Question

Suppose that each of two investments has a 4% chance of loss of $10 million, a 2% chance of loss of $1 million, and a 94% chance of profit of $1 million. They are independent of each other.

a. What is the VaR and CVaR (expected shortfall) at level p = 95% for one of the investments?

b. What is the VaR and CVaR (expected shortfall) at level p = 95% for a portfolio consisting of the two investments?

c. Show that, in this example, VaR does not satisfy the subadditivity condition whereas expected shortfall does.

Answer 1

Solution: (a)

The percentiles of loss distributions are

0 - 94 percentile - ($1) million

94 - 96 percentile $1 million

96 - 100 percentile $10 million

A loss of $1 million extends from the 94% point of the loss distribution to the 96 percentile point. The 95% VaR is the loss at 95 percentile and is therefore $1 million

The expected shortfall for one of the investments is the expected loss conditional that the loss is in the 5 percent tail. Given that we are in the 5% tail, there is,

20% chance that the loss is $1 million

80% chance that the loss is $10 million.
The conditional expected loss in the tail is therefore $8.2 million. This is the expected shortfall.

Solution: (b)
For a portfolio consisting of the two investments the percentiles of loss distribution are:

0.94 x 0.94 = 0.8836 chance (0-88.36 percentile) — loss of (-$2) million.

2 x 0.2 x 0.94 = 0.0376 chance (88.36-92.12 percentile) — loss of zero

0.02 x 0.02 = 0.0004 chance (92.12-92.16 percentile) — loss of $2 million

2 x 0.04 x 0.94 = 0.0752 chance (92.16-99.68 percentile) — loss of $9 million

2 x 0.04 x 0.02 = 0.0016 chance (99.68-99.84 percentile) — loss of $11 million

0.04 x 0.04 = 0.0016 chance (99.84-100 percentile) — loss of $20 million;
It follows that the 95% VaR is $9 million.

The expected shortfall for the portfolio consisting of the two investments is the expected loss conditional that the loss is in the 5% tail. Given that we are in the tail, there is,

0.0016/0.05 = 0.032 chance of a loss of $20 million

0.0016/0.05 = 0.032 chance of a loss of $11 million

0.936 chance of a loss of $9 million.
The expected loss is the mean of these three losses - $9.416 million

Solution: (c)

VaR does not satisfy the subadditivity condition because 9 > 1+1. However expected shortfall does because 9.416 < 8.2 + 8.2