Raising the bar

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

Table 2. Joint table of returns on separate portfolios

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

Table 3. Total return

Table 3 shows that

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

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

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

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

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

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

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

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

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

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

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

(b) The subadditivity definition requires amounts opposite in sign to ours. That is, we define $latex \widetilde{VaR^\alpha}$ from $latex P(X\leq -\widetilde{VaR^\alpha})=\alpha$ and then say that VaR thus defined is sub-additive if $latex \widetilde{VaR^\alpha}(P_1+P_2)\leq \widetilde{VaR^\alpha}(P_1)+\widetilde{VaR^\alpha}(P_2).$ We have been using the definition $latex P(X\leq VaR^\alpha)=\alpha.$ It's easy to see that $latex \widetilde{VaR^\alpha}=-VaR^\alpha.$ Thus, in our case we have $latex \widetilde{VaR_R^{0.05}}=50$ which is not smaller than $latex \widetilde{VaR_{R_1}^{0.05}}+\widetilde{VaR_{R_2}^{0.05}}=0.$ 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: $latex ES^\alpha=E_t[R|R\leq VaR_{t+1}^\alpha].$ Since the setup is static, this is the same as $latex ES^\alpha=E[R|R\leq VaR^\alpha].$ By definition, $latex E(X|A)=\frac{E(X1_A)}{P(A)},$ so $latex ES^\alpha=\frac{E(R1_{\{R\leq VaR^\alpha\}})}{P(R\leq VaR^\alpha)}$.

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

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

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

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

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

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

Therefore $latex ES_R^\alpha=-51.02.$ Converting everything to positive values, we have $latex 51.02>4+4,$ 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.