Binomial Option Pricing — AI Encyclopedia

2.1

No-arbitrage & the law of one price

Everything in option pricing rests on a single principle, and it is almost embarrassingly mundane: two portfolios that pay the same thing in every future state must cost the same today. If they did not, you would buy the cheap one, sell the dear one, pocket the difference, and walk away holding offsetting positions that cancel in every scenario — a riskless profit from nothing. Markets are not perfectly efficient, but they are efficient enough that such free lunches are arbitraged away in seconds. This is the law of one price, and it is the only economic assumption the binomial model needs.

The leap is to notice that an option is such a portfolio in disguise. Over a short enough interval the stock can do only a limited number of things, and we can assemble a position in the stock and a risk-free bond that reproduces the option's payoff exactly. If we can build the option's payoff out of instruments whose prices we already know, then the law of one price hands us the option's price: it must equal the cost of that replicating portfolio. We never forecast the market; we manufacture the option and read off the bill of materials.

KEY

Pricing is replication, not prediction. A binomial model does not ask "will the stock rise?" It asks "what mix of stock and cash exactly mimics this option?" — and then charges what that mix costs. Two traders who violently disagree about the stock's direction must still quote the same option price, because both can build it the same way at the same cost. This is the seed of the entire field; §2.3 makes it arithmetic.

One more piece of furniture. Throughout we discount with a risk-free rate. Over a step of length $\Delta t$ a dollar in the bank grows by the gross factor $e^{r\Delta t}$ (continuous compounding) or $1+r\Delta t$ (simple); a payoff received at the end of the step is worth $e^{-r\Delta t}$ times its face value today. To rule out arbitrage in the stock itself, the up and down moves must straddle the bond: $d < e^{r\Delta t} < u$. If the stock could only ever beat the bond ($d > e^{r\Delta t}$) you would borrow infinitely to buy it; if it could only ever lag ($u < e^{r\Delta t}$) you would short it without limit. The no-arbitrage band is exactly what makes a finite price possible.

2.2

Risk-neutral valuation

The replication argument has a second face that is often easier to compute with. Suppose we insist on writing the option's value today as a discounted expectation of its payoff. The real-world probabilities of up and down moves — driven by the stock's true expected return — are not the right weights; if they were, two traders with different forecasts would price the option differently, contradicting §2.1. Instead there is a single, fictitious set of probabilities under which the discounted price of every traded asset is a fair game (a martingale). These are the risk-neutral probabilities, and under them the stock's expected growth is exactly the risk-free rate.

EQ Q2.1 — RISK-NEUTRAL VALUATION $$ V_0 \;=\; e^{-r\Delta t}\,\mathbb{E}^{\mathbb{Q}}\!\big[\,V_{\Delta t}\,\big] \;=\; e^{-r\Delta t}\big[\,p\,V_u + (1-p)\,V_d\,\big], \qquad \text{where } \; \mathbb{E}^{\mathbb{Q}}[S_{\Delta t}] = e^{r\Delta t} S_0 $$

$p$ is the risk-neutral probability of an up move, $V_u$ and $V_d$ the option's values after up and down. The price is just the discounted, probability-weighted average payoff — but weighted by $\mathbb{Q}$, not by anyone's beliefs. The defining constraint is on the last term: under $\mathbb{Q}$ the stock drifts at the riskless rate, so the expected return $\mu$ of the real world has been engineered out of the problem. Black–Scholes (Quant 03) is the continuous-time limit of exactly this expectation.

This is the most counter-intuitive fact in the chapter, so it is worth stating plainly: the drift vanishes. A stock everyone expects to soar and a stock everyone expects to crash will produce the same option price if they have the same volatility and the same up/down factors, because both can be replicated for the same cost. Risk-neutral valuation is not a claim that investors are indifferent to risk — they are not — but a bookkeeping trick: by discounting at the riskless rate and re-weighting with $\mathbb{Q}$, we absorb the market's risk preferences into the probabilities so they never have to be estimated. The two views — replicate-and-cost (§2.1) and discount-the-Q-expectation (here) — are provably identical; §2.3 derives $p$ from the no-arbitrage condition and they fall out the same.

A caveat experts insist on: a unique $\mathbb{Q}$ exists only when the market is complete — when every payoff can in fact be replicated. The one-step binomial market is complete precisely because two outcomes can be hedged with two instruments (stock and bond). Add a third outcome per step without a third instrument and replication fails, $\mathbb{Q}$ is no longer unique, and prices live in a no-arbitrage interval rather than at a point. The binomial model's elegance is that it stays exactly on the knife-edge of completeness.

2.3

The one-step binomial model

Now the arithmetic. Over one step the stock $S_0$ moves to either $S_0 u$ (up) or $S_0 d$ (down), with $d < u$. An option on it is worth $V_u$ or $V_d$ at the end of the step. Build a portfolio of $\Delta$ shares and $B$ dollars of bond, and choose $\Delta$ and $B$ so it matches the option in both states:

EQ Q2.2 — THE HEDGE RATIO (DELTA) $$ \Delta\,S_0 u + B\,e^{r\Delta t} = V_u, \qquad \Delta\,S_0 d + B\,e^{r\Delta t} = V_d \;\;\Longrightarrow\;\; \Delta = \frac{V_u - V_d}{S_0(u - d)} $$

Subtract the two equations and the bond term cancels: $\Delta$ is the spread of option values divided by the spread of stock values — how many shares hedge the option's exposure over this step. This is the discrete ancestor of the Black–Scholes delta. Because the portfolio reproduces the option in every state, no-arbitrage forces the option's price to equal the portfolio's cost today, $\Delta S_0 + B$.

Substitute $\Delta$ and $B$ back and simplify. The expected-return term collapses and what survives is a clean weighted average — exactly EQ Q2.1, with the weight $p$ determined entirely by the no-arbitrage band:

EQ Q2.3 — RISK-NEUTRAL PROBABILITY & PRICE $$ p = \frac{e^{r\Delta t} - d}{u - d}, \qquad V_0 = e^{-r\Delta t}\big[\,p\,V_u + (1-p)\,V_d\,\big] $$

$p$ is the only weight that makes the stock itself fairly priced under EQ Q2.1: plug $V_u = S_0u,\ V_d = S_0d$ and you recover $\mathbb{E}^{\mathbb{Q}}[S] = e^{r\Delta t}S_0$. The condition $d < e^{r\Delta t} < u$ is exactly what keeps $p \in (0,1)$, a genuine probability. Notice what is absent: no $\mu$, no risk premium, no view on direction. The real-world odds of up versus down never appear.

One step with $ u = 1.1 $, $ d = 0.9 $, and $ r = 0 $ (so $ e^{r\Delta t} = 1 $). What is the risk-neutral probability of an up move, $ p = \dfrac{e^{r\Delta t} - d}{u - d} $?

$ p = \dfrac{1 - 0.9}{1.1 - 0.9} = \dfrac{0.1}{0.2} = $ 0.5. With zero rate and symmetric moves the risk-neutral measure is a fair coin — but note it is set by $u$ and $d$, never by the real-world odds.

For that same call ($ S_0 = 100,\ u = 1.1,\ d = 0.9,\ K = 100 $), the up payoff is $ V_u = 10 $ and the down payoff is $ V_d = 0 $. How many shares does the replicating hedge hold, $ \Delta = \dfrac{V_u - V_d}{S_0(u-d)} $?

$ \Delta = \dfrac{10 - 0}{100\,(1.1 - 0.9)} = \dfrac{10}{20} = $ 0.5. Half a share, financed by borrowing, exactly replicates the call — and replication is the price.

INSTRUMENT Q2.1 — RISK-NEUTRAL PROBABILITY EXPLORERONE STEP · EQ Q2.3 · LIVE

UP FACTOR u 1.10

DOWN FACTOR d 0.90

RATE r/step 0.00

RISK-NEUTRAL WEIGHT p  ·  NO-ARBITRAGE BAND d < e^{rΔt} < u

band OK — p is a valid probability

RISK-NEUTRAL p

—

GROSS RATE e^{rΔt}

—

𝔼ᵠ[S]/S₀ (= gross)

—

Defaults reproduce the worked example: $u=1.1,\ d=0.9,\ r=0 \Rightarrow p = 0.5$. The third readout confirms the defining identity — $p u + (1-p)d$ always equals the gross rate $e^{r\Delta t}$, which is what "risk-neutral" means. Push $d$ above the gross rate and the bar turns red: the band breaks, $p$ leaves $[0,1]$, and the market admits arbitrage. Raise $r$ toward $u$ and watch $p$ climb toward 1.

2.4

Multi-step CRR trees

One step is a cartoon of the market; many steps approximate it. Cox, Ross and Rubinstein's 1979 insight was to chain the one-step rule into a recombining lattice — "up then down" lands on the same node as "down then up", so after $N$ steps there are only $N+1$ terminal nodes instead of $2^N$. The tree is built forward; the option is priced backward, applying EQ Q2.3 at every node from the leaves to the root.

EQ Q2.4 — CRR PARAMETERS & BACKWARD INDUCTION $$ u = e^{\sigma\sqrt{\Delta t}}, \quad d = \frac{1}{u} = e^{-\sigma\sqrt{\Delta t}}, \quad p = \frac{e^{r\Delta t} - d}{u - d}, \qquad V_i^{(n)} = e^{-r\Delta t}\big[\,p\,V_{i+1}^{(n+1)} + (1-p)\,V_{i}^{(n+1)}\,\big] $$

The CRR choice $ud = 1$ makes the lattice recombine and centres it on $S_0$; the $\sqrt{\Delta t}$ scaling matches the variance of log-returns to $\sigma^2\Delta t$ per step. Start with terminal payoffs $V_i^{(N)} = \text{payoff}(S_0 u^i d^{\,N-i})$ and sweep the recursion back to $V_0^{(0)}$. The whole pricer is one loop over the lattice — a dozen lines of NumPy (below).

As the step count $N$ grows, the binomial price marches toward the Black–Scholes value. This is not a coincidence: with the CRR parametrization the multiplicative random walk converges (by the central limit theorem applied to log-returns) to geometric Brownian motion, and the discounted-expectation recursion converges to the Black–Scholes integral. The convergence is famously oscillatory — even and odd $N$ approach from opposite sides because the strike sits differently relative to the terminal grid — so practitioners average adjacent $N$, or reach for smoothing tricks, rather than trusting any single small tree.

True or false: as the number of steps $ N $ increases, the CRR binomial price of a European option converges to the Black–Scholes price. (Answer true or false.)

With $ u = e^{\sigma\sqrt{\Delta t}},\ d = 1/u $, the multiplicative binomial walk converges to geometric Brownian motion and the backward-induction expectation converges to the Black–Scholes integral. So the statement is true — though the approach oscillates with $N$ rather than decreasing monotonically.

PYTHON · RUNNABLE IN-BROWSER

# CRR binomial European call -> converges to Black-Scholes as steps grow
import numpy as np, math

def bs_call(S, K, r, sig, T):                      # closed-form benchmark
    d1 = (math.log(S/K) + (r + 0.5*sig*sig)*T) / (sig*math.sqrt(T))
    d2 = d1 - sig*math.sqrt(T)
    Nf = lambda x: 0.5*(1 + math.erf(x/math.sqrt(2)))
    return S*Nf(d1) - K*math.exp(-r*T)*Nf(d2)

def crr_call(S, K, r, sig, T, N):                  # EQ Q2.4, backward induction
    dt = T/N
    u  = math.exp(sig*math.sqrt(dt)); d = 1/u
    p  = (math.exp(r*dt) - d) / (u - d); disc = math.exp(-r*dt)
    j  = np.arange(N+1)
    V  = np.maximum(S*u**j*d**(N-j) - K, 0.0)       # terminal call payoffs
    for n in range(N, 0, -1):                        # sweep leaves -> root
        V = disc*(p*V[1:n+1] + (1-p)*V[0:n])
    return V[0]

S, K, r, sig, T = 100, 100, 0.05, 0.20, 1.0
exact = bs_call(S, K, r, sig, T)
Ns, errs = [1, 2, 5, 10, 50, 200, 1000], []
print(f"Black-Scholes call = {exact:.4f}\n  N      binomial     error")
for N in Ns:
    c = crr_call(S, K, r, sig, T, N); errs.append(abs(c-exact))
    print(f"{N:5d}   {c:9.4f}   {c-exact:+.4f}")
plot_xy(Ns, errs)                                   # |error| shrinking with N

edits are live — break it on purpose

INSTRUMENT Q2.2 — BINOMIAL-TREE PRICERCRR LATTICE · CONVERGES TO BLACK–SCHOLES · EQ Q2.4

VOL σ 0.20

STEPS N 4

STRIKE K 100

BINOMIAL CALL (N STEPS)

—

BLACK–SCHOLES LIMIT

—

ERROR vs B–S

—

Fixed: $S_0 = 100,\ r = 0.05,\ T = 1$ year. The lattice draws nodes coloured by stock price (brighter = higher); the readout prices a European call by backward induction (EQ Q2.4) and compares it to the Black–Scholes limit. Start at $N = 4$ — a coarse, visibly wrong tree — and drag $N$ up: the error collapses toward zero and visibly oscillates as it does. Raise σ and the lattice fans wider, the call gets dearer.

2.5

American options & early exercise

So far the option could only be exercised at expiry — a European option. An American option may be exercised on any date up to expiry, and that extra freedom is exactly where the binomial tree earns its keep: Black–Scholes has no clean closed form for it, but the lattice handles it with a one-line change. At every node, instead of simply taking the discounted continuation value, you take the larger of continuing and exercising right now:

EQ Q2.5 — AMERICAN BACKWARD INDUCTION $$ V_i^{(n)} = \max\!\Big(\,\underbrace{\text{payoff}\big(S_i^{(n)}\big)}_{\text{exercise now}},\;\; \underbrace{e^{-r\Delta t}\big[\,p\,V_{i+1}^{(n+1)} + (1-p)\,V_i^{(n+1)}\,\big]}_{\text{hold (continuation)}}\Big) $$

The European recursion (EQ Q2.4) is the right-hand branch alone; the American option adds the left branch — the option to stop. Because the holder optimizes at every node, an American option is worth at least as much as its European twin. The gap is the early-exercise premium. The set of nodes where exercising beats holding forms the early-exercise boundary — the curve the instrument below draws.

When does early exercise actually pay? Two classic facts orient the intuition. First, it is never optimal to exercise an American call on a non-dividend-paying stock early — you would throw away remaining time value and the interest you could earn on the strike by waiting, so the American call equals the European call (Merton's result). Dividends break this: a large dividend can make exercising a call just before the ex-date optimal. Second, American puts genuinely can pay to exercise early: deep in the money, the put is worth nearly $K - S$, and exercising now banks that cash to earn interest rather than waiting for a payoff capped at $K$. The higher the rate, the stronger the pull.

EQ Q2.6 — EARLY-EXERCISE PREMIUM $$ \pi_{\text{early}} \;=\; V^{\text{American}} - V^{\text{European}} \;\ge\; 0 $$

The premium is zero for a call on a non-dividend stock and strictly positive for a sufficiently in-the-money put at a positive rate. It is precisely the value of the option to stop early — and it is exactly the quantity the worked example below isolates, node by node.

Two-step tree with $ u = 1.2 $, $ d = 0.8 $, and a per-step gross interest factor $ R = e^{r\Delta t} = 1.05 $. What is the risk-neutral probability $ p = \dfrac{R - d}{u - d} $?

$ p = \dfrac{1.05 - 0.8}{1.2 - 0.8} = \dfrac{0.25}{0.40} = $ 0.625. This is the weight used at every node of the American-put backward induction in the worked example.

PYTHON · RUNNABLE IN-BROWSER

# American put via backward induction; isolate the early-exercise premium
import numpy as np, math

def binom_put(S, K, r, sig, T, N, american):
    dt = T/N
    u  = math.exp(sig*math.sqrt(dt)); d = 1/u
    p  = (math.exp(r*dt) - d) / (u - d); disc = math.exp(-r*dt)
    j  = np.arange(N+1)
    ST = S*u**j*d**(N-j)
    V  = np.maximum(K - ST, 0.0)                     # terminal put payoffs
    for n in range(N, 0, -1):
        V = disc*(p*V[1:n+1] + (1-p)*V[0:n])         # continuation value
        if american:                                # ... or exercise now?
            Sn = S*u**np.arange(n)*d**(n-1-np.arange(n))
            V  = np.maximum(V, K - Sn)               # EQ Q2.5: take the max
    return V[0]

S, K, r, sig, T, N = 100, 110, 0.05, 0.30, 1.0, 500
eu = binom_put(S, K, r, sig, T, N, american=False)
am = binom_put(S, K, r, sig, T, N, american=True)
print(f"European put : {eu:.4f}")
print(f"American put : {am:.4f}   (>= European, EQ Q2.6)")
print(f"early-exercise premium : {am - eu:.4f}")
print("the premium is the value of the right to exercise the put early.")

edits are live — break it on purpose

INSTRUMENT Q2.3 — AMERICAN vs EUROPEAN BOUNDARYPUT · EARLY-EXERCISE FRONTIER · EQ Q2.5

VOL σ 0.30

RATE r 0.08

STRIKE K 100

EUROPEAN PUT

—

AMERICAN PUT

—

EARLY-EXERCISE PREMIUM

—

Fixed: $S_0 = 100,\ T = 1$ year, $N = 80$ steps. The lattice shades each node by its optimal action: mint nodes are where exercising the put beats holding (the early-exercise region), grey nodes are "hold". The jagged frontier between them is the early-exercise boundary. Crank the rate up and the mint region swells — high rates make banking $K-S$ now, to earn interest, increasingly worth it — and the premium readout climbs. Drop the rate to zero and the boundary nearly disappears: with no interest to earn, there is little reason to exercise a put early.

The tree is the same idea Black–Scholes packages in closed form — just discretized. Quant 03 takes $N \to \infty$: the recombining lattice becomes geometric Brownian motion, EQ Q2.3's weighted average becomes a Gaussian integral, and the price collapses to $C = S\,N(d_1) - Ke^{-rT}N(d_2)$. The hedge ratio $\Delta$ you computed here becomes the first of the Greeks — and the whole binomial scaffolding turns into the smooth surface a derivatives desk lives on.

2.R

References

Cox, J. C., Ross, S. A. & Rubinstein, M. (1979). Option Pricing: A Simplified Approach. Journal of Financial Economics 7(3) — the original recombining binomial lattice, the CRR parameters of EQ Q2.4, and the convergence to Black–Scholes.
Merton, R. C. (1973). Theory of Rational Option Pricing. Bell Journal of Economics and Management Science 4(1) — the no-arbitrage bounds and the proof that an American call on a non-dividend stock is never exercised early (§2.5).
Black, F. & Scholes, M. (1973). The Pricing of Options and Corporate Liabilities. Journal of Political Economy 81(3) — the continuous-time closed form the binomial tree converges to as N → ∞ (Quant 03); the lattice is its discrete, fully constructive counterpart.
Hull, J. C. (2021). Options, Futures, and Other Derivatives (11th ed.). Pearson — Ch. 13–21: the standard practitioner treatment of binomial trees, risk-neutral valuation, and American-option pricing.
Shreve, S. E. (2004). Stochastic Calculus for Finance I: The Binomial Asset Pricing Model. Springer — a rigorous, self-contained development of replication, the risk-neutral measure, and market completeness (§2.1–§2.2).