Expected Shortfall for a Discrete Bond Default Portfolio

Summary

The example calculates 95% expected shortfall for a portfolio of two independent bonds, each with a 2% default probability, $100 face value, and zero recovery. The possible losses are $0, $100, and $200, with probabilities obtained from a binomial model. To find expected shortfall, the method averages losses across the worst 5% of outcomes, weighting each loss by the portion of probability mass included in that tail.

The two nonzero loss outcomes together account for 3.96% probability, so they do not fill the full 5% tail. The calculation therefore includes 1.04% probability from the zero-loss outcome, yielding an average tail loss of $80. This illustrates how to handle a discrete loss distribution when the chosen tail probability falls between outcome probabilities. The result relies on the stated independence, default probabilities, and recovery assumption.

Key ideas

• The two independent default indicators produce a binomial distribution of portfolio losses.
• Expected shortfall averages losses over the selected worst-loss tail.
• When severe outcomes have less probability mass than the tail size, include part of the next outcome’s probability mass.
• Under the stated assumptions, the 5% tail includes some probability assigned to zero loss.
• The calculation depends on independent defaults, equal default probabilities, and zero recovery.

Tags

# Calculating expected shortfall


# Calculating expected shortfall












I'm trying to calculate the expected shortfall for the below scenario. I don't understand why the 1.04% probability of 0 bonds defaulting is used as a weight when calculating ES, since the binomial probability was 96.04%.

Problem

Assume a two-bond portfolio where the probability of bond default is 2% for each and independent (i.i.d). The face value of each bond is $100 and recovery is zero. What is the 95% expected shortfall?

Solution Given

- Calculate the binomial probabilities of both bonds defaulting, only one defaulting and 0 bonds defaulting which are: P (0 defaults) = 96.04% P (1 defaults) = 3.92% P (2 defaults) = 0.04%

- The 5% tail therefore contains the following:

0.04% probability of 2 defaults 3.92% probability of 1 default 1.04% probability of 0 defaults

The 95% ES is given by:

((0.04% * 2) + (3.92% * 1) + (1.04% * 0))/5% = 0.8 * 100 = 80.00 The average loss in the 5% tail (of the binomial) is 0.8 defaults or \$80.

## Answer by loxol (score 4)

https://quant.stackexchange.com/a/21649

The 1.04% are used in the calculation because it is 95% expected shortfall so you want to calculate the expectation on the 5% worst loss. In your problem there is 3 possible outcomes: loss of 200, 100 or 0. As the probability of loss of 200 or 100 is 0.04+3.92 = 3.96% < 5%, you need to take account of the loss of 0$ for 1.04% part to reach the 5%.

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