For Example 10.5 with $h=50 %$, show explicitly that CVaR rewards diversification. Example 10.5 (Diversification failure) Consider

For Example 10.5 with $h=50 %$, show explicitly that CVaR rewards diversification.

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Example 10.5 (Diversification failure) Consider the position 1 with probability 1/2 X1 -1 with probability 1/2 } . (10.8) Set h = .50. To compute VaR 50 (X1), we use the shortcut method for discrete prob- abilities. Starting at the value -1, we have exactly .50, so VaR is determined at the next higher node, which is 1. Changing the sign gives Var 50(X1)=-1. Now suppose that X2 has exactly the same structure as X but is independent of X. Consider the diversified position X made up of equal parts of X1 and X2. This has payoff 1 with probability 1/4 X = X+X2 2 O with probability 1/2 -1 with probability 1/4 (10.9) Again, using the shortcut method, we start at the lowest-valued point and move upward accumulating probability. The value .50 is exceeded at 0 with an accumulated probability of 3/4. Since this is greater than .50, it defines VaR as VaR 50 (X) = 0. Thus for h=0.50 we find that VaR. (x1+x) > VaR.(X1) + VaR+(X2). 2 2 which implies that the diversified position is riskier than the weighted sum of each of its component VaR's.