Skip to content

FRM Part I · FRM Exam Part I · Binomial Trees

Which statement about the convergence of a binomial tree to the Black-Scholes-Merton model is correct?

A tree with u and d calibrated to volatility, such as u equal to e to the σ root Δt, converges to the Black-Scholes-Merton price for European options as steps increase. The stock distribution approaches lognormal. Probabilities need not be 0.5, and delta varies across nodes.

  1. AFor a European option, if u and d are calibrated to the volatility (for example u = e^(σ√Δt), d = 1/u), the tree value converges to the BSM price as the number of steps increases.Correct
  2. BConvergence to BSM requires the risk-neutral probability to equal 0.5 at every node.
  3. CThe delta of the option stays fixed across all nodes of the tree, so adding steps does not change the hedge.
  4. DAdding steps makes the tree reflect realized rather than assumed volatility, so the price no longer depends on σ.

Explanation

With u and d tied to σ√Δt, the stock's distribution in the tree approaches lognormal as Δt falls, so European values converge to BSM. The probability p is generally not 0.5 (it depends on r and σ). Delta changes from node to node with the stock price and time. Volatility remains an input.

Did you get it right without looking?

One question tells you little. A timed set on Binomial Trees shows your real accuracy, how long you take and where you lose marks.

Practise Binomial TreesFree diagnostic mock

More Binomial Trees questions