A two-state random walk for the share price, no-arbitrage hedging at every node, and the risk-neutral pricing recipe that becomes Black–Scholes in the limit.
Deck 02 priced contracts that had no probability in them at all (forwards, parity). Once we want the price of a call, we have to commit to some statement about how the underlying might move. The binomial model is the simplest statement that gets us all the way to a hedged price.
"The binomial method is mathematically simple, yet it contains all the important ideas of derivative pricing."
— Wilmott, Ch. 3Suppose at time $0$ the share has price $S$. Over a single time step $\delta t$, the share price can only go to one of two values:
$$S \;\longrightarrow\; \begin{cases} uS & \text{with real probability } p, \\ dS & \text{with real probability } 1-p, \end{cases}$$
with $d < 1 + r\delta t < u$ (the no-arbitrage condition: the bank-account return must lie between the worst-case and best-case stock return, or you could earn a riskless profit by going long or short the stock).
Today: $S$. Tomorrow: $uS$ (up) or $dS$ (down). The probabilities $p, 1-p$ are real-world; we'll see in a moment that they don't matter.
Some function $V(S,t)$ we want to price. At the end of the step it has known values $V^+$ (up state) and $V^-$ (down state) — for a call at expiry, $V^\pm = \max(uS-K,0)$ or $\max(dS-K,0)$.
Two states per step looks absurdly crude; the trick is that with enough steps the multi-step tree converges to the same Black–Scholes price as a continuous random walk. The two-state structure is exactly what lets us hedge perfectly.
Form a portfolio: long one derivative, short $\Delta$ shares.
$$\Pi = V - \Delta S.$$
Over the step the portfolio becomes:
| state | derivative | stock | portfolio |
|---|---|---|---|
| up | $V^+$ | $uS$ | $V^+ - \Delta\, uS$ |
| down | $V^-$ | $dS$ | $V^- - \Delta\, dS$ |
Trick: choose $\Delta$ so that the portfolio is worth the same amount in both states. That kills all the risk.
$$V^+ - \Delta\, uS = V^- - \Delta\, dS \quad\Longrightarrow\quad \boxed{\;\Delta = \dfrac{V^+ - V^-}{uS - dS}.\;}$$
This is the delta-hedge ratio. The number of shares to short per option is the slope of the option payoff with respect to the stock, evaluated discretely across the two branches.
By construction $\Pi$ takes the same value at the end of $\delta t$ no matter which way the stock moves. A riskless portfolio must earn the risk-free rate, or arbitrage exists.
Two states, two unknowns (the option price now, and one share count). With exactly enough degrees of freedom to make the portfolio deterministic, you get a unique price — with no reference to $p$.
Pricing a derivative is not about predicting where the stock will go. It is about hedging — building a portfolio whose return is locked in. The price is whatever makes that locked-in return equal to $r$.
The hedged portfolio $\Pi$ is riskless, so over one time step it earns the risk-free rate:
$$\Pi(0) \cdot (1 + r\delta t) = \Pi(\delta t).$$
Substituting $\Pi = V - \Delta S$ and the up-state value:
$$(V - \Delta S)(1 + r\delta t) = V^+ - \Delta\, uS.$$
Rearranging for $V$ and substituting $\Delta = (V^+ - V^-)/(uS - dS)$ produces the central formula of the chapter:
$$\boxed{\;V = \dfrac{1}{1 + r\delta t}\Bigl(p'\, V^+ + (1-p')\, V^-\Bigr),\quad p' = \dfrac{(1 + r\delta t) - d}{u - d}.\;}$$
The number $p'$ is called the risk-neutral probability. It plays the role of a probability (it lies in $(0,1)$ exactly when the no-arbitrage condition $d < 1 + r\delta t < u$ holds), but it is not the probability anyone believes in. It's the unique number that makes the present value of the stock equal to its current price under discounting at $r$:
$$S = \frac{1}{1+r\delta t}\bigl(p'\, uS + (1-p')\, dS\bigr).$$
Price today = (risk-neutral expectation of price tomorrow) discounted at $r$. The whole rest of the book is one variant or another of this single formula.
$p$ — the actual probability that the stock goes up — never appeared in the price. This is the most surprising thing in the chapter.
Pricing the option = take the expected payoff under the real-world probability $p$, discount it back. That would give
$$V \stackrel{?}{=} \frac{p V^+ + (1-p) V^-}{1+r\delta t}.$$
This is wrong — it's the price two risk-neutral parties would agree on if neither could hedge.
The correct price replaces $p$ with $p'$:
$$V = \frac{p' V^+ + (1-p') V^-}{1+r\delta t}.$$
Same shape, different probability. The hedger doesn't care what $p$ is — their portfolio is risk-free regardless.
$p'$ is the probability under which the stock has expected return $r$ (the risk-free rate) rather than its actual expected return $\mu$. The hedging argument has the effect of changing measure from the real-world probability to one in which everyone is risk-neutral. This is the discrete-time prototype of Girsanov's theorem (Deck 05).
"Construct a hedged portfolio and demand it earns $r$" and "take the risk-neutral expectation and discount at $r$" are the same statement said in different words. Wilmott prefers the hedging view because the calibration to the market is concrete.
Iterate the one-step model over $N$ periods of length $\delta t = T/N$. Insist that an "up then down" gives the same price as "down then up" — i.e. $ud$ in either order. With constant $u, d$ this is automatic and the tree is recombining: after $n$ steps there are only $n+1$ distinct nodes, not $2^n$.
After $n$ steps the share price at node $(n, j)$ — with $j$ counting up-moves, $0 \le j \le n$ — is
$$S_n^{(j)} = S_0\, u^{j} d^{\,n-j}.$$
At maturity ($n=N$) the option's payoff is known:
$$V_N^{(j)} = \mathrm{payoff}(S_0\, u^{j} d^{N-j}).$$
For a European call this is $\max(S_N^{(j)} - K, 0)$; for a put, $\max(K - S_N^{(j)}, 0)$. From there we walk backwards.
A non-recombining tree has $2^N$ leaves — pricing 30 steps deep is already a billion-node calculation. A recombining tree has $N+1$ leaves and $O(N^2)$ total nodes. The whole "lattice methods" industry depends on this.
At every internal node the one-step formula applies:
$$V_n^{(j)} = \frac{1}{1 + r\delta t}\Bigl(p'\, V_{n+1}^{(j+1)} + (1-p')\, V_{n+1}^{(j)}\Bigr).$$
Sweep from $n = N$ down to $n = 0$. At the root you have $V_0$, the price of the option.
Let $S_0 = 100$, $K = 100$, $r = 0.05$, $\sigma = 0.20$, $T = 1$, $N = 3$. Choose CRR: $u = e^{\sigma\sqrt{\delta t}}$, $d = 1/u$.
$$\delta t = 1/3 \approx 0.333,\quad u \approx 1.1224,\quad d \approx 0.8909,\quad p' = \frac{e^{r\delta t} - d}{u-d} \approx 0.5398.$$
| $j$ | $S_3^{(j)}$ | $V_3^{(j)} = (S-K)^+$ |
|---|---|---|
| 3 | 141.4 | 41.4 |
| 2 | 112.2 | 12.2 |
| 1 | 89.1 | 0 |
| 0 | 70.7 | 0 |
Applying the recursion three times gives $V_0 \approx 10.59$. The closed-form Black–Scholes price for the same parameters is $\approx 10.45$ — the binomial price converges to it as $N \to \infty$.
The binomial price oscillates above and below the BS price as $N$ grows, with error $O(1/N)$ in the standard CRR scheme. Refinements (Tian, Leisen–Reimer) smooth this oscillation.
The Greeks are sensitivities of $V$ to the inputs. The tree gives them for free as finite differences across adjacent nodes.
$$\Delta = \frac{\partial V}{\partial S} \approx \frac{V_1^{(1)} - V_1^{(0)}}{uS_0 - dS_0}.$$
The hedge ratio at the root — the slope between the two nodes one step out.
$$\Gamma \approx \frac{\Delta_+ - \Delta_-}{(u-d)S_0/2 \cdot (\text{average})}.$$
Use the two deltas at the up-node and down-node of step 1; differencing them gives gamma at the root. Best taken from step 2 nodes against step 0.
$$\Theta \approx \frac{V_2^{(1)} - V_0}{2\delta t}.$$
The middle node at step 2 is at the same stock price as the root (because $ud=1$ in CRR), so the difference is a clean time-only effect.
Vega ($\partial V/\partial \sigma$) and rho ($\partial V/\partial r$) are computed by rebuilding the tree with bumped inputs and finite-differencing the resulting prices. Slightly more expensive, but still cheap.
Delta and gamma fall out of the same backward sweep that produced the price — no extra trees. That is the main reason traders prefer lattices to Monte Carlo for vanilla products: the Greeks are essentially free.
An American option can be exercised at any node in the tree. At each node compare:
$$V_n^{(j)} = \max\Bigl(\,\underbrace{\mathrm{payoff}(S_n^{(j)})}_{\text{exercise now}}\,,\;\underbrace{\tfrac{1}{1+r\delta t}\bigl(p' V^+ + (1-p') V^-\bigr)}_{\text{continuation}}\,\Bigr).$$
This is dynamic programming — the celebrated Bellman form. Backward induction with a $\max$ at every node.
Never optimal to exercise early: continuation value dominates intrinsic. American call price = European call price for non-dividend stocks.
Can be optimal to exercise early — intrinsic value can exceed continuation value when the stock is deep ITM, because the holder is forgoing further downside but receiving the interest on $K$ now.
The set of $(S, t)$ pairs at which exercise becomes optimal traces out a smooth curve, the early-exercise boundary $S^*(t)$. Below it (for a put), exercise; above it, hold. The boundary is unknown in advance — the lattice finds it implicitly, the PDE picture (later decks) treats it as a free boundary.
Once the underlying pays a non-zero "yield" (dividends, foreign interest rate, futures roll), American calls can rationally be exercised early. The binomial $\max$-step is the simplest way to capture that without solving a free-boundary PDE.
Cox, Ross and Rubinstein (1979) made the standard choice:
$$u = e^{\sigma\sqrt{\delta t}},\qquad d = e^{-\sigma\sqrt{\delta t}} = 1/u,\qquad p' = \frac{e^{r\delta t} - d}{u - d}.$$
One step of this tree, in log space, is a centred jump of $\pm \sigma\sqrt{\delta t}$. Over $N$ steps the log-return is a sum of $N$ such jumps; by the central limit theorem,
$$\log(S_T/S_0) \to \mathcal{N}\!\bigl((r - \tfrac12 \sigma^2) T,\; \sigma^2 T\bigr).$$
This is exactly the distribution of $\log S_T$ under geometric Brownian motion with drift $r$ and volatility $\sigma$. So the binomial-tree price converges to the Black–Scholes price.
Variance is additive in time, standard deviation is the square root. To recover continuous-time vol $\sigma$ in a single step, the jump size must scale like $\sigma\sqrt{\delta t}$ — not $\sigma\delta t$.
Symmetry in log-space. Recombining is automatic; up-then-down brings you back to the start; the tree is centred on $S_0$ at every even step.
The next deck introduces the continuous-time random walk — geometric Brownian motion — whose discrete avatar you have just been pricing on a tree. Same model, different language.
Slide the inputs. The tree redraws live: node radius scales with share price, node colour with option value (yellow = high, blue = low). The metric grid shows the CRR parameters, the option price at the root, the initial delta, and a Black–Scholes reference price.
Watch $p'$ shift as you change $r$ and $\sigma$. Increase $N$ and notice the price oscillating around the BS value. Switch to an American put and (with $r > 0$) see the tree price exceed the European put.