Work out the short-rate change under a time-dependent volatility model and establish how the standard deviation of that change behaves.

FRM® Part II · Market Risk Measurement and Management · The Art of Term Structure Models: Volatility and Distribution · FRM2.MR.15.b

The idea

Over a short interval dt, the model with drift λ(t) and volatility σ(t) gives a change in the short rate written Dr = λ(t) × dt + σ(t) × √dt × ε. Here ε is a standard normal draw. The drift scales with dt and the random part with the square root of dt. The expected change is λ(t) × dt. The standard deviation of the change is σ(t) times the square root of dt, which depends on calendar time but not on the rate level. If volatility decays exponentially at a rate α, the standard deviation shrinks as time passes. Over a longer horizon variances add, not volatilities. The variance of the cumulative change is written Variance = ∫ σ(t)² dt, from 0 to T. Common errors are scaling the shock by dt instead of the square root of dt, and adding or averaging volatilities instead of adding variances. Exam point: convert an annual basis-point volatility to a period by multiplying by the square root of dt.

Formulas

Short-rate change with time-dependent drift and volatility

Dr = λ(t) × dt + σ(t) × √dt × ε

λ(t): the drift per year at time t
σ(t): the basis-point volatility per year at time t
dt: the interval in years
ε: a standard normal draw

Normal changes whose size depends on calendar time only. The trap is scaling the shock by dt instead of √dt.

Variance of the cumulative rate change with time-dependent volatility

Variance = ∫ σ(t)² dt

σ(t): the basis-point volatility per year at time t
dt: the time increment, integrated from today to the horizon T

Variances add over time; adding or averaging volatilities is the usual error. Take the square root for the standard deviation.

A practice question

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

Under dr = λ(t) dt + σ(t) dw, which expression gives the change in the short rate over a short interval dt, with ε a standard normal draw?

Δr = λ(t) dt + σ(t) √dt ε - correct
Δr = λ(t) dt + σ(t) dt ε, with ε standard normalWhy not: Scales the shock by dt rather than √dt; a normal shock dw has variance dt, so its standard deviation is √dt.
Δr = λ(t) √dt + σ(t) dt εWhy not: Swaps the scaling: the drift grows with dt and the random part with √dt, not the reverse.
Δr = λ(t) dt + σ(t) r √dt εWhy not: Multiplies the shock by the rate level; that is a lognormal specification, not a time-dependent normal one.

The drift term is deterministic and proportional to dt. The shock is σ(t) dw with dw = √dt ε, giving Δr = λ(t)dt + σ(t)√dt ε.

A harder one

In a model dr = λ(t) dt + σ(t) dw, λ = 1.20 percent a year and σ = 0.90 percent a year at the date of interest. With dt = 1/12 and normal shocks, the probability that the rate falls by more than 20 basis points over the next month is closest to:

22.1%Why not: Ignores the drift: uses mean zero, giving Φ(-20/25.98) instead of Φ(-30/25.98).
12.4% - correct
35.0%Why not: Subtracts the monthly drift of 10 bp the wrong way (mean of -10 instead of +10).
36.9%Why not: Divides by the annual volatility of 90 bp instead of the monthly 25.98 bp, forgetting √dt.

Mean change = 1.20% x 1/12 = 10 bp; standard deviation = 90 x √(1/12) = 25.98 bp. P(Δr < -20) = Φ((-20 - 10)/25.98) = Φ(-1.155) = 12.4%.

The hardest tier

A model with no drift uncertainty has σ(t) = 100 x e^(-0.20t) basis points a year. The standard deviation of the cumulative change in the short rate between now and five years is closest to:

147 basis points - correct
224 basis pointsWhy not: Ignores the decay in volatility: 100 x √5.
82 basis pointsWhy not: Applies the year-5 volatility 100 x e^(-1) to the whole horizon: 100 x e^(-1) x √5.
178 basis pointsWhy not: Uses (1 - e^(-αT))/α inside the root instead of (1 - e^(-2αT))/(2α); the integral is of σ², which decays at 2α.

Variance = ∫ σ(t)² dt = 100² x (1 - e^(-2 x 0.2 x 5))/(2 x 0.2) = 10,000 x 0.8647/0.4 = 21,617. The square root is 147 bp.

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