Binomial trees and backward induction

Definition A binomial tree with $N$ periods models a stock price at the dates $n = 0, 1, \dots, N$. It starts at $S_0 > 0$, and in each period the price is multiplied by an up factor $u$ or a down factor $d$,

and both moves have positive physical probability at every node. The stock pays no dividends. A cash account starts at $B_0 = 1$ and has the same gross return $R > 0$ in every period, so $B_n = R^n$. Trading is frictionless and happens only at the dates $n$, and at date $n$ the first $n$ moves are known. Because $ud = du$, the order of the moves does not affect the price, and after $n$ periods with $k$ up moves the price is $S_0 u^k d^{\,n-k}$. The tree therefore recombines, with $n + 1$ price nodes at date $n$, although there are $2^n$ paths and $\binom{n}{k}$ of them lead to the node with $k$ up moves.

Definition The local no-arbitrage condition of the binomial tree is

It is necessary. If $R \ge u$, selling one share short at a node with price $S$ and placing $S$ in the cash account costs nothing and pays $RS - uS \ge 0$ after an up move and $RS - dS > 0$ after a down move, which is an arbitrage. If $R \le d$, borrowing $S$ to buy one share costs nothing and pays $dS - RS \ge 0$ and $uS - RS > 0$, again an arbitrage. The inequalities must be strict, since at $R = u$ or $R = d$ one state still has a positive payoff. Under the condition, the number

satisfies $0 < q < 1$ and $qu + (1 - q)d = R$. It is the up weight of the pricing measure studied in Risk neutral probabilities and physical probabilities. Here $q$ is this weight, not a dividend yield. Conversely, when the condition holds, the whole multiperiod tree is free of arbitrage, a consequence of The fundamental theorem of asset pricing.

Definition At a node where the stock price is $S$, let $V_u$ and $V_d$ be the claim values at the two successor nodes, in currency units. The one-period hedge holds $\Delta$ shares and the amount $b$ in the cash account, chosen so that

Subtracting the equations gives the hedge ratio, and either equation then gives $b$,

The solution is unique because $R > 0$, $S > 0$ and $u \ne d$, so a hedge exists at every node for any successor values. The hedge ratio is measured in shares. It is the change in claim value per unit change in stock price between the two successors, and it generally differs from node to node.

Definition Let a European claim pay $V_N$ at date $N$, where $V_N$ may depend on the whole path, and let $d < R < u$. Backward induction gives every final node the payoff as its value, using the tree of paths when the payoff depends on the path, and then moves back one date at a time, giving each node the cost $\Delta S + b$ of the hedge that delivers the values of its two successors. Inserting the formulas for $\Delta$ and $b$, the coefficient of $V_u$ in this cost is $(R - d)/(R(u - d)) = q/R$ and the coefficient of $V_d$ is $(u - R)/(R(u - d)) = (1 - q)/R$, so

where $V^u_{n+1}$ and $V^d_{n+1}$ are the values at the two successors of the node. The holdings chosen at a node use only the moves observed up to that node, so the strategy is predictable, and it is self-financing, because on arrival at each node the old portfolio is worth exactly the cost of the new hedge. Started with $V_0$, it therefore delivers the payoff on every path, and $V_0$ is the only price of the claim consistent with no arbitrage. For a payoff $g(S_N)$ that depends only on the final price, the induction collapses to

where the binomial coefficient counts the paths that reach each node. A path-dependent payoff, such as one on the running maximum or average, needs that extra path information as a state variable even though the price nodes recombine.

Definition To approximate a model in continuous time on $[0, T]$, split it into $N$ steps of length $\Delta t = T/N$ and let the tree depend on $\Delta t$. The Cox-Ross-Rubinstein scaling rule, for a volatility $\sigma > 0$ and a constant continuously compounded rate $r$, sets

The local condition $d < R < u$ reads $-\sigma\sqrt{\Delta t} < r\Delta t < \sigma\sqrt{\Delta t}$, which holds exactly when $\lvert r \rvert \sqrt{\Delta t} < \sigma$, so the step must be small enough. Let $\mathbb{Q}$ make the moves independent, each up with probability $q$. Each log increment $Y_n = \ln(S_{n+1}/S_n)$ equals $\pm\sigma\sqrt{\Delta t}$, and expanding $q$ in powers of $\sqrt{\Delta t}$ gives

Over the $N = T/\Delta t$ steps the log return $\ln(S_T/S_0)$ then has mean $(r - \tfrac12\sigma^2)T + O(\Delta t)$ and variance $\sigma^2 T + O(\Delta t)$. With $u$ and $d$ fixed while $N$ grows, this variance would grow without bound, and with log moves of order $\Delta t$ it would vanish.

Definition Fix $T$, $\sigma > 0$ and $r$, use the scaling rule and let $N \to \infty$. Under $\mathbb{Q}$ the terminal log return $\ln(S_T/S_0)$ is a sum of $N$ independent, identically distributed increments with the moments above. A central limit theorem for triangular arrays applies, because the increments are bounded by $\sigma\sqrt{\Delta t} \to 0$, and gives convergence in distribution of $\ln(S_T/S_0)$ to a normal law with mean $(r - \tfrac12\sigma^2)T$ and variance $\sigma^2 T$. Equivalently,

with $Z$ standard normal, which is the lognormal law of the Black-Scholes model under its pricing measure. This concerns the distribution at one date. Option values need more, and the course states the following conditions without proof. Bounded continuous payoffs converge by the definition of convergence in distribution, and the discount factor $R^{-N} = e^{-rT}$ is exact. A put payoff is bounded and continuous. Unbounded payoffs need uniform integrability, which holds for a call because $\mathbb{E}^{\mathbb{Q}}[S_T^2]$ stays bounded as $N$ grows. A bounded payoff with jumps, such as a digital option, converges when its jump points have probability zero in the limit, as they do under a lognormal law, though slowly and unevenly. American options add an exercise decision that also changes with the grid and need a separate convergence argument. The functional limit theorem, under which whole paths converge to geometric Brownian motion, is not proved here.