# Exercise Solution 10.3

- We calculate
^{0}*p*as follows[s1]

[s2]

[s3]

- We can easily write out a portfolio mapping off the top of our heads, much as we wrote out a formula for
^{0}*p*in part (a). Formally, though, we first define assets:[s4]

and portfolio holdings

[s5]

We set

[s6]

and define a mapping

^{1}= φ(**S**^{1}) with**R**[s7]

Since θ = ωφ, we obtain

[s8]

Our mapping function is a quadratic polynomial.

- We express our mapping from part (b) as
[10.42]

where

[s9]

[s10]

[s11]

- As described in Section 2.7.3, we construct the Cholesky matrix of
^{1|0}**Σ**:[s12]

- Multiplying the respective matrices, we obtain
[s13]

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

[s14]

- We set
[s15]

[s16]

[s17]

- 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]}^{0}*E*(^{1}*P*^{ k})0 282128 1 1.216 x 10 ^{7}282128 2 5.828 x 10 ^{8}7.961 x 10 ^{10}3 5.289 x 10 ^{10}2.247 x 10 ^{16}4 3.970 x 10 ^{1}6.341 x 10 ^{21}5 1.790 x 10 ^{27}Exhibit s1: Values for*g*^{[k]}and^{0}*E*(^{1}*P*^{ k}), calculated in items (h) and (i).

- Based upon its first two moments, we calculate the standard deviation of
^{1}*P*using the formula for variance derived in in Exercise 3.15:[s22]

[s23]

[s24]

- Calculations are performed with a spreadsheet. Results are
central

momentnormalized

central

momentnormalized

cumulant1 0.000 0.0000 0.000000 2 1.216×10 ^{7}1.0000 1.000000 3 5.828×10 ^{8}0.0137 0.013745 4 4.436×10 ^{14}3.0004 0.000358 5 7.087×10 ^{16}0.1375 0.000011 Exhibit s2: Results for item (k)—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 .05-quantile of
^{1}*P** as –1.6409. By [3.195], the .05-quantile of^{1}*P*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:
[s25]

[s26]

We have expressed

^{1}*P*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
^{1}*p*^{[1]}= 280000 and^{1}*p*^{[2]}= 290000. We use that 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 .05-quantile of^{1}*P*:*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^{1}*P*is obtained using the secant method.The .05-quantile of

^{1}*P*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).