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FRM Part II · FRM Exam Part II · The Art of Term Structure Models: Volatility and Distribution

A risk manager uses a model where dr = λ(t)dt + σ r dw with σ = 20%. A second model uses dr = λ(t)dt + σ√r dw. Currently r = 4%. Instead of the 20% lognormal volatility, the second model's σ is calibrated so both have the same absolute volatility at r = 4%. If rates later rise to 9%, how does the absolute (bp) volatility of the second model compare with that of the first?

At 9% the lognormal model's absolute volatility is 1.80% while the square-root model's is 1.20%. Both start at 0.80% at 4%, but the first scales with r and the second only with the square root of r, so it rises less.

  1. ASecond model: 0.80% vs first model: 1.80%, so the second is lower
  2. BSecond model: 1.20% vs first model: 1.80%, so the second is lowerCorrect
  3. CBoth are 1.80% since they were calibrated equally
  4. DSecond model: 1.20% vs first model: 0.80%, so the second is higher

Explanation

At r = 4%, the first model's absolute volatility is 0.20×4% = 0.80%. For the second, σ√0.04 = 0.80% gives σ = 0.04. At r = 9%, first: 0.20×9% = 1.80%. Second: 0.04×√0.09 = 0.04×0.30 = 1.20%. So the second is lower, because absolute volatility grows with √r rather than r.

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