When Recombining Binomial Trees Need the Product of Up and Down Factors to Equal One
Summary
The document examines whether a recombining binomial option-pricing tree must use up and down factors whose product is one. One response says this condition is not necessary: recombination follows when the factors remain constant across steps, while risk-neutral pricing requires the expected growth factor to lie between the down and up factors. It also notes that multiple tree constructions can converge to Black–Scholes.
A second response attributes a stronger role to the product-one condition, connecting it to convergence of the tree’s mean and standard deviation and describing the Cox–Ross–Rubinstein parameterization, where the factors are reciprocal. The two answers therefore present different claims about whether that condition is required, without reconciling the meaning or assumptions behind them. The excerpt supplies no derivation or comparative evidence to settle the disagreement. Its useful lesson is to distinguish recombination, risk-neutral probability admissibility, and convergence conditions when evaluating a binomial tree.
Key ideas
- A constant up factor and down factor across steps allow a binomial tree to recombine.
- Risk-neutral pricing requires the growth factor to lie between the down and up factors.
- The Cox–Ross–Rubinstein construction uses reciprocal up and down factors.
- The answers disagree on whether the product-one condition is necessary for convergence, and the excerpt does not resolve the distinction.
Tags
Full text
# Is it fair to assume $(ud=1)$ in the binomial tree option pricing model?
# Is it fair to assume $(ud=1)$ in the binomial tree option pricing model?
I have discussion with my colleague on why a general assumption $$ud=1$$ in binomial tree option pricing model would be necessary?
I take it a simplification of the problem, otherwise, there will be more intermediate nodes in the tree, which will be hard to compute. While my colleague insists that since the underlying is often lognormal with 0 mean, $ud=1$ were a fair assumption.
I think his explanation is acceptable. The question, is there any other reason to make $ud=1$ assumption?
## Answer by Mark Joshi (score 5)
https://quant.stackexchange.com/a/16271
you don't need $ud=1.$ In fact, there are now about 30 binomial trees which converge to Black--Scholes in the large step limit. Most of them do not have $ud=1.$ All you need is
$$ d < e^{r \Delta t} < u $$
The tree recombines provided $u$ and $d$ don't change from step to step.
See my book More Mathematical Finance for a comprehensive review and classification of binomial trees.
## Answer by emcor (score 4)
https://quant.stackexchange.com/a/14909
The condition
$$ud=1\text{, or equivalently }u=1/d$$
is necessary to ensure convergence of the Binomial tree's mean $\mu$ and standard deviation $\sigma$ to nonfinite values when $n$ (number of steps) goes to infinity.
Cox-Rubinstein-Ross showed in their famous paper, that to achieve this, we must have:
$$u=e^{\sigma\sqrt{t/n}}\text{, }d=e^{-\sigma\sqrt{t/n}}$$ respectively, which holds exactly: $$ud=1.$$ We can aswell choose $\mu$ and $\sigma$ to fit $u$ and $d$ respectively, so $ud=1$ is the actual condition.
Paper excerpt:Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.