14
May 19

## 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

 $R_1$$R_1$ $0$$0$ $-100$$-100$ $R_2$$R_2$ $0$$0$ $0.96^2$$0.96^2$ $0.04\cdot 0.96$$0.04\cdot 0.96$ $-100$$-100$ $0.04\cdot 0.96$$0.04\cdot 0.96$ $0.04^2$$0.04^2$

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

Table 3. Total return

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

$F_R^{-1}(y)=-50$ for $0.0016

$F_R^{-1}(y)=-100$ for $0 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.