FRM Part I · FRM Exam Part I · Measures of Financial Risk
A portfolio's one-day profit and loss is normally distributed with mean zero and standard deviation of USD 2.00 million. Using the normal distribution, what is the 99% one-day expected shortfall (ES)? Use z(0.99) = 2.326 and the standard normal density at 2.326 of 0.0267.
Expected shortfall equals the standard deviation times the normal density at the cutoff divided by the tail probability: 2 x 0.0267 / 0.01 = USD 5.34 million. This exceeds the USD 4.65 million VaR because ES averages losses beyond the VaR level.
- AUSD 5.34 millionCorrect
- BUSD 4.65 million
- CUSD 2.67 million
- DUSD 2.33 million
Explanation
For a normal distribution with zero mean, ES = sigma x phi(z)/(1-c) = 2.00 x 0.0267/0.01 = 2.00 x 2.67 = USD 5.34 million. The VaR is 2.326 x 2 = 4.65 million, so ES must exceed it. 2.67 omits the sigma scaling.
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