FRM Part II · FRM Exam Part II · Future Value and Exposure
A bank sets credit limits for a counterparty using PFE. The counterparty's exposure is modelled as normally distributed at a six-month horizon with a mean of USD 4 million and a standard deviation of USD 5 million. Using a 95% one-sided confidence level (z = 1.645), what is the approximate PFE, ignoring truncation at zero?
PFE for a normal exposure equals the mean plus z times the standard deviation: 4 + 1.645 x 5 = about USD 12.2 million. Leaving out the mean gives 8.2 million, which understates the exposure at the 95% quantile.
- AUSD 8.2 million
- BUSD 12.2 millionCorrect
- CUSD 9.0 million
- DUSD 16.5 million
Explanation
PFE = mean + z x standard deviation = 4 + 1.645 x 5 = 4 + 8.225 = 12.225, about USD 12.2 million. USD 8.2 million omits the mean (z x sigma only). USD 16.5 million uses z x 10 wrongly.
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