The Black–Scholes formula and binomial convergence

Price a call on a binomial tree with more and more steps. Where do the prices go, and why?

They go to the Black–Scholes value. Under risk-neutral pricing the call is worth its discounted expected payoff. As the steps shrink, the log of the tree's final stock price becomes normal by the central limit theorem, and the discounted expected payoff of a lognormal stock price is the Black–Scholes formula.

Before you start · 3 quick checks on risk-neutral probabilities, the central limit theorem and the lognormal mean, about 3 min

Setup: the market, risk-neutral pricing and the tree

Market. A stock with price \(S_0 \gt 0\) pays no dividends. The interest rate \(r\) is constant (any real number) and continuously compounded, and the volatility \(\sigma \gt 0\) is constant. Trading is continuous and frictionless: no costs or taxes, any amount can be bought, sold short, borrowed or lent at \(r\), and the option trades at a single price. The market has no arbitrage. The option is a European call with strike \(K \gt 0\) and maturity \(T \gt 0\); it pays \((S_T - K)^+\) at \(T\) and nothing before.
Risk-neutral pricing (step 17). Under the risk-neutral measure \(\mathbb Q\), the discounted stock price \(e^{-rt}S_t\) is a martingale. The stock follows a geometric Brownian motion with volatility \(\sigma\) (step 14), and the martingale property fixes its drift under \(\mathbb Q\) at \(r\), so

\[ S_T = S_0\,e^{(r - \frac12\sigma^2)T + \sigma\sqrt T\,Z}, \]

where \(Z\) is standard normal under \(\mathbb Q\).

A European payoff \(X\) paid at \(T\) that is determined by the stock prices up to \(T\), with \(\mathbb E^{\mathbb Q}\lvert X\rvert \lt \infty\), can be replicated in this model, and its no-arbitrage price at time 0 is \(e^{-rT}\,\mathbb E^{\mathbb Q}[X]\), the risk-neutral pricing formula of step 17. The replication argument behind this price, delta hedging and the Black–Scholes PDE, is on step 16. Write \(N\) and \(\varphi\) for the standard normal CDF and density, and

\[ \begin{aligned} d_2 &= \frac{\ln(S_0/K) + (r - \frac12\sigma^2)T}{\sigma\sqrt T}, \\ d_1 &= d_2 + \sigma\sqrt T. \end{aligned} \]

The Black–Scholes value of the call is \(C = S_0N(d_1) - Ke^{-rT}N(d_2)\); stage 3 proves that it equals \(e^{-rT}\,\mathbb E^{\mathbb Q}[(S_T - K)^+]\).

The tree (Cox, Ross and Rubinstein). Split \([0, T]\) into \(n\) steps of length \(\Delta t = T/n\). Each step multiplies the stock by \(u = e^{\sigma\sqrt{\Delta t}}\) or by \(d = e^{-\sigma\sqrt{\Delta t}}\), and the bond grows by \(e^{r\Delta t}\). The tree's risk-neutral probability of an up move is \(q_n = (e^{r\Delta t} - d)/(u - d)\), which lies strictly between 0 and 1 exactly when \(d \lt e^{r\Delta t} \lt u\), that is when \(n \gt r^2T/\sigma^2\). Every example on this page meets this for every \(n \ge 1\).
Tree price. Backward induction (step 6, with \(e^{r\Delta t}\) in place of \(1 + r\)) gives the call the price

\[ \begin{aligned} C_n = e^{-rT}\sum_{k=0}^{n}\binom{n}{k}&q_n^k(1 - q_n)^{n-k} \\ &\times\big(S_0u^kd^{n-k} - K\big)^+. \end{aligned} \]

Fine print: the two models and the random numbers

The tree and the continuous model are two different models of the same stock. The tree is used here as an approximation: its step sizes are chosen so that over each step the log price has variance close to \(\sigma^2\Delta t\), and its probabilities \(q_n\) make the discounted stock price a martingale on the tree, as \(\mathbb Q\) does in the continuous model.

The Monte Carlo estimates on this page draw \(Z\) from a seeded random number generator and use \(S_T\) from the formula above directly, so they need no time steps. An estimate from \(M\) draws is the average of the \(M\) discounted payoffs; its standard error is the sample standard deviation divided by \(\sqrt M\). The approximate 95% interval is the estimate plus or minus 1.96 standard errors.

1 See

Tree, simulation and formula side by side

Take \(S_0 = 100\), \(K = 100\), \(T = 1\) year, \(r = 5\%\) and \(\sigma = 20\%\). The page can price the call on a tree with any number of steps \(n\) up to 500. Make your prediction first; the explorer opens after it.

2 Try

How fast each method closes the gap

Monte Carlo: how many more draws?

Goals on the tree

The same stock as in stage 1. Move \(n\) and \(K\) until the goal turns green. The tiles show the tree prices at \(n\) and at \(n + 1\), their average and the Black–Scholes value.

3 Prove

The formula, then the convergence

The idea in three steps

Split the payoff at the strike: the call pays \(S_T\) and owes \(K\) exactly on the event \(S_T \gt K\), which is an event about \(Z\) alone.
The strike part is a normal probability. The stock part becomes one after completing the square, which shifts the normal density by \(\sigma\sqrt T\).
On the tree, the log price is a sum of many small independent steps, so it becomes normal. The put payoff is bounded, so its tree price converges, and put–call parity carries this to the call.

Theorem A (the Black–Scholes formula)

Assumptions. The setup: \(S_0, K, T, \sigma \gt 0\), \(r\) any real number, and \(S_T = S_0e^{(r - \frac12\sigma^2)T + \sigma\sqrt T Z}\) with \(Z \sim N(0, 1)\) under \(\mathbb Q\).

Conclusion. \(C_0 = e^{-rT}\,\mathbb E^{\mathbb Q}\big[(S_T - K)^+\big] = S_0N(d_1) - Ke^{-rT}N(d_2)\).

Each line of the proof is one claim about every admissible \(S_0, K, T, r, \sigma\) and every outcome of \(Z\). Line 4 is an identity of functions, stated for every real \(a\) and \(z\).

Scratch work, not part of the proof. For line 4, \(az - \tfrac12 a^2 - \tfrac12 z^2 = -\tfrac12\big(z^2 - 2az + a^2\big) = -\tfrac12(z - a)^2\), so \(e^{az - \frac12a^2}\,\tfrac{1}{\sqrt{2\pi}}e^{-z^2/2} = \tfrac{1}{\sqrt{2\pi}}e^{-(z-a)^2/2}\).

The same steps with the event \(Z \lt -d_2\) give the put, \(P_0 = e^{-rT}\,\mathbb E^{\mathbb Q}\big[(K - S_T)^+\big] = Ke^{-rT}N(-d_2) - S_0N(-d_1)\), and so \(C_0 - P_0 = S_0 - Ke^{-rT}\).

Theorem B (the tree price converges)

Assumptions. The setup, with \(n \gt r^2T/\sigma^2\) so that \(0 \lt q_n \lt 1\). Under the tree's probabilities, let \(J_n\) be the number of up moves, so \(J_n\) is binomial with parameters \(n\) and \(q_n\), and let \(X_n = \ln(S_n/S_0) = \sigma\sqrt{\Delta t}\,(2J_n - n)\). It is the sum of the steps \(\xi_1, \dots, \xi_n\), where \(\xi_i\) is \(\sigma\sqrt{\Delta t}\) if step \(i\) is up and \(-\sigma\sqrt{\Delta t}\) if it is down; the steps are independent. Write \(C_n, P_n\) for the tree prices of the call and the put with strike \(K\), and \(C, P\) for \(C_0, P_0\) of Theorem A.

Facts used, stated without proof. (i) Central limit theorem for small steps (Lindeberg–Feller): if each \(X_n\) is a sum of \(n\) independent steps, each at most \(c_n\) in size with \(c_n \to 0\), and \(\mathbb E[X_n] \to m\), \(\operatorname{Var}(X_n) \to v \gt 0\), then \(X_n\) converges in distribution to \(N(m, v)\). (ii) Convergence in distribution \(X_n \to X\) means \(\mathbb E[g(X_n)] \to \mathbb E[g(X)]\) for every bounded continuous function \(g\).

Conclusion. \(P_n \to P\) and \(C_n \to C\) as \(n \to \infty\).

Theorem B says nothing about how fast the tree price converges. For a call or put the error is of order \(1/n\), with a coefficient that depends on where the strike sits between the tree's nodes; that finer result is in Diener and Diener (2004), listed below.

Key point The Black–Scholes formula is the discounted risk-neutral expectation of the call payoff, worked out for a lognormal stock price. The tree price converges to it because the tree's log price becomes normal, a bounded payoff then carries the convergence to the put, and exact put–call parity on both sides carries it to the call.

4 Apply

Problems

All prices are hypothetical, and every stock pays no dividends.