Characterise lognormal rate models extended with a drift that is deterministic or with mean-reverting behaviour.

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

The idea

A lognormal model can be given a drift in two ways. Let Y stand for ln r, the log of the short rate. With a deterministic drift, the function a(t) is chosen to fit today's term structure of rates. There is no mean reversion, so the dispersion of Y grows with the square root of time. In symbols, dY = a(t) dt + σ dw. With mean reversion, as in the Black-Karasinski model, Y is pulled toward ln θ at speed k. In symbols, dY = k(ln θ − Y) dt + σ dw. Since Y is normal, r is lognormal and always positive. The median of the long-run rate is θ, while its mean is larger. The gap between them depends on the long-run variance of Y, written Var, which settles at Var = σ² ÷ (2k). The mean is the median scaled up by a factor that depends on that variance: Mean = θ × exp(Var ÷ 2). The usual mistake is applying mean reversion to r instead of ln r, or treating θ as the mean. Exam point: median versus mean of a lognormal rate.

Formulas

Long-run variance of ln r with mean reversion

Var = σ² ÷ (2 × k)

σ: the volatility of ln r
k: the speed of mean reversion

Applies to ln r, not r, in the Black-Karasinski form. It sets the gap between the mean and the median of the long-run rate.

Long-run mean of a mean-reverting lognormal rate

Mean = θ × e^(Var ÷ 2)

θ: the long-run median rate
Var: the long-run variance of ln r

θ is the median, not the mean; the mean is always higher because the rate is lognormal.

A practice question

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

In the lognormal model d ln r = a(t) dt + σ dw, what is the role of the deterministic function a(t)?

It is chosen to fit today's term structure of rates; the model has no mean reversion. - correct
It pulls ln r back toward a long-run level at a speed given by a(t).Why not: Describes a mean-reversion term k(ln θ - ln r), which this model does not have.
It sets the basis-point volatility of the rate at each date.Why not: Volatility is set by σ; a(t) is the drift of ln r.
It keeps the rate positive, which would otherwise be possible to violate.Why not: The rate is positive because ln r is normal; a(t) plays no role in this.

In d ln r = a(t) dt + σ dw, a(t) is a time-varying drift used to match the initial curve, analogous to Ho-Lee for a lognormal model. There is no force pulling ln r back to a long-run level.

A harder one

For d ln r = k(ln θ - ln r) dt + σ dw with k = 0.5, θ = 5 percent, a current rate of 3 percent, the median rate in two years is closest to:

4.26%Why not: Applies mean reversion to r itself: 5 - 2 x e^(-1), rather than to ln r.
4.14% - correct
3.62%Why not: Weights the starting log-rate by (1 - e^(-kT)) and ln θ by e^(-kT), the wrong way round.
5.00%Why not: Assumes the median reaches θ within two years, ignoring that reversion is gradual.

E[ln r_2] = ln θ + (ln r_0 - ln θ) e^(-kT) = -2.9957 + (-0.5108)(0.3679) = -3.1836. The median is e^(-3.1836) = 4.14%.

The hardest tier

In d ln r = k(ln θ - ln r) dt + σ dw with k = 0.4, θ = 5 percent and σ = 20 percent, the long-run distribution of the short rate has a mean closest to:

5.13% - correct
5.00%, equal to θWhy not: Treats θ as the mean; for a lognormal rate θ is the median.
5.26%Why not: Uses exp(σ²/(2k)) as the mean factor, which forgets to halve the variance in the lognormal mean adjustment.
4.88%Why not: Applies the convexity adjustment with the wrong sign.

The long-run variance of ln r is σ²/(2k) = 0.05, with median θ. The mean of a lognormal rate is θ x exp(variance/2) = 5% x e^(0.025) = 5.13%.

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