FRM Exam Part I · Binomial Trees

Delta and Binomial Tree Convergence to Black-Scholes

Q: How do I calculate delta in a binomial tree?

Divide the difference between the option values at the up and down nodes by the difference between the stock prices at those nodes. Use the nodes that follow the point where you want delta. Probabilities and discounting are not needed.

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

With u = e^(σ√Δt) and d = 1/u, the tree matches the volatility of the lognormal stock model. As the number of steps grows, the many small two-outcome moves approximate that continuous process. For European options the tree price then tends to the Black-Scholes price.

Q: How does the number of steps affect the binomial option price?

More steps give a price closer to the Black-Scholes value for a European option. The error shrinks overall but the tree price oscillates above and below the true value. More steps also take more computation.

Q: Does binomial delta equal Black-Scholes delta?

Not exactly for a finite number of steps. The tree delta is an approximation that tends to N(d1) for a call on a non-dividend-paying stock as steps increase.

Updated 11 October 2026 · Fact-checked

Delta in a binomial tree is the change in option value divided by the change in stock price across one step: (f_u − f_d) ÷ (S_u − S_d). It is the share count that hedges the option. As steps increase with CRR u and d, the tree price converges to the Black-Scholes value.

Understand Delta and Binomial Tree Convergence to Black-Scholes

A binomial tree lets the stock move to one of two prices over each short step. Because only two outcomes exist, you can build a portfolio of shares and cash that copies the option exactly. The number of shares in that portfolio is the delta.

The logic is simple. The option payoff changes from f_d to f_u when the stock moves from S_d to S_u. A position of Δ shares moves by Δ × (S_u − S_d). Set the two equal and you get Δ = (f_u − f_d) ÷ (S_u − S_d). Hold Δ shares against one short option and the value is the same in both nodes, so the position is riskless over that step.

Delta is not constant. It differs at each node, so a hedger must rebalance after every move. In a multi-step tree, compute delta at each node using only the two nodes that follow it. A call delta lies between 0 and 1 and a put delta lies between −1 and 0.

As you split the life of the option into more steps, each step gets smaller. With CRR parameters u = e^(σ√Δt) and d = 1/u, the tree matches the volatility of the lognormal process behind Black-Scholes. The tree price for a European option then approaches the Black-Scholes price, and the tree delta approaches N(d1) for a call. The convergence is not smooth. The price oscillates around the Black-Scholes value, and the error shrinks as steps increase. American options have no closed form, so trees with many steps are the standard tool for them.

Key formulas to remember

Tree delta (one step): Δ = (f_u − f_d) ÷ (S_u − S_d); Shares held per option. Use the two nodes that follow the node you are at.
Up and down factors (CRR): u = e^(σ√Δt); d = 1 ÷ u; Δt is the step length in years. Matches the volatility σ.
Risk-neutral probability: p = (e^(rΔt) − d) ÷ (u − d); With a continuous yield q, use e^((r − q)Δt) in the numerator.
Option value at a node: f = e^(−rΔt) × [p × f_u + (1 − p) × f_d]; Work backwards from maturity. For American options, compare with the exercise value at each node.
Replicating portfolio: f = Δ × S − B, where B is cash borrowed; Borrowing equals the PV of Δ × S_d − f_d.
Black-Scholes call delta: Δ_call = N(d1); Δ_put = N(d1) − 1; The tree delta converges to this as steps increase (no dividends).

How to solve Delta and Binomial Tree Convergence to Black-Scholes questions

Use this method for any question on tree delta, hedging or convergence.

1Write down S0, K, r, σ or u and d, the step length Δt and the number of steps. Check if the option is European or American and a call or put.
2If only σ is given, compute u = e^(σ√Δt) and d = 1 ÷ u.
3Compute the risk-neutral probability p = (e^(rΔt) − d) ÷ (u − d). Check that 0 < p < 1.
4Build the stock prices at maturity and compute the option payoff at each end node.
5Roll back one step at a time with f = e^(−rΔt) × [p f_u + (1 − p) f_d]. For American options, replace f by the larger of f and the exercise value.
6Compute delta at the node asked for: (f_u − f_d) ÷ (S_u − S_d), using the two nodes that follow it.
7For a hedge, buy Δ shares per short call. Hedge a short put by shorting |Δ| shares, where Δ is the negative put delta. Scale by the number of options. For convergence questions, state that the price tends to the Black-Scholes value and delta tends to N(d1), with oscillating error.

Quickest way: Delta from the payoffs only

When to use it: When the question asks only for delta, not for the option price.

Find the two option values (or payoffs) at the next step.
Find the two stock prices at the next step.
Divide the difference in option values by the difference in stock prices.
Do not discount and do not use p. Delta does not depend on probabilities or on r.
Sanity check: a call delta must be in [0, 1] and a put delta in [−1, 0].

Common mistakes in Delta and Binomial Tree Convergence to Black-Scholes

Dividing by S0 or by u − d instead of S_u − S_d.
Students mix up the stock-price difference with the factor difference.
Fix: Delta is a ratio of two dollar differences: option values over stock prices at the same step.
Using the initial delta for the whole tree.
Delta at time 0 is the one most often asked, so it feels like the only one.
Fix: Delta changes at every node. Recompute it at the node you are asked about and say the hedge needs rebalancing.
Discounting the payoffs before computing delta.
Mixing up the pricing formula with the delta formula.
Fix: Delta uses undiscounted values at the same date. Discounting applies only to the option price.
Getting the hedge direction wrong.
Delta is positive for a call, which is read as 'sell the stock'.
Fix: A short call is hedged by buying Δ shares. A short put is hedged by shorting |Δ| shares, where Δ is the negative put delta.
Saying the binomial price rises steadily toward Black-Scholes.
Students assume convergence is one-directional.
Fix: The tree price oscillates around the Black-Scholes value for European options, and the error shrinks as steps increase.
Using the wrong Δt in u and p.
Students plug in the full maturity T for a multi-step tree.
Fix: Δt = T ÷ n. Use it in u = e^(σ√Δt) and in e^(rΔt).

Worked examples

Example 1

A stock is at $50. Over one year it moves to $60 or $40. The continuously compounded risk-free rate is 5% a year. A one-year European call has strike $50. Find the delta, the risk-neutral probability and the call price in a one-step tree. How many shares hedge a short position of 1,000 calls?

Show the solution

Payoffs: f_u = 60 − 50 = 10 and f_d = 0.
Delta = (10 − 0) ÷ (60 − 40) = 0.5.
u = 60 ÷ 50 = 1.2 and d = 40 ÷ 50 = 0.8. e^0.05 = 1.05127.
p = (1.05127 − 0.8) ÷ (1.2 − 0.8) = 0.25127 ÷ 0.4 = 0.62818.
Price = e^(−0.05) × 0.62818 × 10 = 0.95123 × 6.2818 = 5.975.
Check by replication: 0.5 × 50 = 25. Borrowing = 0.5 × 40 × 0.95123 = 19.025. Price = 25 − 19.025 = 5.975.
Hedge for 1,000 short calls: 1,000 × 0.5 = 500 shares.

Answer: Delta = 0.5, p ≈ 0.628, call price ≈ $5.98. Buy 500 shares to hedge 1,000 short calls.

Example 2

A stock is at 100. In each of two steps it rises by a factor 1.1 or falls by a factor 0.9 (d = 0.9). These factors are given directly in the question. They are not CRR parameters, because u and d are not reciprocals. The risk-free rate is zero. A European call has strike 100. Find the delta at time 0 and at the up node after step one, and the call price.

Show the solution

The factors u = 1.1 and d = 0.9 are given, not derived from σ, so do not apply the CRR formulas here.
With r = 0: p = (1 − 0.9) ÷ (1.1 − 0.9) = 0.5.
Stock prices: step one 110 and 90. Step two: 121, 99 and 81.
Payoffs at maturity: 21, 0 and 0.
Up node (110): f_u = 0.5 × 21 + 0.5 × 0 = 10.5. Down node (90): f_d = 0.
Price at time 0 = 0.5 × 10.5 + 0.5 × 0 = 5.25.
Delta at time 0 = (10.5 − 0) ÷ (110 − 90) = 0.525.
Delta at the up node = (21 − 0) ÷ (121 − 99) = 21 ÷ 22 = 0.9545.
Delta at the down node is 0, because both payoffs are 0.

Answer: Call price = 5.25. Delta at time 0 = 0.525. Delta at the up node ≈ 0.955. The hedge must be rebalanced after the first step.

Exam tips

Questions often ask for delta only. Skip discounting and go straight to the payoff and price differences.
If the stock goes up, a call delta rises toward 1. Use this to sense-check a multi-step answer.
For convergence questions, remember three facts: CRR u = e^(σ√Δt), error oscillates, and delta tends to N(d1).
Check the sign of the hedge in the question: shorting a call needs a long stock position.
Keep four decimals for p and e^(rΔt). Rounding early can move you to a neighbouring option.

Practice questions from Binomial Trees