Economic Modelling · Binomial option-pricing model

Calibrating u and d in the Binomial Model and the Link to Black-Scholes

Q: What are the Cox-Ross-Rubinstein formulas for u and d?

u = e^(σ√Δt) and d = 1/u = e^(−σ√Δt). Here σ is the annual volatility and Δt is the length of one step in years. The choice makes the tree recombine.

Q: Why does the binomial price converge to Black-Scholes?

Each step moves the log price up or down by σ√Δt. As the steps get smaller and more numerous, the sum of these moves behaves like a normal distribution with variance σ²T. That is the lognormal share price used in Black-Scholes.

Q: Do I use the real-world probability to calibrate the tree?

No. The volatility fixes u and d. The probability used for pricing is the risk-neutral p, which comes from the risk-free rate. The real-world drift of the share does not affect the option price.

Q: What if p comes out above 1 or below 0?

Then the condition d < e^(rΔt) < u has failed and the tree allows arbitrage. This usually happens when the step is too long for the given volatility. Use more, shorter steps.

Updated 11 October 2026 · Fact-checked

In the Cox-Ross-Rubinstein (CRR) model you set u = e^(σ√Δt) and d = 1/u = e^(−σ√Δt), where σ is annual volatility and Δt is the step length in years. Then use p = (e^(rΔt) − d) ÷ (u − d). As the number of steps grows, the price converges to Black-Scholes.

Understand Calibrating u and d and Link to Black-Scholes

A binomial tree needs two numbers at each step: the up factor u and the down factor d. They decide how widely share prices spread. If you pick them badly, the tree gives the wrong volatility and the wrong option price.

The Black-Scholes model assumes the share price follows geometric Brownian motion. Over a short time Δt, the log of the price ratio then has standard deviation σ√Δt. The CRR choice copies this. You move the log price up or down by exactly σ√Δt each step. That gives u = e^(σ√Δt) and d = e^(−σ√Δt) = 1/u.

Because d = 1/u, an up move followed by a down move returns the price to where it started. The tree recombines. After n steps there are only n + 1 final prices, not 2^n. This keeps the tree small and easy to work through.

You do not choose the probability from your view of the share. The risk-neutral probability is p = (e^(rΔt) − d) ÷ (u − d), where r is the continuously compounded risk-free rate. For no-arbitrage you need d < e^(rΔt) < u. If this fails, p falls outside 0 to 1 and the tree allows arbitrage. This can happen if Δt is too large for the given σ and r.

As you split a fixed time T into more steps (Δt = T ÷ n, n → ∞), the binomial price of a European option converges to the Black-Scholes price. The convergence is not smooth. The binomial price often swings above and below the Black-Scholes value as n changes, but the error shrinks. The same tree also handles American options, which Black-Scholes does not price in closed form.

Key rules to remember

CRR up factor: u = e^(σ√Δt); σ is annual volatility. Δt is the step length in years, so Δt = T ÷ n.
CRR down factor: d = 1/u = e^(−σ√Δt); This makes the tree recombine, so an up then a down move returns the price to S₀.
Risk-neutral probability of an up move: p = (e^(rΔt) − d) ÷ (u − d); r is the continuously compounded risk-free rate. The probability of a down move is 1 − p.
No-arbitrage condition: d < e^(rΔt) < u; Equivalent to 0 < p < 1. Check it when Δt is large.
Option value by backward induction: V = e^(−rΔt) × [p × V_up + (1 − p) × V_down]; Apply at each node, working back from the final payoffs.
Step length: Δt = T ÷ n; T is the option term in years and n is the number of steps.
Convergence statement: binomial price → Black-Scholes price as n → ∞ (Δt = T ÷ n → 0); Holds for European options with u and d calibrated to σ as above.

How to solve Calibrating u and d and Link to Black-Scholes questions

Use this method for any question that asks you to build or calibrate a binomial tree from volatility, or to link it to Black-Scholes.

1Write down S₀, strike K, term T, volatility σ, risk-free rate r and the number of steps n. Check whether r is continuous or annual effective. Convert to continuous if needed.
2Calculate Δt = T ÷ n in years.
3Calculate σ√Δt, then u = e^(σ√Δt) and d = 1/u.
4Calculate e^(rΔt). Check d < e^(rΔt) < u. Then compute p = (e^(rΔt) − d) ÷ (u − d).
5Build the share price nodes: S₀uʲd^(i−j) at step i after j up moves. Use recombination to keep only distinct nodes.
6Write the option payoff at the final nodes, for example max(S − K, 0) for a call.
7Work backwards using V = e^(−rΔt) [p V_up + (1 − p) V_down]. For an American option, compare with the exercise value at every node and take the larger.
8If asked about Black-Scholes, state that the tree price converges to it as n → ∞ and say what assumptions both models share (lognormal prices, constant σ and r, no arbitrage).

Quickest way: Three-line calibration shortcut

When to use it: Use when the question gives σ, r and Δt and only asks for u, d or p, or when you have little time in the multiple-choice section.

Compute x = σ√Δt. Then u = e^x and d = e^(−x).
Compute g = e^(rΔt). Then p = (g − d) ÷ (u − d).
Sanity check: p should be a little above or below 0.5 and always between 0 and 1. If not, recheck Δt and the units of r and σ.

Common mistakes in Calibrating u and d and Link to Black-Scholes

Using u = e^σ instead of e^(σ√Δt).
Students forget that σ is an annual figure and the step is shorter than a year.
Fix: Always compute Δt first, then σ√Δt. Write the units beside each number.
Using Δt in months or using n instead of T ÷ n.
The question gives the term in months or quotes a number of steps rather than a step length.
Fix: Convert everything to years. Δt = T (years) ÷ n.
Using a real-world probability such as 0.5 or the investor's view when discounting.
Students treat the tree like an expected-value problem.
Fix: Use the risk-neutral p = (e^(rΔt) − d) ÷ (u − d) and discount at the risk-free rate. The share's real-world drift does not enter the price.
Using the annual effective rate directly as e^(rΔt) without converting.
The question gives r as an effective annual rate but the formula assumes a continuous rate.
Fix: If the rate is effective annual i, use (1 + i)^Δt in place of e^(rΔt), or convert with r = ln(1 + i).
Not checking that p lies between 0 and 1.
Students assume the formula always gives a valid probability.
Fix: Check d < e^(rΔt) < u. If it fails, the tree has arbitrage and you need smaller steps.
Saying the binomial price rises steadily towards Black-Scholes as n increases.
Students overstate convergence as smooth and one-directional.
Fix: Say the price converges, but it can oscillate above and below the Black-Scholes value as n changes.

Worked examples

Example 1

A share has volatility σ = 20% a year. A European option has a term of 1 year and is modelled with a CRR tree of 4 equal steps. The continuously compounded risk-free rate is 5% a year. Find u, d and the risk-neutral probability p.

Show the solution

Δt = 1 ÷ 4 = 0.25 years.
√Δt = 0.5, so σ√Δt = 0.20 × 0.5 = 0.10.
u = e^0.10 = 1.105171 and d = e^(−0.10) = 0.904837.
e^(rΔt) = e^(0.05 × 0.25) = e^0.0125 = 1.012578.
Check: 0.904837 < 1.012578 < 1.105171, so no arbitrage.
p = (1.012578 − 0.904837) ÷ (1.105171 − 0.904837) = 0.107741 ÷ 0.200334 = 0.5378.

Answer: u = 1.1052, d = 0.9048 and p ≈ 0.538 (probability of a down move ≈ 0.462).

Example 2

A share is priced at ₹200. Its volatility is 25% a year and the continuously compounded risk-free rate is 6% a year. Using a single CRR step of one year, price a European call with strike ₹200 and expiry in one year. Then say what happens to the price if you use more steps.

Show the solution

Δt = 1, so σ√Δt = 0.25. u = e^0.25 = 1.284025 and d = e^(−0.25) = 0.778801.
e^(rΔt) = e^0.06 = 1.061837. Check d < 1.061837 < u. This holds.
p = (1.061837 − 0.778801) ÷ (1.284025 − 0.778801) = 0.283036 ÷ 0.505224 = 0.5602.
Share prices at expiry: up = 200 × 1.284025 = ₹256.81 and down = 200 × 0.778801 = ₹155.76.
Call payoffs: up = 256.81 − 200 = ₹56.81 and down = 0 (since 155.76 < 200).
Value = e^(−0.06) × [0.5602 × 56.81 + 0.4398 × 0] = 0.941765 × 31.82 = ₹29.97.
With more steps (n → ∞, Δt = T ÷ n), the tree price converges to the Black-Scholes price. A single step is a rough approximation.

Answer: The one-step call value is about ₹29.97. Using more steps with the same CRR calibration moves the price towards the Black-Scholes value.

Exam tips

Show Δt, σ√Δt, u, d and e^(rΔt) as separate lines. Marks are often given for each of these, even if a later step has an error.
Keep at least five or six decimal places for u, d and e^(rΔt), as p depends on small differences between them.
In written answers on convergence, name the conditions: Δt = T ÷ n → 0, u and d calibrated to σ, constant σ and r, and lognormal prices. Add that the convergence can oscillate.
If a question asks why the tree needs d < e^(rΔt) < u, explain it in plain words: otherwise one asset always beats the other and arbitrage is possible.
In the computer-based paper, build the tree with a loop over n and compare the price with Black-Scholes. Set u and d from σ√Δt and compute p once outside the loop.

Practice questions from Binomial option-pricing model

Calibrating u and d and Link to Black-Scholes: frequently asked questions

What are the Cox-Ross-Rubinstein formulas for u and d?

u = e^(σ√Δt) and d = 1/u = e^(−σ√Δt). Here σ is the annual volatility and Δt is the length of one step in years. The choice makes the tree recombine.

Why does the binomial price converge to Black-Scholes?

Each step moves the log price up or down by σ√Δt. As the steps get smaller and more numerous, the sum of these moves behaves like a normal distribution with variance σ²T. That is the lognormal share price used in Black-Scholes.

Do I use the real-world probability to calibrate the tree?

No. The volatility fixes u and d. The probability used for pricing is the risk-neutral p, which comes from the risk-free rate. The real-world drift of the share does not affect the option price.

What if p comes out above 1 or below 0?

Then the condition d < e^(rΔt) < u has failed and the tree allows arbitrage. This usually happens when the step is too long for the given volatility. Use more, shorter steps.