How the Binomial Tree Encodes Up and Down Stock Moves

Summary

The document explains how a binomial stock-price model represents an up move and a down move using one expression. A binary variable selects between the two outcomes, while the exponential terms set their proportional size from volatility and the time step. The answer connects this notation to the familiar tree multipliers, which use positive and negative volatility-scaled exponents. It also discusses a drift or discounting term and says the two formulations can be written equivalently with that adjustment.

The evidence is a short algebraic comparison of the two forms, rather than a derivation from a continuous-time process or a calibration procedure. The post does not clarify its assumptions about the meaning of the drift parameter, probability measure, or whether prices are discounted. As a result, it is useful for reading the notation but insufficient as a complete account of how to construct or price a binomial model.

Key ideas

• A binary variable lets one expression represent either an up move or a down move.
• The tree multipliers use opposite volatility-scaled exponents for the two branches.
• The answer relates the exponential notation to standard binomial up and down factors.
• The role of drift and discounting depends on how the model is formulated.

Tags

# Stock price modelling under binomial tree model?


# Stock price modelling under binomial tree model?












In binomial tree model, the stock price is modelled in the form of $S_{k\delta}=S_{(k-1)\delta}\exp(\mu\delta+\sigma\sqrt\delta Z_k)$, where $\delta$ is time invertal between two observations $S_{k\delta},S_{(k-1)\delta}$, $Z_k=1,-1$ for upward and downward scenarios of the stock price change.

I noted some illustrations of variance and mean to explain why the model is set in the form, but I cannot find more explicit explanation. Could someone help?

## Answer by THATS MY QUANT MY QUANTITATIVE (score 0)

https://quant.stackexchange.com/a/77596

Typically the binomial tree is written as:

$$S_{t+\Delta t} = uS_t$$ $$S_{t+\Delta t} = dS_t$$

where $u=e^{\sigma\sqrt{\Delta t}}$ and $d=e^{-\sigma\sqrt{\Delta t}}$.

There is $Z$ instead to eloquently write the $u$ and $d$ in the same line. Then you now have to include discounting, $$e^{-\mu\Delta t}S_{t+\Delta t} = uS_t$$ $$e^{-\mu\Delta t}S_{t+\Delta t} = dS_t$$ which is the same as what you have - there is no difference.

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.