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FRM Part II · FRM Exam Part II · The Evolution of Stress Testing Counterparty Exposures

At a future date, the mark-to-market value of a netting set is normally distributed with mean 0 and standard deviation 20 (USD millions). No collateral is held. Using z-values of 1.645 for 95% and 0.399 for the standard normal density at zero, the expected exposure (EE = E[max(V,0)]) and the 95% PFE are closest to:

With zero mean and standard deviation 20, expected exposure is 20 times 0.399, about USD 7.98 million, because only positive values count. The 95% PFE is 1.645 times 20, about USD 32.9 million. Other answers use the wrong z-value or ignore truncation at zero.

  1. AEE = 7.98; PFE = 32.9Correct
  2. BEE = 7.98; PFE = 39.2
  3. CEE = 16.0; PFE = 32.9
  4. DEE = 0; PFE = 32.9

Explanation

For V~N(0,σ), E[max(V,0)] = σ·φ(0) = 20×0.399 = 7.98. The 95% PFE is the 95th percentile of V = 1.645×20 = 32.9. EE of 16 doubles the correct figure; 39.2 would use z=1.96 (97.5%); EE of 0 ignores that only positive values count.

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