MR 15 – The Art of Term Structure Models – Volatility and Distribution | FRM Part 2

This lesson covers short rate models whose volatility changes with time or with the level of the rate, the CIR and lognormal models, the Salomon Brothers and Black-Karasinski trees, and closed-form spot rates for these models. Check-yourself items are study aids.

A time-varying drift lets a short rate model fit a curve of bond or swap rates; a time-varying volatility lets it fit many option prices. The general form makes both functions of calendar time only:

Short rate process with time-dependent drift and volatility (equation 15.1).

Volatility now depends on when the change happens, not on the rate level. Suppose the function gives σ(1) = 1.26% and σ(2) = 1.20%. Then a year from today the short rate moves with an annual basis-point volatility of 126 basis points, while two years from today it moves with 120 basis points per year.

The Model 3 specification

A convenient special case lets volatility start at a constant σ and decay exponentially toward zero at rate α:

Model 3 (equation 15.2).

The decay could equally stop at a positive floor. With σ = 126 basis points and α = 0.025, the instantaneous volatility falls from 126 basis points today to about 98 basis points at year 10 and about 60 basis points at year 30, as Figure 1 shows.

Figure 1: Instantaneous volatility in Model 3 (sigma 126 bp, alpha 0.025)

Each new year of the horizon adds less variance than the one before it, because the shocks arriving later are smaller.

Horizon standard deviation accumulates these declining shocks, so it climbs quickly over the first few years and then flattens, reaching roughly 5% at a 30-year horizon here. Other choices of σ(t) give very different shapes.

Why option market makers like deterministic volatility

A caplet pays, on some notional amount, the excess of the short rate over a strike at expiry, or nothing if that excess is negative. So its value hinges on how the rate is distributed on that date, and a dealer can tune σ(t) date by date until the model reprices caplets across many expiries.

Example 1 · Worked

Use Model 3 with σ = 126 basis points and α = 0.025. Ten years from today the drift is λ = 0.30% per year. Take a monthly step, dt = 1/12, and a realised shock dw = 0.12.

1. What is the instantaneous volatility at year 10, the change in the short rate over the month, and the monthly standard deviation of that change? How does it compare with a step taken today?

Solution. Volatility at year 10 is 0.0126 × e^(−0.025 × 10) = 0.0126 × 0.7788 = 0.98129%, or 98.13 basis points per year. The drift term is 0.0030 × 1/12 = 0.00025, which is 2.50 basis points. The random term is 0.0098129 × 0.12 = 0.0011775, which is 11.78 basis points. So dr = 2.50 + 11.78 = 14.28 basis points. The monthly standard deviation is 98.13 × √(1/12) = 28.33 basis points. A step taken today would have a standard deviation of 126 × √(1/12) = 36.37 basis points, so the rate change becomes less volatile further out.

Check yourself

In Model 3, what happens to the instantaneous volatility of the short rate as t grows large?

It decays exponentially toward zero, since σe^(−αt) falls as t rises for any positive α.

Why does a caplet price depend on the volatility function at its own expiry horizon?

The caplet payoff depends on the short rate at expiry, so its value depends on the distribution of the rate on that date, which is shaped by σ(t) up to then.

Model 3’s rising then flattening horizon standard deviation mimics what mean reversion does in the Vasicek model, and under one parameter choice the match is exact. If the initial volatility of Model 3 equals the Vasicek volatility, and the decay rate α equals the Vasicek mean reversion speed, the two horizon standard deviations coincide. If, in addition, the Model 3 drift λ(t) is chosen to trace the expected path of the rate in the Vasicek model, the full terminal distributions match.

The models still differ. Model 3 has no mean reversion, so a shock today moves every rate on the curve by the same amount: it is a parallel shift model. Its term structure of volatility, meaning the volatility of spot rates of different maturities at a given moment, is flat. Its level drifts down as σ(t) decays, but on any given date it is flat. The Vasicek model instead produces factor loadings and volatilities that slope down with maturity.

Assessing the efficacy of time-dependent volatility

The case for and against σ(t) mirrors the debate about a time-dependent drift.

Quoting options that are hard to observe. If the aim is to price a fixed income option with no liquid quote, a time-dependent volatility function interpolates sensibly between the option prices that are known.
Valuing and hedging. For valuing and hedging bonds and options, mean reversion may be preferred, for two reasons.

First, mean reversion rests on a plausible economic story, whereas a fitted σ(t) implicitly claims the market can forecast short-term volatility many years ahead. A common compromise lets volatility vary over the near future and then settle at a constant.

The second reason is empirical: rates rarely move in parallel and long rates tend to be less volatile than short ones, which a mean-reverting model captures better. Vasicek may still fall short for some purpose, but a caplet-calibrated parallel shift model is unlikely to do better.

Check yourself

Model 3 and a Vasicek model are calibrated to identical terminal distributions. Name one way they still differ.

Model 3 is a parallel shift model with a flat term structure of volatility at each date, while the Vasicek model has a downward-sloping factor structure and term structure of volatility.

What is the main economic objection to fitting a long time-dependent volatility function?

It assumes the market can forecast short-term volatility far into the future.

The models above give the short rate the same basis-point volatility at every rate level. That is unlikely to hold at the extremes. High inflation, high-rate periods are unstable and rates swing more; near zero, the floor limits how far rates can fall. So it makes sense to let basis-point volatility rise with the rate.

Square-root volatility: the Cox-Ingersoll-Ross model

The risk-neutral dynamics of the CIR model are

Cox-Ingersoll-Ross (equation 15.3).

The drift is not random, and the standard deviation of dw is √dt by construction. So the annualised standard deviation of dr, which is the basis-point volatility, equals σ√r. The parameter σ is constant, but basis-point volatility grows with the square root of the rate.

Proportional volatility: Courtadon and Model 4

A second family makes basis-point volatility proportional to the rate itself, and σ is then called the yield volatility. Two examples are the Courtadon model and the simplest lognormal model, called Model 4:

Equations 15.4 and 15.5: yield volatility σ is constant; basis-point volatility is σr.

Calibrating the three specifications at a rate of 8%

To compare like with like, each σ is chosen so that basis-point volatility equals 100 basis points (0.01) when the short rate is 8%:

Equations 15.6 to 15.8.

Note the units: basis-point volatility is an interest rate (100 basis points), yield volatility a percentage of the rate (12.5%). Table 1 applies the calibrated parameters at other rates.

Table 1: Annual basis-point volatility by short rate level (basis points)

MR 15 – The Art of Term Structure Models – Volatility and Distribution — table 1
Short rate	Proportional (12.5% × r)	Square root (0.0354 × √r)	Constant
2%	25.0	50.1	100
4%	50.0	70.8	100
8%	100.0	100.1	100
12%	150.0	122.6	100
16%	200.0	141.6	100
20%	250.0	158.3	100

Original calculation. The square-root column at 8% shows 100.1 only because 0.0354 is rounded from 0.035355.

Figure 2: Basis-point volatility against the short rate

Both rate-dependent curves pass through zero at a zero rate; the proportional curve rises in a straight line and overtakes the square-root curve above 8%.

Because basis-point volatility vanishes as the rate reaches zero, and the drift is positive at zero, the CIR and proportional models keep the short rate from going negative. That helps in some uses, but purpose still decides. A trader who thinks constant volatility suits current conditions, that negative rates barely affect the securities at hand, and that computational simplicity matters may sensibly accept some chance of negative rates.

Example 2 · Worked

The short rate is 4%. The CIR model has k = 0.25, θ = 6% and σ = 0.0354. Model 4 has a = 2% and σ = 12.5%. Take dt = 1/12 and a realised shock dw = 0.20.

1. Compute the change in the short rate and the annual basis-point volatility under each model, and compare with a constant 100 basis point volatility.

Solution. CIR drift: 0.25 × (6% − 4%) × 1/12 = 0.000417, or 4.17 basis points. CIR basis-point volatility: 0.0354 × √0.04 = 0.0354 × 0.2 = 0.00708, or 70.8 basis points. Random term: 0.00708 × 0.20 = 0.001416, or 14.16 basis points. So dr = 4.17 + 14.16 = 18.33 basis points.
Model 4 drift: 0.02 × 0.04 × 1/12 = 0.0000667, or 0.67 basis points. Basis-point volatility: 12.5% × 4% = 0.50%, or 50 basis points. Random term: 0.0050 × 0.20 = 10.00 basis points. So dr = 10.67 basis points.
With a constant 100 basis point volatility, the same shock alone moves the rate 0.01 × 0.20 = 20 basis points. Below the 8% calibration point, the square-root model is calmer than the normal one and the proportional model calmer still.

Check yourself

At a short rate of 16%, what basis-point volatility does each calibrated specification give?

Constant 100 bp; square root 0.0354 × √0.16 = 141.6 bp; proportional 12.5% × 16% = 200 bp.

Why is yield volatility quoted in percent of the rate rather than in basis points?

Because in a proportional model σ multiplies the rate, so it measures the size of rate changes relative to the rate level, not in absolute basis points.

A lognormal model with deterministic drift, credited to a Salomon Brothers research team in the ’80s, has the dynamics

Equation 15.9.

Ito’s Lemma, which is taken here as given, turns the process for r into a process for its natural logarithm:

Equation 15.10.

Substituting the dynamics into this expression gives d[ln r] = [ã(t) − ½σ²]dt + σdw. Folding the constant ½σ² into a relabelled drift, a(t) = ã(t) − ½σ², leaves

Equations 15.11 and 15.12.

This says ln r is normally distributed. A variable is lognormal when its logarithm is normal, so r itself is lognormal. The equation is simply the Ho-Lee model written for ln r instead of r, which means the Ho-Lee tree can be reused with the logarithm of the rate on each node. Each branch has probability 1/2:

Ho-Lee tree on the natural logarithm of the rate.

Exponentiating every node gives the tree in rates, for example r₀e^((a₁+a₂)dt + 2σ√dt) at the top of date 2. Shocks now multiply the rate rather than add to it. Since e raised to any power is positive, every node stays positive as long as r₀ is positive, which is why the model cannot produce negative rates.

Figure 3: Two-date lognormal tree (r0 = 5%, sigma 12.5%, zero drift, dt = 1)

Up and down moves are percentages of the current rate, so the gaps between nodes shrink as rates fall and the tree never reaches zero.

Why lognormal volatility is a percentage of the rate

For small x, e^x is close to 1 + x. With a₁ = 0 and dt = 1, the top node of date 1 is approximately

Equation 15.13.

So σ reads directly as a percentage move: with σ = 12.5%, the up-state rate sits about 12.5% above the starting rate. Starting from 5%, the exact up node is 5% × e^0.125 = 5.666%, against the approximation 5% × 1.125 = 5.625%. As in the Ho-Lee model, the drift constants a₁ and a₂ are then chosen so that the tree reprices market bonds.

Check yourself

Starting at 5% with σ = 12.5%, zero drift and dt = 1, what is the rate at the down node of date 1?

5% × e^(−0.125) = 4.412%.

What are the drift constants a₁ and a₂ used for in this tree?

They are set so that the tree matches market bond prices, just as in the Ho-Lee model.

The Black-Karasinski model adds mean reversion to a lognormal short rate. Volatility, the speed of mean reversion and the central tendency may all depend on time, which places it among arbitrage-free models; a user can switch off any of that time dependence. Its dynamics are

Equation 15.14.

or, using the same Ito step as in the previous section,

Equation 15.15.

So ln r is normal, reverting toward ln θ(t) at speed k(t) with volatility σ(t): a time-dependent Vasicek model for the log rate.

Building the tree and the recombination problem

Over the first date, writing r₁ = r₀e^(k(1)(ln θ(1) − ln r₀)dt) for readability, the up and down rates are r₁e^(σ(1)√dt) and r₁e^(−σ(1)√dt), with logarithms

Equations 15.16 and 15.17.

The up-then-down and down-then-up paths are

Equations 15.18 and 15.19.

Setting the exponents equal shows the two paths meet only if

Equation 15.20.

That restriction ties mean reversion completely to the volatility function, yet the two do different jobs: mean reversion sets how volatility varies across maturities, while σ(t) sets future short rate volatility and so the prices of options with different expiries. The fix is to let the time step change. With steps dt₁ and dt₂, the same algebra gives

Equation 15.23, from the variable-step versions 15.21 and 15.22.

The first step can be any length; the second is then solved from this condition, leaving the user free to choose mean reversion and volatility separately.

Example 3 · Worked

A Black-Karasinski tree uses σ(1) = 20%, σ(2) = 18% and dt₁ = 1 year.

1. With equal steps, what mean reversion speed does recombination force? If the desired speed is k(2) = 0.25, how long must the second step be?

Solution. With equal steps of 1 year, k(2) = (0.20 − 0.18) / (0.20 × 1) = 0.10, regardless of what the user wants. For k(2) = 0.25, solve 0.25 × dt₂ = 1 − (0.18/0.20)√(dt₂/1) = 1 − 0.9√dt₂. Numerically, dt₂ = 0.7934 years. Check: 0.25 × 0.7934 = 0.1983, and 1 − 0.9 × √0.7934 = 1 − 0.9 × 0.8907 = 0.1983. A shorter second step lets the tree recombine with the stronger mean reversion.

Check yourself

If σ(t) is constant and the steps are equal, what does recombination imply for k(2)?

k(2) = (σ − σ)/(σ dt) = 0, so no mean reversion is allowed unless the step length changes.

Which variable does the Black-Karasinski tree adjust to keep mean reversion and volatility independent?

The length of the time step, dt, from one date to the next.

MR 15 – The Art of Term Structure Models – Volatility and Distribution

Overview

What is a time-dependent volatility short rate model (Model 3)

The Model 3 specification

Why option market makers like deterministic volatility

Time-dependent volatility vs mean reversion: which model to use

Assessing the efficacy of time-dependent volatility

CIR model vs lognormal model: basis-point volatility and yield volatility

Square-root volatility: the Cox-Ingersoll-Ross model

Proportional volatility: Courtadon and Model 4

Calibrating the three specifications at a rate of 8%

Terminal distribution of the short rate: CIR vs normal vs lognormal

How to build a lognormal tree for the Salomon Brothers model

Why lognormal volatility is a percentage of the rate

Black-Karasinski model: lognormal short rate with mean reversion

Building the tree and the recombination problem

Closed-form spot rate formulas for term structure models