The One-Step Binomial Model

Everything you know about pricing so far — the no-arbitrage bounds on an option's price and the risk-neutral pricing machinery itself — was built in continuous time, with stochastic calculus doing the heavy lifting. That machinery is powerful, but it is also a black box: type in S_0, K, r, \sigma, T and a closed-form number falls out, with the reasoning buried inside a PDE or a change of measure. A trading desk that only ever plugs numbers into Black–Scholes has no lever for the cases the formula can't reach — American exercise, a payoff that depends on the whole path, a basket of ten correlated names. This module builds those levers. We start with the simplest possible market: one asset, one time step, two possible outcomes — and rebuild risk-neutral pricing from the ground up, by hand, so that every later generalisation (many steps, early exercise, simulation, grids) is just "do this again, more times."

Building a market with exactly two futures

Fix one period — a day, a month, whatever the model's clock ticks in. A stock worth S_0 today will be worth exactly one of two things at the end of the period: S_0 u ("up") or S_0 d ("down"), with d < R < u where R = 1+r is one dollar's growth in the risk-free bank account over the period (the inequality rules out arbitrage: if u \le R the stock never beats cash, and if R \le d it always beats cash — either way one asset dominates the other for free). A derivative on this stock — a call, a put, anything — pays C_u in the up state and C_d in the down state, both known once we fix the payoff rule. The question: what is the derivative worth today?

Follow the logic of replication: build a portfolio of \Delta shares of stock and B dollars in the bank that reproduces the payoff in both states at once.

\begin{aligned} \Delta\,S_0 u + R\,B &= C_u \\ \Delta\,S_0 d + R\,B &= C_d \end{aligned}

Two equations, two unknowns. Subtract them and the bank position cancels, leaving \Delta directly:

\Delta = \frac{C_u - C_d}{S_0 u - S_0 d}, \qquad B = \frac{C_u - \Delta\,S_0 u}{R}.

By no-arbitrage, a portfolio that matches a payoff exactly in every future state must cost the same as that payoff today — two ways to own the identical thing can't have different prices. So the option's value is simply the replicating portfolio's cost today:

V_0 = \Delta\,S_0 + B.

Notice what never appeared anywhere in this derivation: the probability of the up-move. Hedging pins the price with nothing but algebra.

The same price, read off as an expectation

Substitute the formulas for \Delta and B into V_0 = \Delta S_0 + B and simplify — it's routine algebra, done once in the risk-neutral pricing page — and the price rearranges into a discounted expectation:

V_0 = \frac{q\,C_u + (1-q)\,C_d}{R}, \qquad q = \frac{R - d}{u - d}.

Check that q \in (0,1) whenever d < R < u — it behaves exactly like a probability, even though nobody asserted it was one. Call q the risk-neutral probability: the (fictional) chance of the up-move that would make the stock's expected return equal the risk-free rate, S_0 = (q\,S_0 u + (1-q)\,S_0 d)/R. Solve that one equation and you get exactly the q above.

Worked example

A stock trades at S_0 = 50. Over one period it will move up by a factor u = 1.2 to 60, or down by d = 0.8 to 40. One dollar in the bank grows to R = 1.10. Price a European call struck at K = 55.

The payoffs are C_u = \max(60-55,0) = 5 and C_d = \max(40-55,0) = 0. Replicating:

\Delta = \frac{5-0}{60-40} = 0.25, \qquad B = \frac{5 - 0.25\cdot 60}{1.10} = -9.09, V_0 = 0.25\cdot 50 - 9.09 = 3.41.

Check it the second way. The risk-neutral probability is q = (1.10 - 0.8)/(1.2-0.8) = 0.75, so

V_0 = \frac{0.75\cdot 5 + 0.25\cdot 0}{1.10} = \frac{3.75}{1.10} = 3.41.

Same number, from two different-looking calculations — because they are the same calculation. Step through the figure to watch the replication and the risk-neutral roads meet at the price.

Because hedging doesn't erase the up/down uncertainty — it just moves who bears it. The replicating portfolio still ends up worth C_u in the up state and C_d in the down state; someone (the writer, or the hedger's counterparty) is still exposed to which one happens. What no-arbitrage guarantees is only that today's price doesn't depend on whose estimate of the up-probability you use. Two traders who violently disagree about the real chance of the stock rising — one says 10%, the other says 90% — will still agree, to the penny, on V_0 = 3.41, because both can build the identical hedge and neither wants to sell it for less (or pay more) than it costs to build. The disagreement about p shows up in what each trader thinks their own expected profit is after hedging — never in the price they're willing to trade at.