Characterise the evolution of the short-term rate in the CIR model of Cox, Ingersoll and Ross, and in lognormal models.

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

The idea

The CIR model of Cox, Ingersoll and Ross specifies the process below. dr = k(θ − r) dt + σ√r dw. The drift pulls the rate toward the long-run level θ at speed k. The volatility term makes the basis-point volatility proportional to the square root of the rate, so it is high when rates are high and shrinks as the rate approaches zero. The rate therefore cannot go negative, and when 2kθ is at least σ² it cannot even reach zero. A lognormal model of the short rate instead follows the process written below. dr = a r dt + σ r dw. Here the logarithm of the rate is normal, the rate is always positive, and the basis-point volatility is proportional to the rate itself. Without an added term, the lognormal model has no mean reversion. The usual mistake is mixing up the three volatility rules: constant for normal models, square root for CIR, proportional to the rate for lognormal. Exam point: match each model to its volatility dependence.

Formulas

Basis-point volatility in the CIR model

VolCIR = σ × √r

σ: the CIR volatility parameter
r: the short rate, as a decimal

Enter r as a decimal, not in percent. Doubling the rate multiplies this volatility by √2 only.

Basis-point volatility in the lognormal model

VolLog = σ × r

σ: the proportional (percentage) volatility
r: the short rate, as a decimal

Volatility is proportional to the rate, so it falls to zero as the rate does; doubling the rate doubles it.

A practice question

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

In the Cox-Ingersoll-Ross model, dr = k(θ - r) dt + σ√r dw, which description is correct?

The rate is pulled toward θ, and basis-point volatility is proportional to √r, so it falls as the rate falls. - correct
The rate is pulled toward θ, but the basis-point volatility is the same constant at all rate levels.Why not: Describes the Vasicek model, which has constant volatility.
There is no mean reversion, and the basis-point volatility is proportional to the rate.Why not: Describes a plain lognormal model, not CIR.
The rate is pulled toward θ, and the basis-point volatility is proportional to the rate r.Why not: Takes the volatility proportional to r (lognormal) instead of √r.

CIR keeps the Vasicek-style drift but the diffusion term is σ√r, so basis-point volatility is σ√r: it shrinks as the rate approaches zero, which keeps rates non-negative.

A harder one

Rates follow dr = a r dt + σ r dw with a = 2 percent, σ = 20 percent, and a current rate of 5 percent. The expected value of the rate in 10 years is closest to:

6.00%, from simple growthWhy not: Grows the rate linearly: 5 x (1 + 0.02 x 10) instead of compounding at e^(aT).
6.11% - correct
5.00%Why not: Uses the median, 5 x e^((a - σ²/2)T), which here equals the starting rate since a = σ²/2.
7.46%Why not: Adds the Ito term instead of leaving it out: 5 x e^((a + σ²/2)T).

For dr = a r dt + σ r dw, E[r_T] = r_0 e^(aT) = 5% x e^(0.2) = 6.11%. The median is lower, 5% x e^((0.02 - 0.02) x 10) = 5%.

The hardest tier

The log of a short rate follows d ln r = (a - σ²/2) dt + σ dw, with a = 1 percent and σ = 30 percent a year. The probability that the rate in four years is below today's rate is closest to:

59.2% - correct
47.3%Why not: Omits the Ito term: uses a drift of +a for ln r instead of a - σ²/2, so the probability falls below one half.
40.8%Why not: Computes the probability the rate is above, rather than below, today's level.
68.0%Why not: Divides by σ instead of σ√T when standardising, overstating the z-score.

ln(r_4/r_0) is normal with mean (0.01 - 0.045) x 4 = -0.14 and standard deviation 0.30 x √4 = 0.60. P(below) = Φ(0.14/0.60) = Φ(0.233) = 59.2%.

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

Start free →