From Binomial Trees to the Black-Scholes PDE: A Limit Argument

The previous article derived the Black-Scholes PDE by constructing a replicating portfolio and arguing about its return. That argument is clean, but it leaves an uncomfortable question: where does the continuous-time SDE for $S_t$ come from in the first place? If markets move in discrete ticks, what justifies a model where they move continuously?

This article answers that question by constructing the Black-Scholes PDE from the ground up. We start with a one-period binomial model, extend it to $n$ periods, and take the limit as $n \to \infty$. The limit is geometric Brownian motion, and the discrete replicating portfolio converges to the Black-Scholes PDE.

The One-Period Binomial Model

Cox, Ross, and Rubinstein (1979) introduced the binomial model that bears their names. In the one-period version, the stock can move up by a factor $u$ or down by a factor $d$ over time $\Delta t$:

with $u > 1 > d > 0$. A derivative with payoff $f(S)$ at time $\Delta t$ has value $V$ today and pays $f_u$ or $f_d$ depending on which branch the stock took.

Construct a portfolio with $\Delta$ shares of stock and $-V$ short the derivative. The portfolio is worth $\Delta S - V$ today. It ends up worth $\Delta u S_0 - f_u$ on the up branch and $\Delta d S_0 - f_d$ on the down branch.

Choose $\Delta$ to make these equal:

Now the portfolio is riskless, so it must earn the risk-free rate. Pricing the bond leg at $1 + r\Delta t$:

Solving for $V$:

Define the risk-neutral probability:

Then:

This is the central insight: option price equals the discounted expectation of the payoff under the risk-neutral measure. The actual probability of an up move never appears (Cox & Ross, 1976).

The Multi-Period Binomial Tree

Now stack $n$ periods, each of length $\Delta t = T/n$. At each step, the stock moves up by $u$ or down by $d$, and the same risk-neutral probabilities $p, 1-p$ apply. The terminal stock price is:

where $j$ is the number of up moves. By the central limit theorem, as $n \to \infty$, $S_T$ converges in distribution to a lognormal variable. In other words, $\ln S_T$ becomes normally distributed.

The price at node $(i, j)$, meaning $i$ up moves and $j$ down moves, with $i + j = k$, is:

with $V_n^{i,j} = f(S_0 u^i d^j)$ at the terminal nodes.

This is just dynamic programming applied backward from the leaves (Press et al., 2007, Ch. 20.0). It is the original “Monte Carlo backwards” long before anyone used that phrase.

Choosing $u$ and $d$ the Right Way

We need the moments of $S_T$ in the limit to match those of geometric Brownian motion with volatility $\sigma$. With $\Delta t = T/n$, the standard Cox-Ross-Rubinstein (CRR) parameterization is:

This gives $ud = 1$ (the “recombining tree” property), and in the limit the tree prices converge to Black-Scholes (Hull, 2018, Ch. 21.5).

Other choices exist, including Jarrow-Rudd and Tian, but they all converge to the same limit because that limit depends only on the first two moments of the log-return distribution. Specifically, you need:

Working through the algebra, the risk-neutral probability becomes:

(Here I’m using the symmetric CRR parameterization.) The important point: as $\Delta t \to 0$, the drift term vanishes and $p \to \frac{1}{2}$. In the limit the random walk is symmetric.

The Limit as $n \to \infty$

Take logs of the terminal price. With $j$ up moves and $n-j$ down moves:

With $ud = 1$ and $2\ln u = 2\sigma\sqrt{\Delta t} + O(\Delta t^{3/2})$:

Define $X = (j - n/2)/\sqrt{n}$. This is approximately standard normal by the central limit theorem:

So $\ln S_T$ is asymptotically normal with mean $\ln S_0$ and variance $\sigma^2 T$. That’s geometric Brownian motion with zero drift (under the risk-neutral measure).

The full limit theorem is stronger than one-step marginal convergence. Donsker’s invariance principle says the rescaled random-walk path converges in distribution to Brownian motion, with the stock price process converging to log-Brownian motion (Wilmott, 2007, Vol. 1, Ch. 5). Convergence is in $D[0,T]$ under the Skorokhod topology.

How the PDE Emerges

The PDE emerges by expanding the option values at neighboring nodes. Consider three consecutive nodes in a recombining tree:

              (i+1, j+1)
            /
    (i, j). (i+1, j)
            \
              (i, j+1)

The option values at these nodes are $V_{i,j}$, $V_{i+1,j}$ (up from center), and $V_{i,j+1}$ (down from center). The center node sits at stock price $S$, the up node at $S u = S + \Delta S$ where $\Delta S = S(u-1)$, and the down node at $S - \Delta S (d-1)/...$, with the sign written as $-S(1-d)$.

In time $\Delta t = T/n$, the option value moves backward from $V_{i+1, j}$ and $V_{i, j+1}$ to $V_{i, j}$:

Taylor expand everything to first order in $\Delta S$ and $\Delta t$. The two relevant expansions are:

Here $V$ and its derivatives are evaluated at the center node. Plug in:

Now expand the discount factor $e^{-r\Delta t} \approx 1 - r\Delta t$ and use the risk-neutral expectation formula $2p - 1 \approx (r - \tfrac{1}{2}\sigma^2) \cdot 2\sqrt{\Delta t}/\sigma$ (this comes from $p$ being close to $\frac{1}{2}$). Substituting and collecting terms up to $O(\Delta t)$:

The bracket must equal zero. That’s the Black-Scholes PDE. The discrete binomial sum has, in the limit, become a derivative.

This is also why binomial model prices converge to Black-Scholes for any smooth payoff. Every smooth function satisfies a Taylor expansion, and the second-order term in $V$ survives only if the variance of the price process scales correctly, as it does in CRR.

The Martingale Argument

There is a parallel derivation, more elegant in some ways and uglier in others, that does not require a Taylor expansion. It uses the idea that the discounted stock price under the risk-neutral measure is a martingale (Hull, 2018, Ch. 15.6; Wilmott, 2007, Vol. 1, Ch. 7).

The discounted stock price $D_t = e^{-rt} S_t$ under the risk-neutral measure $\mathbb{Q}$ satisfies:

where $W^{\mathbb{Q}}$ is a $\mathbb{Q}$-Brownian motion. Equivalently, $\mathbb{E}^{\mathbb{Q}}[D_T \mid \mathcal{F}_t] = D_t$.

For any contingent claim paying $V_T$ at $T$, the no-arbitrage price is:

This is the risk-neutral pricing formula. It is an if-and-only-if: in a complete market, this is the unique arbitrage-free price. In an incomplete market, you get a range of no-arbitrage prices, but the upper and lower bounds are still given by the supremum and infimum over equivalent martingale measures.

Apply Itô’s lemma to $D_t = e^{-rt} S_t$ and $V_t$ to derive the PDE:

For the discounted price $V^*_t = e^{-rt}V_t$ to be a martingale, its drift must vanish:

This is almost the Black-Scholes PDE, but $\mu$ appears. We need the Girsanov change of measure to remove it. The Radon-Nikodym derivative:

is the transformation that converts the real-world Brownian motion $W^{\mathbb{P}}$ into the risk-neutral Brownian motion $W^{\mathbb{Q}}$. The drift correction absorbs the $\mu - r$ term, leaving the Black-Scholes PDE with $r$ in place of $\mu$.

This is the rigorous statement of why drift does not enter option prices. The mechanism is the change of measure, not the absence of drift from the start.

Why Binomial Trees Still Matter

If you can write down the closed form, why bother with trees? Three reasons.

American options. An American put can be exercised early. The closed-form Black-Scholes formula assumes European exercise. The binomial tree handles American features naturally: at each node, compare continuation value with immediate exercise value.

Closed-form American pricing does not exist except for perpetual options and a few specials. Trees handle them in $O(n^2)$ operations for $n$ time steps, which is fast enough that high-five-figure binomial trees still ship to production.

Path-dependent options. Asian and barrier options need the stock path, which a tree can track explicitly. Re-strike options and lookbacks also have state variables, so a tree is a natural method.

Bias diagnostics. The binomial price has a known convergence rate: $O(1/n)$ for the standard parameterization, $O(1/n^2)$ for the Leisen-Reizer parameterization with Peizer-Pratt inversion (Hull, 2018, Ch. 21.6). If your finite-difference or PDE solver gives a different bias, you know your solver is wrong.

Numerical check: price a 1-year ATM call with $r=0.05$, $\sigma=0.20$, $S=K=100$. Black-Scholes gives $C = 10.4506$. A 50-step binomial gives $C \approx 10.45$ plus or minus a basis point. A 500-step tree gets you to four-decimal accuracy.

import math

def crr_call(S, K, r, sigma, T, n):
    dt = T / n
    u  = math.exp(sigma * math.sqrt(dt))
    d  = 1.0 / u
    p  = (math.exp(r * dt) - d) / (u - d)
    disc = math.exp(-r * dt)
    # terminal values
    prices = [max(S * u**n * d**(2*i - n) - K, 0.0) for i in range(n + 1)]
    # backward induction
    for k in range(n - 1, -1, -1):
        prices = [disc * (p * prices[i] + (1 - p) * prices[i+1])
                  for i in range(len(prices) - 1)]
    return prices[0]

This is roughly 30 lines, no SciPy required, and gets you four-digit agreement with Black-Scholes for $n$ in the few hundreds. Most production desks still have a copy of this stashed somewhere for when a quant librarian leaves and no one knows how the fancy library works.

The Takeaway

The Black-Scholes PDE is the continuous-time shadow of the binomial model. The replicating argument, the martingale argument, and the PDE are the same result expressed in different languages. Once you see them as perspectives on one fact, namely that derivative prices in a complete market are discounted expectations under the risk-neutral measure, you can use whichever form suits the problem.

For closed-form European vanilla, the Taylor-expansion or the characteristic-function argument wins. For American features or path dependence, the tree wins. For large books or non-vanilla payoffs, you end up doing Monte Carlo, which we cover in another module. But everything starts here: a one-step binomial, stacked to $n$, taken to the limit.

A Theorem Worth Knowing: Donsker’s Invariance

The limit from binomial to geometric Brownian motion is not just hand-waving. It is Donsker’s invariance principle, which says the rescaled random walk converges in distribution to Brownian motion on the space of continuous functions $C[0, T]$ with the sup-norm topology.

Applied here: if you stack $n$ binomial steps each of size $\Delta t = T/n$, the cumulative log-return after $n$ steps divided by $\sqrt{n}$ has the joint distribution converging to Brownian motion. Specifically, define:

where $R_i$ is the log-return in step $i$, and the drift subtraction makes this a zero-mean random walk. As $n \to \infty$, $W^{(n)}$ converges in distribution to $W$ a Brownian motion.

This is the deeper reason the binomial model converges to Black-Scholes for terminal prices and entire paths. The implied vol on a path-dependent option under the binomial model converges to the Black-Scholes implied vol with the appropriate corrections.

An Important Generalization: Incomplete Markets

The martingale argument presumes a complete market: the option’s risk can be perfectly hedged with the underlying. In reality, markets are incomplete. Sources of incompleteness:

• Stochastic volatility. Once $\sigma$ is itself random, you have more risk than the underlying covers. The volatility is an additional non-hedgeable source of risk.
• Jumps. Between trades, the stock can gap. Your hedge is broken.
• Multiple underlyings with non-traded correlations. You can trade each underlying but cannot trade the correlation.
• Transaction costs. Hedging at high frequency is limited by cost.

In an incomplete market, multiple risk-neutral measures are consistent with the same observed drift and volatility of the underlying. Each measure gives a different option price. The set of admissible prices is bounded:

where $V_{\min}$ is the price under the “most pessimistic” measure (worst-case future of the underlying), and $V_{\max}$ is the price under the “most optimistic” measure. For European options on liquid single stocks with non-stochastic volatility, the bounds are tight, within a few percent. For more exotic payoffs, they can be 50% or more wide.

The natural remedy is to add a model: assume something about the unobserved part (stochastic vol, jump structure) and you get a single price back. This is the topic of several other articles in this module.

An Imperfect Market Reality: Transaction Costs

Every rebalance costs money in the real world. The optimal hedging strategy with transaction costs isn’t delta-hedge-every-tick; it’s delta-hedge-with-some-buffer (Whalley & Wilmott, 1997).

Define the half-spread $\delta_b$ per share traded, $w$ as the per-trade friction. The optimal hedging width around the target delta is:

where $\Delta_G$ is the unconstrained delta. This is the “no-trade band”: when delta is close enough to target, you do not trade. The intuition is that every trade has a fixed cost, but reduces variance quadratically in trade size. Trading too often costs more than it saves; trading too rarely leaves hedge error.

Practical numbers: for SPX with $\sigma = 0.20$, $T = 0.25$, and round-trip transaction cost of 1bp, the no-trade band is roughly 1-2% of position notional. Below that band, no trades; above it, partial rebalances.

This is one of the reasons retail “delta-hedged” strategies don’t quite work: their transaction costs are 30-50x institutional rates, so the no-trade band is 5-7% wider.

Numerical Convergence of the Binomial

The CRR binomial tree converges to Black-Scholes at rate $O(1/n)$ for the standard parameterization. With the Leisen-Reizer tree using a Peizer-Pratt inversion, you can get $O(1/n^2)$ convergence.

Worked out: pricing an ATM call with $S_0 = K = 100$, $r = 0.05$, $\sigma = 0.20$, $T = 0.5$. Black-Scholes gives $C = 6.8887$:

The CRR error decreases by roughly a factor of four as $n$ doubles, which is the expected $O(1/n)$ behavior. Leisen-Reizer improves on this, typically halving the error at the same $n$ for OTM options, and converges at $O(1/n^2)$. The improvement is most pronounced for OTM strikes, where CRR’s oscillation between in- and out-of-the-money branches is most acute. For ATM options, both methods converge quickly and the difference is small.

For accuracy requirements of 1bp, you typically need $n \sim 1000$ for CRR and $n \sim 500$ for L-R. CPU cost on a modern machine: microseconds per option. Memory scales as $n^2$ for non-recombining trees; recombining trees are $O(n^2)$ time and $O(n)$ memory.

Why Some Books Still Use the PDE Form

If you have an analytical formula, why bother with finite differences on the PDE? Several reasons:

1. Stability analysis. The PDE exposes singularities and boundary issues more cleanly than the tree. Studying the boundary behavior near expiry tells you where finite differences will have problems.
2. Higher dimensions. Multi-asset options (basket, rainbow) need a 2D or 3D PDE; trees become computationally infeasible. PDEs scale as $n_1 \cdot n_2 \cdot n_3$ in computational cost for the grid, which is manageable up to 4 dimensions.
3. American features. LCP (linear complementarity problem) formulations give clean numerical schemes for American options via PDE. Trees handle American features but with different numerical artifacts.

The PDE formulation also connects to the entire optimal-stopping literature: American options are optimal stopping problems, and the PDE gives you the value function satisfying the variational inequality:

with appropriate boundary conditions. That’s a max/min problem (variational inequality, not just PDE) at each point of the domain, and projects the option price onto the larger of the two at every time step.

Reading list. Hull, Options, Futures, and Other Derivatives, 10th ed., Ch. 13, 21 (2018). Wilmott, Paul Wilmott on Quantitative Finance, 2nd ed., Vol. 1, Ch. 5-7 (2007). Cox, Ross & Rubinstein, “Option Pricing: A Simplified Approach,” Journal of Financial Economics 7 (1979). Luenberger, Investment Science, 2nd ed., Ch. 12 (2013). Numerical Recipes, 3rd ed., Ch. 20 (2007) for the PDE side of the same story.