Exercise Solution 10.3
- We calculate 0p as follows
- 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:
and portfolio holdings
and define a mapping 1S = φ(1R) with
Since θ = ωφ, we obtain
Our mapping function is a quadratic polynomial.
- We express our mapping from part (b) as
- As described in Section 2.7.3, we construct the Cholesky matrix of 1|0Σ:
- Multiplying the respective matrices, we obtain
We take normalized orthogonal eigenvectors and use them as the rows of
- We set
- First, we calculate
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 1027Exhibit s1: Values for g[k] and 0E(1P k), calculated in items (h) and (i).
- Based upon its first two moments, we calculate the standard deviation of 1P using the formula for variance derived in in Exercise 3.15:
- Calculations are performed with a spreadsheet. Results are
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 0.000011Exhibit 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.
- 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.
- We multiply out our result from part (g) and complete the squares:
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.
- 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 = 280000 and 1p = 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 0.05000Exhibit 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).