Charta is on Google Play: practise on the go. Get the app
Charta

Characterise how the short rate evolves in a model whose volatility depends on time.

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

The idea

In a model with time-dependent volatility the short rate follows the process below, where dw is a normal shock with variance dt. dr = λ(t) dt + σ(t) dw. The drift λ(t) is a function of calendar time, usually chosen so the model reproduces today's term structure of rates. The volatility σ(t) is also a function of calendar time only. It is known today for every future date and does not depend on the rate level or on the path the rate has taken. Rate changes are therefore normally distributed with a basis-point volatility that can rise or fall as time passes. An exponentially decaying schedule, with decay rate α, is the example below. Volatility = σ × e^(−αt). Because volatility does not shrink at low rates, negative rates remain possible. A common mistake is to treat this as stochastic or rate-dependent volatility, but it is neither. Exam point: time dependence means a deterministic schedule of basis-point volatility, nothing more.

A practice question

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

A model specifies the short rate as dr = λ(t) dt + σ(t) dw, where λ and σ are deterministic functions of calendar time and dw is a normal shock with variance dt. Which statement characterises the process?

  1. Rate changes are lognormally distributed, so basis-point volatility is proportional to the level of the rate, as in lognormal models.Why not: Confuses a time-dependent normal model with a lognormal model, where volatility is proportional to the rate.
  2. Rate changes are normal, with basis-point volatility depending on calendar time but not on the rate level. - correct
  3. Basis-point volatility is proportional to the square root of the rate, which keeps the rate non-negative.Why not: Describes the CIR model, where the volatility term is σ√r, not a time-dependent normal model.
  4. Basis-point volatility is the same constant at every date; only the drift changes with time.Why not: Ignores σ(t): the whole point of the model is that volatility, not just drift, varies with calendar time.

With dr = λ(t)dt + σ(t)dw the shock is normal and its size is σ(t), a deterministic function of time only. It does not depend on r, so it is neither lognormal nor square-root; and since σ varies with t, volatility is not constant.

A harder one

The short rate is currently 0.40 percent. A risk manager uses dr = λ(t) dt + σ(t) dw with σ(t) = 90 basis points a year over the next two years. Which statement about the model is correct?

  1. Volatility stays at 90 bp however low the rate falls, so negative rates are likely. - correct
  2. Volatility shrinks automatically as the rate approaches zero, so the rate cannot become negative.Why not: Attributes the volatility scaling of CIR or lognormal models to a time-dependent normal model, whose volatility does not depend on r.
  3. A suitably chosen drift λ(t) guarantees the rate stays positive, whatever the volatility is.Why not: A deterministic drift cannot offset a normal shock of fixed size; negative outcomes remain possible.
  4. Rate changes are lognormal, so only the proportional size of moves, 90 percent, is fixed.Why not: Reads σ(t) as a proportional volatility; in this model it is an absolute basis-point amount and rate changes are normal.

Volatility is a function of time only, so at a 0.40 percent rate the standard deviation of annual changes is still 90 bp. A normal shock of that size makes negative rates likely over the horizon.

The hardest tier

A bank uses dr = k(θ - r) dt + σ(t) dw with k = 0.30 and σ(t) = 100 basis points a year in the next period. A junior analyst says mean reversion must lower the standard deviation of the next month's rate change below the unadjusted figure. Which statement is correct?

  1. It is about 28.9 x e^(-0.30) = 21.4 basis points, because mean reversion damps each shock.Why not: Applies a decay factor to the one-step standard deviation; k only enters the drift, not the size of the shock.
  2. It is about 100 basis points, because the volatility is quoted annually and does not scale with the period.Why not: Forgets that the standard deviation of a one-month change is σ(t)√dt = 100 x √(1/12), not the annual figure.
  3. It exceeds 28.9 basis points, because the drift k(θ - r) adds variance to the shock.Why not: The deterministic drift adds to the mean of the change, not to its variance.
  4. The standard deviation of the next month's change is about 28.9 basis points, unaffected by k; mean reversion works through the drift and thereby limits long-horizon dispersion of rate levels. - correct

Standard deviation of dr over dt is σ(t)√dt = 100 x √(1/12) = 28.9 bp whatever k is. Mean reversion changes the expected change and, accumulated over many steps, keeps the rate from wandering as far.

In the app: 6 practice questions on this outcome, spaced review that brings it back before you forget it.

Start free →

More in The Art of Term Structure Models: Volatility and Distribution