FRM Exam Part II · The Art of Term Structure Models: Volatility and Distribution

Lognormal Short Rate Models: Black-Karasinski and Salomon Brothers

Updated 11 October 2026 · Fact-checked

A lognormal short-rate model makes the log of the rate normally distributed, so the rate stays positive and its basis-point volatility equals σ × the rate level. A stylized form is dr = a·r·dt + σ·r·dw (Salomon Brothers). Black-Karasinski adds mean reversion to ln r. To solve questions, multiply σ by r, or scale tree nodes by e^(2σ√dt).

Understand Lognormal Models: Black-Karasinski and Salomon Brothers

Start with the problem in the normal models. In Ho-Lee and Vasicek, the short rate is normally distributed, with the same basis-point volatility at every rate level. That means rates can go negative, and a rate of 1% moves as much in basis points as a rate of 10%. Many practitioners find this unrealistic when rates are low.

The lognormal fix is to model the percentage change in the rate rather than the change in basis points. This page uses a stylized lognormal form, called the Salomon Brothers form, which is dr = a·r·dt + σ·r·dw. It is a simple illustration, not an official FRM definition. Both the drift and the volatility are proportional to r. Here σ is a percentage (proportional) volatility. Because ln r is normally distributed, r itself is lognormal: it can never fall below zero.

The key link is volatility. Basis-point volatility = σ × r. If σ is 20% and the rate is 5%, the annual volatility is 100 bps. If the rate falls to 2%, volatility falls to 40 bps. Volatility shrinks as rates fall and grows as rates rise. The distribution is also right-skewed: large upward moves are possible, but the rate cannot go below zero.

Black-Karasinski is the mean-reverting lognormal model. Its form is d(ln r) = k(t)·[ln θ(t) − ln r]·dt + σ(t)·dw. The log of the rate is pulled toward a long-run level, with a time-dependent speed and volatility. It has no closed-form bond price, so it is implemented on a tree or by numerical methods and calibrated to the term structure and volatilities.

On a recombining binomial tree, you build the tree in ln r. Adjacent nodes at the same date differ by a constant multiple: r(up) = r(down) × e^(2σ√dt). Rates are never negative, and the gaps between nodes widen as you move up the tree.

Key formulas to remember

Salomon Brothers (lognormal) model: dr = a·r·dt + σ·r·dw; Drift and volatility are both proportional to r. The rate stays positive. There is no mean reversion in this form.
Black-Karasinski model: d(ln r) = k(t)·[ln θ(t) − ln r]·dt + σ(t)·dw; Mean-reverting lognormal model. ln r reverts to ln θ at speed k. It needs a tree or numerical calibration.
Basis-point volatility: Annual bp vol = σ × r; σ is a percentage volatility, for example 20%. Convert to bps by multiplying by the rate level.
Tree node spacing: r(up) = r(down) × e^(2σ√dt); Applies to adjacent nodes on the same date. Use the same σ and dt for that step.
Expected rate with proportional drift: E[r(T)] = r(0) × e^(aT); Applies to dr = a·r·dt + σ·r·dw. The drift multiplies the rate, not a basis-point amount.
Log-rate dynamics: ln r is normally distributed, so r > 0; This is why lognormal models cannot produce negative rates.

How to solve Lognormal Models: Black-Karasinski and Salomon Brothers questions

Use this method for any question on lognormal short-rate models, whether it asks for volatility, tree nodes or model comparison.

1Identify the model. Rate-proportional volatility (σ·r) means lognormal. Constant bp volatility means a normal model. σ√r means CIR.
2Note whether the question gives σ as a percentage of the rate (such as 20%) or in bps. Lognormal σ is always a percentage.
3For volatility at a given rate, compute σ × r. Convert to bps at the end.
4For a tree, compute the factor e^(2σ√dt) for one step. Multiply to go up a node and divide to go down. Check that the step uses the right dt.
5For drift or expected rate, apply the proportional drift, such as e^(aT), to the starting rate. Do not add a basis-point amount.
6Check sense: rates must be positive, and volatility must be higher at higher nodes.
7Interpret in words: say whether volatility rises or falls with the rate, and whether mean reversion is present (Black-Karasinski) or not (basic Salomon form).

Quickest way: Three-check shortcut

When to use it: Use this on multiple-choice questions where options differ in volatility scale or distribution shape.

Ask: does volatility change with the rate level? If yes, it is lognormal or CIR, not Ho-Lee or Vasicek.
Ask: can rates go negative? If not, and volatility is proportional to r, choose lognormal.
Compute bp vol as σ × r for the number asked. For tree gaps, multiply by e^(2σ√dt). Eliminate options that break positivity or the proportional pattern.

Common mistakes in Lognormal Models: Black-Karasinski and Salomon Brothers

Treating σ as basis points, for example reading σ = 20% as 20 bps.
Students carry over the normal-model habit where σ is in bps.
Fix: In lognormal models σ is a percentage of the rate. Basis-point volatility = σ × r.
Adding drift in bps to the rate instead of scaling it.
Ho-Lee and Vasicek drifts are additive.
Fix: With drift a·r·dt, the expected rate is r(0)·e^(aT). Multiply, do not add.
Using e^(σ√dt) instead of e^(2σ√dt) for the gap between adjacent nodes.
Students confuse a move from the centre with the full distance between the up and down nodes.
Fix: The ratio of the up node to the down node from the same starting point is e^(2σ√dt).
Saying a lognormal model has constant basis-point volatility.
Students mix it up with Model 1 or Ho-Lee.
Fix: Lognormal bp volatility moves with the rate. Constant percentage volatility means bp volatility rises and falls with r.
Claiming Black-Karasinski has closed-form bond prices like Vasicek.
Both are described as mean-reverting.
Fix: Black-Karasinski is solved on a tree or numerically. Closed forms belong to the normal-type models such as Vasicek.
Assuming lognormal models are better because they never go negative.
The positivity point is overstated.
Fix: Positivity is a feature, but it brings right skew, no closed form and sensitivity to the σ you calibrate. Say these trade-offs when comparing models.

Worked examples

Example 1

In a lognormal tree with annual steps (dt = 1), σ = 15%. At date 2, the middle node has a rate of 5%. Assuming the tree recombines with node spacing e^(2σ√dt) between adjacent nodes at the same date, find the rates at the upper and lower nodes, and the annual bp volatility at the upper node.

Show the solution

Adjacent nodes at the same date differ by the ratio e^(2σ√dt) = e^(2 × 0.15 × 1) = e^(0.30) = 1.349859. This ratio applies between the middle node and the upper node, and between the lower node and the middle node.
Upper node = 5% × 1.349859 = 6.7493%.
Lower node = 5% ÷ 1.349859 = 5% × 0.740818 = 3.7041%.
Volatility at the upper node = 0.15 × 6.7493% = 1.012%, which is about 101 bps.
For comparison, volatility at the middle node = 0.15 × 5% = 0.75%, or 75 bps. At the lower node = 0.15 × 3.7041% = 0.556%, or about 56 bps.

Answer: Upper node about 6.75%, lower node about 3.70%. Annual volatility at the upper node is about 101 bps, compared with about 75 bps at the middle node and about 56 bps at the lower node.

Example 2

A short rate follows dr = a·r·dt + σ·r·dw with r(0) = 4%, a = 2% per year and σ = 25%. Find (1) the current annual basis-point volatility, (2) the expected rate after 5 years under the model, and (3) the volatility if the rate falls to 2%.

Show the solution

(1) bp volatility = σ × r = 0.25 × 4% = 1.00% = 100 bps.
(2) Expected rate = 4% × e^(0.02 × 5) = 4% × e^0.10.
e^0.10 = 1.105171, so the expected rate = 4.4207%.
(3) At r = 2%, volatility = 0.25 × 2% = 0.50% = 50 bps.

Answer: Current volatility is 100 bps. The expected 5-year rate is about 4.42%. If the rate falls to 2%, volatility falls to 50 bps, which shows volatility proportional to the rate.

Exam tips

Look for the volatility clue first. Rate-proportional volatility points to lognormal. Constant bp volatility points to Ho-Lee or Vasicek.
Always convert σ to bps by multiplying by the rate. Check that your answer scales with the rate level.
Remember Black-Karasinski is the lognormal model with mean reversion, and it needs a tree. Do not credit it with a closed-form bond price.
For tree questions, write the step factor e^(2σ√dt) first and check dt matches the step length.
When asked to compare models, give one benefit (positive rates, realistic volatility at low rates) and one cost (no closed form, right skew, heavier calibration).

Practice questions from The Art of Term Structure Models: Volatility and Distribution

Lognormal Models: Black-Karasinski and Salomon Brothers: frequently asked questions

What is the difference between Black-Karasinski and the Salomon Brothers model?

Both make the log of the rate normally distributed, so rates stay positive. The basic Salomon Brothers form is dr = a·r·dt + σ·r·dw, with proportional drift and no mean reversion. Black-Karasinski models d(ln r) with mean reversion toward a long-run level and time-dependent parameters.

How does a lognormal distribution of rates affect volatility?

Basis-point volatility equals σ × r, so it is higher when rates are high and lower when rates are low. The distribution is also right-skewed and bounded below by zero. This matches the idea that rate changes are smaller at low rate levels.

Why can lognormal models not produce negative rates?

Because ln r is normally distributed, r = e^(ln r) is always positive. The rate can get very close to zero but never crosses it. This is a limitation when negative rates actually occur.

How is a lognormal model built on a tree?

You model ln r on a recombining binomial tree. The rate at an up node equals the rate at the adjacent down node times e^(2σ√dt). Drift and mean reversion adjust the centre of the nodes, and calibration fits the tree to today's curve.