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FRM Part II · FRM Exam Part II · Future Value and Exposure

A bank's netting set at time t has a normally distributed value V with mean USD 0 and standard deviation USD 10 million. The bank computes EE(t) = E[max(V,0)]. Using the fact that for a zero-mean normal variable E[max(V,0)] = σ/√(2π), and taking √(2π) ≈ 2.5, which is the closest value of EE, and how does it compare with the 97.5% PFE (z ≈ 1.96)?

EE is about USD 4.0 million and the 97.5% PFE is about USD 19.6 million. For a zero-mean normal value, EE equals σ divided by √(2π), roughly 10/2.5 = 4, while the PFE is the 1.96 standard deviation quantile, 19.6 million, far above the average exposure.

  1. AEE ≈ USD 4.0 million; PFE ≈ USD 19.6 millionCorrect
  2. BEE ≈ USD 4.0 million; PFE ≈ USD 9.8 million
  3. CEE ≈ USD 10 million; PFE ≈ USD 19.6 million
  4. DEE ≈ USD 5.0 million; PFE ≈ USD 4.0 million

Explanation

EE = 10/2.5 = 4.0 million. The 97.5% PFE is the quantile of V, which is 1.96 × 10 = 19.6 million, and since this is positive it equals the exposure quantile. Option with 9.8 halves the quantile mistakenly, and EE of 10 confuses EE with the standard deviation.

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