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Work out the short-rate change and the standard deviation of that change under a normal-rates model with zero drift.

FRM® Part II · Market Risk Measurement and Management · The Art of Term Structure Models: Drift · FRM2.MR.14.b

The idea

Under dr = σ × dw with no drift, the change in the short rate over a step dt is normally distributed with mean zero. Write its standard deviation as SD. The standard deviation over one step is then given by SD = σ × √dt. Sigma is quoted per year in basis points. A monthly step is one twelfth of a year, so dt = 1 ÷ 12. A daily step uses one 252nd of a year, taking 252 trading days. Independent changes add in variance, not in standard deviation. So over a horizon of T years the standard deviation grows with the square root of time, giving SDT = σ × √T. A 100 bp annual volatility therefore means about 28.9 bp per month and 223.6 bp over five years. The usual errors are scaling by dt instead of its square root, quoting the variance as the standard deviation, and mixing daily and annual volatility. Exam point: the expected change is zero, so quadrupling the horizon only doubles the standard deviation.

Formulas

Standard deviation of a short-rate change with zero drift

SD = σ × √dt

σ
the annual basis-point volatility
dt
the step length in years, such as 1 ÷ 12 for a month

Normal model with independent changes. Scale by the square root of dt, not dt; over a horizon T the same rule gives σ × √T.

A practice question

From the question bank, with the reason each wrong option is wrong.

In a normal short-rate model with zero drift and annual volatility sigma, what are the mean and standard deviation of the change in the short rate over a short interval dt?

  1. Mean zero, standard deviation sigma x dt.Why not: Scales the standard deviation by dt instead of the square root of dt.
  2. Mean zero, standard deviation sigma x sqrt(dt). - correct
  3. Mean zero, standard deviation sigma² x dt.Why not: Quotes the variance as though it were the standard deviation.
  4. Mean sigma x dt, standard deviation sigma x sqrt(dt).Why not: Gives the change a nonzero mean; with zero drift the expected change is zero.

With dr = sigma x dw the expected change is zero and the variance over dt is sigma² x dt, so the standard deviation is sigma x sqrt(dt).

A harder one

Daily changes in the short rate are modelled as independent with a standard deviation of 6.5 bp and zero drift. Assuming 252 trading days per year, the standard deviation of the change in the rate over 10 trading days is closest to:

  1. 20.6 bp - correct
  2. 65.0 bpWhy not: Adds the daily standard deviation 10 times instead of scaling by the square root of 10.
  3. 103.2 bpWhy not: Reports the annualised volatility rather than the 10-day figure.
  4. 1.3 bpWhy not: Applies the annual-to-period conversion to a number that is already a daily figure.

Independent daily changes add in variance: 6.5 x sqrt(10) = 20.6 bp. Multiplying by 10 would assume perfectly correlated changes.

The hardest tier

Monthly changes in a short rate have historically had a standard deviation of 28.9 bp, which a zero-drift normal model matches by setting annual volatility to 100 bp. The historical standard deviation of five-year changes is 150 bp. What does the model imply for five-year changes, and what does the comparison suggest?

  1. 500 bp, above the historical 150 bp, which suggests the model has too little volatility.Why not: Scales volatility linearly with time, and misreads a model that overshoots as one that undershoots.
  2. 223.6 bp, above the historical 150 bp, which suggests mean reversion that the zero-drift model lacks. - correct
  3. 144.3 bp, close to the historical 150 bp, so the model is adequate.Why not: Multiplies the monthly standard deviation by the number of years, mixing units.
  4. 100 bp, below the historical 150 bp, which suggests volatility rises with horizon.Why not: Treats annual volatility as the standard deviation at every horizon, ignoring the square-root-of-time growth.

Model standard deviation over 5 years = 100 x sqrt(5) = 223.6 bp. Observed five-year dispersion of 150 bp is smaller than the random-walk implication, evidence of mean reversion, which motivates the Vasicek model.

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