Scaling Binomial Tree Moves to Converge to Black–Scholes
Summary
The document addresses why holding the up and down multipliers fixed while adding steps to a binomial tree can inflate an option value. With maturity held constant, fixed moves allow the underlying to reach an increasingly wide range, changing the model’s implied volatility instead of refining the same process. The question describes an example in which a call value rises as the step count increases.
The answer gives a time-step-dependent choice of up and down factors, centered on the risk-adjusted drift and separated by a volatility-scaled move proportional to the square root of the time step. As the time step shrinks, the modeled return has approximately the risk-free drift and variance equal to volatility squared times the step length. This moment matching is the stated intuition for convergence to Black–Scholes. The note provides no derivation, numerical validation, or discussion of alternative tree parameterizations.
Key ideas
- Fixed up and down factors can change the tree’s implied volatility as the number of steps grows.
- The up and down moves should shrink with the time step when refining a fixed-maturity tree.
- The proposed factors match the local return’s drift to the risk-free rate and its variance to volatility squared times the step length.
- The answer presents this moment matching as a route to Black–Scholes convergence.
Tags
Full text
# Should U and D change with the number of steps in a Binomial Tree?
# Should U and D change with the number of steps in a Binomial Tree?
In everyone's binomial trees online I see constant U and D. Even when I read Option Volatility and Pricing by Natenburg, all his diagrams use a constant U and D (where U is the upwards magnitude from one step to the next. For example 1.05). The whole premise of a binomial model is that, as the amount of steps increase, the option price derived from the model should get closer and closer to Black Scholes' (see below).
Binomial Vs Black Scholes
The problem is that, with a constant U and D, it doesn't. The value of the option continues to grow indefinitely with the number of steps. If we fix the time to maturity, and increase the number of steps in between now and maturity, we increase the range of potential prices the underlying can reach. This increases volatility and hence increases the value of the option.
Holding all else constant, increasing the number of steps only increases the value of the option, which makes me believe that I'm missing some other adjustment.
My question in: What adjustments to U and D, or to the model, do I need to make such that I can approximate the BS Model (as per the image above).
A constant U and D does not seem correct. I've attached another picture from the book by Natenburg below for reference. If I increase the number of steps from 3 to 20, the value of the option goes from 5.22 to 9.24.
## Answer by Antoine Conze (score 3)
https://quant.stackexchange.com/a/50880
You can choose $U=e^{(r-\sigma^2/2)\delta t + \sigma \sqrt{\delta t}}$ and $D=e^{(r-\sigma^2/2)\delta t - \sigma \sqrt{\delta t}}$ to have the binomial model converge to the BS model when $\delta t=T/N \rightarrow 0$. The intuition is that in the binomial model as $\delta t\rightarrow 0$ you have $E[(S_{t+\delta t} - S_t)/S_t] \approx r \delta t$ and $\text{VAR}[(S_{t+\delta t} - S_t)/S_t] \approx \sigma^2 \delta t$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.