Exercise Solution 10.3

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 = ω φ, 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).
  1. 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]

  2. 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.
  3. 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.
  4. 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.

  5. 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).

 
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