky matrix of ^1|0Σ:

   [s12]

   
5. Multiplying the respective matrices, we obtain

   [s13]

   

   We take normalized orthogonal eigenvectors and use them as the rows of

   [s14]

   
6. We set

   [s15]

   

   [s16]

   

   [s17]

   
7. First, we calculate

   [s18]

   

   Then

   [10.43]

   

   where, based upon [3.161], [3.162] and [3.163],

   [s19]

   

   [s20]

   

   [s21]

   

   h, i. Calculations are performed with a spreadsheet. Results are

   k	g[k]	0E(1P k)
   0	282128
   1	1.216 x 107	282128
   2	5.828 x 108	7.961 x 1010
   3	5.289 x 1010	2.247 x 1016
   4	3.970 x 101	6.341 x 1021
   5		1.790 x 1027

   Exhibit s1: Values for g^[k] and ⁰E(¹P ^k), calculated in items (h) and (i).

10. Based upon its first two moments, we calculate the standard deviation of ¹P using the formula for variance derived in in Exercise 3.15:

    [s22]

    

    [s23]

    

    [s24]

    
11. Calculations are performed with a spreadsheet. Results are

    	central moment	normalized central moment	normalized cumulant
    1	0.000	0.0000	0.000000
    2	1.216×107	1.0000	1.000000
    3	5.828×108	0.0137	0.013745
    4	4.436×1014	3.0004	0.000358
    5	7.087×1016	0.1375	0.000011

    Exhibit s2: Results for item (k)—conditional central moments of ¹P as well as the conditional central moments and cumulants of the normalization ¹P * of ¹P.
12. Using a spreadsheet, we use the Cornish-Fisher expansion to calculate the .05-quantile of ¹P* as –1.6409. By [3.195], the .05-quantile of ¹P is 276406. Because the portfolio’s current value is 282130, the .95-quantile of loss is 282130 – 276406 = 5724 EUR.
13. We multiply out our result from part (g) and complete the squares:

    [s25]

    

    [s26]

    

    We have expressed ¹P as a linear polynomial of two independent chi-squared random variables, each with 1 degree of freedom. Non-centrality parameters are 1724.54 and 32044.

14. This item is best performed by coding a simple program that applies the inversion theorem by using the trapezoidal rule. We select seed values of ¹p^[1] = 280000 and ¹p^[2] = 290000. We use that 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 .05-quantile of ¹P:

    k	p[k]	Φ(p[k])
    1	280000	0.26528
    2	290000	0.98711
    3	277018	0.06858
    4	276755	0.05911
    5	276502	0.05100
    6	276471	0.05006
    7	276469	0.05000

    Exhibit s3: The conditional .05-quantile of ¹P is obtained using the secant method.

    The .05-quantile of ¹P is 276469. Because the portfolio’s current value is 282130, the .95-quantile of loss is 282130 – 276469 = 5661 EUR. Compare this with the approximate result of 5724 EUR obtained in part (l).