In a Cox-Ross-Rubinstein tree with u = e^{sigma*sqrt(dt)} and d = 1/u, show what the risk-neutral probability tends to as dt goes to zero and explain why the price oscillates as the number of steps grows rather than converging smoothly.

In a Cox-Ross-Rubinstein tree with u = e^{sigma*sqrt(dt)} and d = 1/u, show what the risk-neutral probability tends to as dt goes to zero and explain why the price oscillates as the number of steps grows rather than converging smoothly.

Approach: Expand the up and down factors to second order in sqrt(dt), substitute into the risk-neutral probability, then think about where the terminal nodes sit relative to the strike as the step count changes.

The risk-neutral probability tends to 0.5 + 0.5*sqrt(dt)*(r - sigma^2/2)/sigma, and the tree price converges to Black-Scholes at order 1/n while oscillating because the terminal grid straddles the strike differently for each n. Expanding u = 1 + sigma*sqrt(dt) + sigma^2*dt/2 and d = 1 - sigma*sqrt(dt) + sigma^2*dt/2 in p = (e^{r*dt} - d)/(u - d) gives that limit, so the log return has mean (r - sigma^2/2)*dt and variance sigma^2*dt per step and the central limit theorem delivers the lognormal law. The oscillation comes from the payoff kink: for one step count a terminal node lands almost exactly on the strike and for the next it lands well away, so the error alternates in sign as n increases. Averaging the n and n+1 step prices, or Richardson extrapolation on 1/n, removes most of it, and placing the strike midway between two terminal nodes is the cheaper fix used in production trees.

Follow-up: How would you adapt the tree so an American put on a discrete dividend payer converges without the dividend date falling between two levels?

Key concepts: risk-neutral probability, convergence, Black-Scholes, strike straddling, Richardson extrapolation.