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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

  1. 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%

  2. 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.

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    $\begingroup$ To compute the 5% tail, you string events together (from worst outcome to best outcome) until they add up to 5% probability, so 3.92+0.04+1.04=5.00. We only take a small part (namely 1.04) of the no default case because that is all we need to get to our desired 5%. $\endgroup$
    – noob2
    Nov 9 '15 at 14:13
  • $\begingroup$ How would I construct the probability weights if my binomial probabilities were each > 5%? $\endgroup$ Nov 9 '15 at 16:06
  • $\begingroup$ Then you take your 5% from the worst case scenario, I believe. $\endgroup$
    – noob2
    Nov 9 '15 at 16:27
  • $\begingroup$ @noob2, I don't follow sorry. Assuming I have a similar 2 bond i.i.d. portfolio, but with a PD of 65%. My binomial probabilities for 0,1 and 2 defaults would 12.25%, 45.50% 42.25% respectively. Given each of these is well over 5%, how would I calculate a 95% ES as a probability weighted average? $\endgroup$ Nov 9 '15 at 19:23
  • $\begingroup$ In English, the 5% worst outcomes all involve double default so the loss is 200. Using math we have (5%*2)/5% multiplied by 100 is 200. In fact all K% ES for K from 100 to 57.75 are equal to 200 in this case. $\endgroup$
    – noob2
    Nov 9 '15 at 19:43
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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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  • $\begingroup$ Welcome to Quant Stackexchange loxol! $\endgroup$
    – noob2
    Nov 9 '15 at 14:57

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