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

Model 1: Normal Model with Constant Volatility

Q: What is Model 1 in Tuckman's term structure models?

Model 1 is the normal model with constant volatility and no drift: dr = σ dw. The short rate has a normal distribution centred on today's rate. It is the base case, and later models add drift, mean reversion or changing volatility.

Q: Why can Model 1 produce negative interest rates?

Rate changes are normally distributed, and a normal distribution has no lower limit. Any positive σ gives some probability that the rate falls below zero, and it rises with the horizon. This is a weakness when rates are high, but less so in markets that have seen negative policy rates.

Q: How do I build the binomial tree for Model 1?

Compute σ√dt, then at each step move up or down by that amount with probability 0.5. The tree recombines, so step n has n + 1 nodes, with rates r0 + (n − 2k) × σ√dt where k is the number of down moves.

Q: What is the difference between Model 1 and Model 2?

Model 1 has zero drift, so the expected rate stays at r0. Model 2 adds a constant drift term, so the expected rate shifts steadily over time. Volatility is constant in both.

Updated 11 October 2026 · Fact-checked

Model 1 is the simplest term structure model: the short rate changes by dr = σ dw, with constant volatility and no drift. Rates are normally distributed, so they can go negative. You build a recombining tree where each step moves up or down by σ√dt with probability 0.5, then price by discounting backwards.

Understand Normal Model with Constant Volatility (Model 1)

A term structure model describes how the short-term interest rate moves over time. Model 1 is the starting point. It says the change in the short rate is pure randomness: dr = σ dw. Here σ is the volatility of rate changes, quoted in basis points per year, and dw is a normally distributed shock with mean zero and variance dt.

Two features define the model. First, drift is zero: the rate has no expected direction, so the expected future rate equals today's rate. Second, volatility is constant: σ does not depend on time or on the level of the rate. Model 2 later adds a constant drift term. Do not mix the two.

Because the shocks are normal, the rate after time T is normally distributed: r(T) ~ N(r0, σ²T). The standard deviation grows with √T. Normal distributions have no floor, so there is always a positive probability that the rate is below zero. This was seen as a flaw when rates were far above zero. After years of negative policy rates in the euro area and Japan, it is seen as less of a flaw, but it still matters for realism.

To price securities you turn the model into a binomial tree. Each time step of length dt, the rate goes up by σ√dt or down by σ√dt, each with probability 0.5. An up move followed by a down move gives the same rate as a down move followed by an up move, so the tree recombines. Step n has n + 1 nodes, which keeps the tree small. You then value a security by working backwards from maturity, discounting the average of the two next-step values at the rate in the current node.

Key formulas to remember

Model 1 dynamics: dr = σ dw; No drift term. σ is the annual volatility of rate changes, in basis points per year. Model 2 would add + λ dt.
Distribution of the rate: r(T) ~ Normal(mean = r0, standard deviation = σ√T); Mean stays at r0 because drift is zero. Standard deviation grows with the square root of time.
Tree step size: Up or down move per step = σ√dt; Use dt in years. For quarterly steps dt = 0.25 and √dt = 0.5. For monthly steps dt = 1/12.
Node rates: r(n, k) = r0 + (n − 2k) × σ√dt; n = number of steps, k = number of down moves. Gives n + 1 nodes at step n.
Node probabilities: Each branch probability = 0.5; P(k downs in n steps) = C(n, k) ÷ 2ⁿ; Use for the chance of reaching a specific node.
Backward induction: V(node) = 0.5 × [V(up) + V(down)] ÷ (1 + r(node) × dt); Discount with the rate at the current node, over one step. Cash flows at that node are added after discounting.
Probability of a negative rate: P(r(T) < 0) = Φ(−r0 ÷ (σ√T)); Φ is the standard normal cumulative distribution. Keep r0 and σ in the same units.

How to solve Normal Model with Constant Volatility (Model 1) questions

Use this order for any Model 1 question, whether it asks for tree values, probabilities or a price.

1Confirm the model: dr = σ dw means zero drift and constant volatility. If a drift term appears, it is Model 2.
2Put all inputs in the same units. Convert basis points to percent or decimals, and state dt in years.
3Compute the step size σ√dt. This is the move up or down in rate at each step.
4Build the nodes with r0 + (n − 2k) × σ√dt. Check that step n has n + 1 nodes and that the middle node equals r0 for even n.
5If the question is about probabilities, use 0.5 per branch for the tree, or the normal distribution N(r0, σ²T) for a time horizon.
6If the question is about price, set the terminal payoffs, then discount backwards one step at a time using 0.5 × [Vu + Vd] ÷ (1 + r × dt).
7For negative-rate questions, find the lowest node or compute Φ(−r0 ÷ (σ√T)). State clearly that the model allows negative rates.
8Sanity check: with zero drift the average rate at each step should equal r0.

Quickest way: Step size first, then nodes

When to use it: Use this when the question gives σ, dt and r0 and asks for a node rate, the lowest rate, or whether rates can turn negative.

Work out σ√dt in the same unit as r0. For dt = 0.25 just halve σ.
Lowest node at step n is r0 − n × σ√dt. Highest is r0 + n × σ√dt.
Set the lowest node below zero to find the first step where a negative rate is possible: n > r0 ÷ (σ√dt).
The chance of the lowest node is (1/2)ⁿ.
For a long horizon, skip the tree: use z = −r0 ÷ (σ√T) and read the normal table.

Common mistakes in Normal Model with Constant Volatility (Model 1)

Using σ as the step size instead of σ√dt.
Students forget that σ is an annual figure and the tree step is shorter than a year.
Fix: Always multiply σ by √dt first. For quarterly steps the move is σ × 0.5. For monthly steps it is σ ÷ √12.
Mixing basis points and percent, for example subtracting 120 from 3%.
Questions give σ in bp and r0 in percent.
Fix: Convert σ to percent first: 120 bp = 1.20%. Write units next to every number.
Saying Model 1 cannot produce negative rates.
Students confuse it with lognormal models, where rates stay positive.
Fix: Remember normal shocks have no floor. Model 1 always gives a positive probability of negative rates, which grows with σ and T.
Adding a drift to Model 1 or treating the expected rate as moving.
Students mix up Model 1 and Model 2, or think rates should trend.
Fix: Model 1 has zero drift, so the expected short rate stays at r0. Only Model 2 has a constant drift.
Discounting with the wrong rate in backward induction, for example using r0 at every step.
The tree is read as a single rate rather than a different rate at each node.
Fix: At each node, discount the average of the two next values with that node's own rate over one step: divide by (1 + r × dt).
Counting 2ⁿ nodes at step n.
Students treat the tree as non-recombining.
Fix: The tree recombines, so step n has n + 1 nodes. The 2ⁿ figure counts paths, not nodes.

Worked examples

Example 1

The current three-month rate is 3.00%. Model 1 applies with σ = 1.20% per year and quarterly steps (dt = 0.25). (a) Give the rates at step 2. (b) Find the first step where a negative rate is possible and its probability at that step. (c) Price a zero-coupon bond with face value 100 that pays at step 2, using backward induction.

Show the solution

Step size = σ√dt = 1.20% × √0.25 = 1.20% × 0.5 = 0.60%.
(a) Step 1 rates: 3.60% and 2.40%. Step 2 rates: 3.00% + 1.20% = 4.20%, 3.00%, and 3.00% − 1.20% = 1.80%.
(b) Lowest node at step n is 3.00% − 0.60% × n. This is below zero when n > 5, so n = 6 gives −0.60%. Step 5 gives exactly 0.00%.
The only negative node at step 6 is the all-down path. Probability = (1/2)⁶ = 1/64 = 1.5625%. The next node up is 0.60%, which is positive.
(c) Terminal value is 100 at all step 2 nodes. Step 1 up node (3.60%): 100 ÷ (1 + 0.036 × 0.25) = 100 ÷ 1.009 = 99.1080. Step 1 down node (2.40%): 100 ÷ 1.006 = 99.4036.
Average = (99.1080 + 99.4036) ÷ 2 = 99.2558. Discount at the root rate 3.00%: 99.2558 ÷ (1 + 0.03 × 0.25) = 99.2558 ÷ 1.0075 = 98.5173, which is about 98.52.

Answer: (a) 4.20%, 3.00%, 1.80%. (b) The first negative rate is possible at step 6, with a rate of −0.60% and probability 1.5625%. (c) The zero-coupon bond price is about 98.52.

Example 2

Under Model 1 the short rate today is 2.00% and σ = 90 bp per year. What is the distribution of the short rate in 4 years, and what is the approximate probability that it is negative?

Show the solution

Mean = r0 = 2.00%, because drift is zero.
Standard deviation = σ√T = 90 bp × √4 = 180 bp = 1.80%.
z = (0 − 2.00%) ÷ 1.80% = −1.111.
P(r < 0) = Φ(−1.111) = 1 − Φ(1.111). Φ(1.11) is 0.8665 and Φ(1.12) is 0.8686, so Φ(1.111) ≈ 0.8667.
P(r < 0) ≈ 1 − 0.8667 = 0.1333.

Answer: The rate in 4 years is normal with mean 2.00% and standard deviation 1.80%. The probability that it is negative is about 13.3%.

Exam tips

Read the units first. Many wrong answers come from bp versus percent and from forgetting that dt is in years.
Questions often ask what the model implies, not for a number. Know the three facts: zero drift, constant volatility, negative rates possible.
On tree questions, compute σ√dt once and write it down. Every node then follows by adding or subtracting.
Quarterly steps give √dt = 0.5 and monthly steps give 1/√12 ≈ 0.2887. These two appear often.
If an option offers a drift term or a rate that cannot go negative, it describes a different model. Check which model the stem names.

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

Normal Model with Constant Volatility (Model 1): frequently asked questions

What is Model 1 in Tuckman's term structure models?

Model 1 is the normal model with constant volatility and no drift: dr = σ dw. The short rate has a normal distribution centred on today's rate. It is the base case, and later models add drift, mean reversion or changing volatility.

Why can Model 1 produce negative interest rates?

Rate changes are normally distributed, and a normal distribution has no lower limit. Any positive σ gives some probability that the rate falls below zero, and it rises with the horizon. This is a weakness when rates are high, but less so in markets that have seen negative policy rates.

How do I build the binomial tree for Model 1?

Compute σ√dt, then at each step move up or down by that amount with probability 0.5. The tree recombines, so step n has n + 1 nodes, with rates r0 + (n − 2k) × σ√dt where k is the number of down moves.

What is the difference between Model 1 and Model 2?

Model 1 has zero drift, so the expected rate stays at r0. Model 2 adds a constant drift term, so the expected rate shifts steadily over time. Volatility is constant in both.