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Outline how a binomial model's computed value converges as further time steps are included.

FRM® Part I · Valuation and Risk Models · Binomial Trees · FRM1.VR.14.c

The idea

For a fixed option life T, a tree with n steps has a step length of dt = T ÷ n, so each step covers a shorter time as n rises. The up and down factors u and d also move closer to 1 as n rises. A recombining tree has n + 1 terminal nodes and (n + 1)(n + 2) ÷ 2 nodes in total, so work grows roughly with n². The terminal prices approach a lognormal distribution, and the tree value of a European option converges to the Black-Scholes-Merton value. Convergence is not monotone. Values often oscillate around the limit as n changes, but the error shrinks. An American option has no closed-form limit, so the tree is run with many steps until the value stabilises. Do not mistake oscillation for divergence, and do not treat a coarse tree value as the answer. Relationships that hold in the limit, such as put-call parity, can serve as a check.

Formulas

Total nodes in a recombining binomial tree

Nodes = (n + 1) × (n + 2) ÷ 2

n
the number of time steps

Counts the starting node and the n + 1 terminal nodes. A tree that did not recombine would need 2 to the power n paths, which is why recombining matters.

A practice question

From the question bank, with the reason each wrong option is wrong.

As the number of time steps in a Cox-Ross-Rubinstein tree is increased, the value of a European call on a stock that pays no dividends approaches which value?

  1. The intrinsic value, max(S0 - K, 0)Why not: Confuses the limit with the exercise value, which leaves out time value.
  2. The Black-Scholes-Merton value for the same inputs - correct
  3. The value from the one-step tree, which is already exactWhy not: A one-step tree is the coarsest approximation, not an exact value.
  4. The undiscounted risk-neutral expected payoff at expiryWhy not: Leaves out discounting at the risk-free rate.

As dt shrinks the tree's terminal prices approach a lognormal distribution and the European option value converges to the Black-Scholes-Merton formula. Intrinsic value omits time value, and a coarse tree is only an approximation.

A harder one

A six-month option is valued on a recombining binomial tree with weekly steps, 52 steps per year. How many distinct terminal stock prices does the tree contain?

  1. 26Why not: Counts the steps rather than the n + 1 terminal nodes.
  2. 53Why not: Uses the 52 weekly steps of a full year instead of the 26 steps in a six-month life.
  3. 27 - correct
  4. 67,108,864Why not: Counts 2^26 paths, as if the tree did not recombine.

Six months at one step per week is n = 26 steps. A recombining tree has n + 1 = 27 distinct terminal prices, because the order of up and down moves does not matter, even though there are 2^26 paths.

The hardest tier

A risk manager checks convergence by running a 100-step recombining binomial tree and must compute a value at every node. How many nodes does the tree contain in total, counting the initial node and all terminal nodes?

  1. 5,050Why not: Sums 1 + 2 + ... + 100, leaving out the terminal level, which has 101 nodes.
  2. 5,151 - correct
  3. 10,201Why not: Counts 101 nodes at each of 101 time points, but time point i has only i + 1 nodes.
  4. about 2.5 x 10^30Why not: Counts 2^101 - 1 nodes, as if the tree did not recombine.

Time point i has i + 1 nodes, so the total is 1 + 2 + ... + 101 = 101 x 102 / 2 = 5,151. Work therefore grows roughly with the square of the number of steps, while the full set of paths would grow as 2^n.

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