May 19

Question 1 from UoL exam 2016, Zone B, Post 2

Question 1 from UoL exam 2016, Zone B, Post 2

For the problem statement and first part of the solution see Question 1 from UoL exam 2016, Zone B, Post 1.

Let R denote the return on P_1+P_2. From Table 1 we can derive the probabilities table for this return:

Table 2. Joint table of returns on separate portfolios

0 -100
R_2 0 0.96^2 0.04\cdot 0.96
-100 0.04\cdot 0.96 0.04^2

From Table 2 we conclude that the return on the combined portfolio looks as follows:

Table 3. Total return

R \text{Prob}
0 0.96^2
-50 2\cdot 0.04\cdot 0.96
-100 0.04^2

Table 3 shows that

F_R(x)=0 for x<-100,

F_R(x)=0.04^2=0.0016 for -100\leq x<-50,

F_R(x)=2\cdot 0.04\cdot 0.96+0.0016=0.0784 for -50\leq x<0 and

F_R(x)=0.96^2+0.0784=1 for x\geq 0.

Try to follow the procedure used in Post 1 and you will see that

F_R^{-1}(y)=+\infty for y>1,

F_R^{-1}(y)=0 for 0.0784<y\leq 1,

F_R^{-1}(y)=-50 for 0.0016<y\leq 0.0784,

F_R^{-1}(y)=-100 for 0<y\leq 0.0016 and

F_R^{-1}(y)=-\infty for y\leq 0.

This implies VaR_R^\alpha=F_R^{-1}(0.05)=-50.

(b) The subadditivity definition requires amounts opposite in sign to ours. That is, we define \widetilde{VaR^\alpha} from P(X\leq -\widetilde{VaR^\alpha})=\alpha and then say that VaR thus defined is sub-additive if \widetilde{VaR^\alpha}(P_1+P_2)\leq \widetilde{VaR^\alpha}(P_1)+\widetilde{VaR^\alpha}(P_2). We have been using the definition P(X\leq VaR^\alpha)=\alpha. It's easy to see that \widetilde{VaR^\alpha}=-VaR^\alpha. Thus, in our case we have \widetilde{VaR_R^{0.05}}=50 which is not smaller than \widetilde{VaR_{R_1}^{0.05}}+\widetilde{VaR_{R_2}^{0.05}}=0. Sub-additivity does not hold in this example. Absence of sub-additivity means that riskiness of the whole portfolio, as measured by VaR, may exceed riskiness of the sum of the portfolio parts.

(c) The problem uses the definition of the expected shortfall that yields positive values. I use everywhere the definition that gives negative values: ES^\alpha=E_t[R|R\leq VaR_{t+1}^\alpha]. Since the setup is static, this is the same as ES^\alpha=E[R|R\leq VaR^\alpha]. By definition, E(X|A)=\frac{E(X1_A)}{P(A)}, so ES^\alpha=\frac{E(R1_{\{R\leq VaR^\alpha\}})}{P(R\leq VaR^\alpha)}.

In Post 1 we found that VaR^\alpha=0 for each of R_1,R_2. The condition R_i\leq VaR^\alpha=0 places no restriction on R_i, so from Table 1

E(R_i1_{\{R_i\leq VaR^\alpha \}})=ER_i=0\cdot 0.96-100\cdot 0.04=-4,\ P(R_i\leq VaR^\alpha)=1.

As a result, ES_i^\alpha=-4\%.

Since VaR_R^\alpha=F_R^{-1}(0.05)=-50, from Table 3

E(R1_{\{R\leq -50\}})=-50\cdot 2\cdot 0.04\cdot 0.96-100\cdot 0.04^2=-4,

P(R\leq -50)=1-0.9216=0.0784.

Therefore ES_R^\alpha=-51.02. Converting everything to positive values, we have 51.02>4+4, so that sub-additivity does not hold. In fact, there is a theoretical property that it should hold. In this example it does not because of the bad behavior of the generalized inverse for distribution functions of discrete random variables.

The returns in percentages can be easily converted to those in dollars.

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